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Chapter 1 Algebra Fundamentals

 

1.1 Review of Real Numbers and Absolute Value

Learning Objectives

  1. Review the set of real numbers.
  2. Review the real number line and notation.
  3. Define the geometric and algebraic definition of absolute value.

Real Numbers

Algebra is often described as the generalization of arithmetic. The systematic use of variablesLetters used to represent numbers., letters used to represent numbers, allows us to communicate and solve a wide variety of real-world problems. For this reason, we begin by reviewing real numbers and their operations.

A setAny collection of objects. is a collection of objects, typically grouped within braces { }, where each object is called an elementAn object within a set.. When studying mathematics, we focus on special sets of numbers.

={1,2,3,4,5,}             NaturalNumbersW={0,1,2,3,4,5,}              WholeNumbers={,3,2,1,0,1,2,3,}Integers

The three periods (…) are called an ellipsis and indicate that the numbers continue without bound. A subsetA set consisting of elements that belong to a given set., denoted , is a set consisting of elements that belong to a given set. Notice that the sets of naturalThe set of counting numbers: {1, 2, 3, 4, 5, …}. and whole numbersThe set of natural numbers combined with zero: {0, 1, 2, 3, 4, 5, …}. are both subsets of the set of integers and we can write:

andW

A set with no elements is called the empty setA subset with no elements, denoted Ø or { }. and has its own special notation:

{}=ØEmptySet

Rational numbersNumbers of the form ab, where a and b are integers and b is nonzero., denoted , are defined as any number of the form ab where a and b are integers and b is nonzero. We can describe this set using set notationNotation used to describe a set using mathematical symbols.:

={ab|a,b,b0}RationalNumbers

The vertical line | inside the braces reads, “such that” and the symbol indicates set membership and reads, “is an element of.” The notation above in its entirety reads, “the set of all numbers ab such that a and b are elements of the set of integers and b is not equal to zero.” Decimals that terminate or repeat are rational. For example,

0.05=5100and0.6=0.6666=23

The set of integers is a subset of the set of rational numbers, , because every integer can be expressed as a ratio of the integer and 1. In other words, any integer can be written over 1 and can be considered a rational number. For example,

7=71

Irrational numbersNumbers that cannot be written as a ratio of two integers. are defined as any numbers that cannot be written as a ratio of two integers. Nonterminating decimals that do not repeat are irrational. For example,

π=3.14159and2=1.41421

Finally, the set of real numbersThe set of all rational and irrational numbers., denoted , is defined as the set of all rational numbers combined with the set of all irrational numbers. Therefore, all the numbers defined so far are subsets of the set of real numbers. In summary,

The set of even integersIntegers that are divisible by 2. is the set of all integers that are evenly divisible by 2. We can obtain the set of even integers by multiplying each integer by 2.

{,6,4,2,0,2,4,6,}EvenIntegers

The set of odd integersNonzero integers that are not divisible by 2. is the set of all nonzero integers that are not evenly divisible by 2.

{,5,3,1,1,3,5,}OddIntegers

A prime numberInteger greater than 1 that is divisible only by 1 and itself. is an integer greater than 1 that is divisible only by 1 and itself. The smallest prime number is 2 and the rest are necessarily odd.

{2,3,5,7,11,13,17,19,23,}PrimeNumbers

Any integer greater than 1 that is not prime is called a composite numberIntegers greater than 1 that are not prime. and can be uniquely written as a product of primes. When a composite number, such as 42, is written as a product, 42=221, we say that 221 is a factorizationAny combination of factors, multiplied together, resulting in the product. of 42 and that 2 and 21 are factorsAny of the numbers that form a product.. Note that factors divide the number evenly. We can continue to write composite factors as products until only a product of primes remains.

Therefore, the prime factorizationThe unique factorization of a natural number written as a product of primes. of 42 is 237.

Example 1

Determine the prime factorization of 210.

Solution:

Begin by writing 210 as a product with 10 as a factor. Then continue factoring until only a product of primes remains.

210=1021=2537=2357

Since the prime factorization is unique, it does not matter how we choose to initially factor the number; the end result will be the same.

Answer: 2357

A fractionA rational number written as a quotient of two integers: ab, where b0. is a rational number written as a quotient, or ratio, of two integers a and b where b0.

The integer above the fraction bar is called the numeratorThe number above the fraction bar. and the integer below is called the denominatorThe number below the fraction bar.. Two equal ratios expressed using different numerators and denominators are called equivalent fractionsTwo equal fractions expressed using different numerators and denominators.. For example,

50100=12

Consider the following factorizations of 50 and 100:

50=225100=425

The numbers 50 and 100 share the factor 25. A shared factor is called a common factorA factor that is shared by more than one real number.. Making use of the fact that 2525=1, we have

50100=225425=241=24

Dividing 2525 and replacing this factor with a 1 is called cancellingThe process of dividing out common factors in the numerator and the denominator.. Together, these basic steps for finding equivalent fractions define the process of reducingThe process of finding equivalent fractions by dividing the numerator and the denominator by common factors.. Since factors divide their product evenly, we achieve the same result by dividing both the numerator and denominator by 25 as follows:

50÷25100÷25=24

Finding equivalent fractions where the numerator and denominator are relatively primeNumbers that have no common factor other than 1., or have no common factor other than 1, is called reducing to lowest termsFinding equivalent fractions where the numerator and the denominator share no common integer factor other than 1.. This can be done by dividing the numerator and denominator by the greatest common factor (GCF).The largest shared factor of any number of integers. The GCF is the largest number that divides a set of numbers evenly. One way to find the GCF of 50 and 100 is to list all the factors of each and identify the largest number that appears in both lists. Remember, each number is also a factor of itself.

{1,2,5,10,25,50}Factorsof50{1,2,4,5,10,20,25,50,100}Factorsof100

Common factors are listed in bold, and we see that the greatest common factor is 50. We use the following notation to indicate the GCF of two numbers: GCF(50, 100) = 50. After determining the GCF, reduce by dividing both the numerator and the denominator as follows:

50÷50100÷50=12

Example 2

Reduce to lowest terms: 10872.

Solution:

A quick way to find the GCF of the numerator and denominator requires us to first write each as a product of primes. The GCF will be the product of all the common prime factors.

108=2233372=22233}GCF(108, 72) = 2233=36

In this case, the product of the common prime factors is 36.

10872=108÷3672÷36=32

We can convert the improper fraction 32 to a mixed number 112; however, it is important to note that converting to a mixed number is not part of the reducing process. We consider improper fractions, such as 32, to be reduced to lowest terms. In algebra it is often preferable to work with improper fractions, although in some applications, mixed numbers are more appropriate.

Answer: 32

Recall the relationship between multiplication and division:

dividenddivisor126=2quotientbecause62=12

In this case, the dividendA number to be divided by another number. 12 is evenly divided by the divisorThe number that is divided into the dividend. 6 to obtain the quotientThe result of division. 2. It is true in general that if we multiply the divisor by the quotient we obtain the dividend. Now consider the case where the dividend is zero and the divisor is nonzero:

06=0since60=0

This demonstrates that zero divided by any nonzero real number must be zero. Now consider a nonzero number divided by zero:

120=?or0?=12

Zero times anything is zero and we conclude that there is no real number such that 0?=12. Thus, the quotient 12÷0 is undefinedA quotient such as 50 is left without meaning and is not assigned an interpretation.. Try it on a calculator, what does it say? For our purposes, we will simply write “undefined.” To summarize, given any real number a0, then

0÷a=0a=0zeroanda÷0=a0undefined

We are left to consider the case where the dividend and divisor are both zero.

00=?or0?=0

Here, any real number seems to work. For example, 05=0 and also, 03=0. Therefore, the quotient is uncertain or indeterminateA quotient such as 00 is a quantity that is uncertain or ambiguous..

0÷0=00indeterminate

In this course, we state that 0÷0 is undefined.

The Number Line and Notation

A real number lineA line that allows us to visually represent real numbers by associating them with points on the line., or simply number line, allows us to visually display real numbers by associating them with unique points on a line. The real number associated with a point is called a coordinateThe real number associated with a point on a number line.. A point on the real number line that is associated with a coordinate is called its graphA point on the number line associated with a coordinate.. To construct a number line, draw a horizontal line with arrows on both ends to indicate that it continues without bound. Next, choose any point to represent the number zero; this point is called the originThe point on the number line that represents zero..

Positive real numbers lie to the right of the origin and negative real numbers lie to the left. The number zero (0) is neither positive nor negative. Typically, each tick represents one unit.

As illustrated below, the scale need not always be one unit. In the first number line, each tick mark represents two units. In the second, each tick mark represents 17:

The graph of each real number is shown as a dot at the appropriate point on the number line. A partial graph of the set of integers , follows:

Example 3

Graph the following set of real numbers: {52,0,32,2}.

Solution:

Graph the numbers on a number line with a scale where each tick mark represents 12 unit.

Answer:

The oppositeReal numbers whose graphs are on opposite sides of the origin with the same distance to the origin. of any real number a is −a. Opposite real numbers are the same distance from the origin on a number line, but their graphs lie on opposite sides of the origin and the numbers have opposite signs.

Given the integer −7, the integer the same distance from the origin and with the opposite sign is +7, or just 7.

Therefore, we say that the opposite of −7 is −(−7) = 7. This idea leads to what is often referred to as the double-negative propertyThe opposite of a negative number is positive: −(−a) = a.. For any real number a,

(a)=a

Example 4

Calculate: ((38)).

Solution:

Here we apply the double-negative within the innermost parentheses first.

((38))=(38)=38

Answer: 38

In general, an odd number of sequential negative signs results in a negative value and an even number of sequential negative signs results in a positive value.

When comparing real numbers on a number line, the larger number will always lie to the right of the smaller one. It is clear that 15 is greater than 5, but it may not be so clear to see that −1 is greater than −5 until we graph each number on a number line.

We use symbols to help us efficiently communicate relationships between numbers on the number line.

EqualityRelationshipsOrderRelationships=   "isequalto"<   "islessthan"   "isnotequalto">   "isgreaterthan"   "isapproximatelyequalto"   "islessthanorequalto"    "isgreaterthanorequalto"

The relationship between the integersThe set of positive and negative whole numbers combined with zero: {…, −3, −2, −1, 0, 1, 2, 3, …}. in the previous illustration can be expressed two ways as follows:

5<1"Negativefiveislessthannegativeone."or1>5"Negativeoneisgreaterthannegativefive."

The symbols < and > are used to denote strict inequalitiesExpress ordering relationships using the symbol < for “less than” and > for “greater than.”, and the symbols and are used to denote inclusive inequalitiesUse the symbol to express quantities that are “less than or equal to” and for quantities that are “greater than or equal to” each other.. In some situations, more than one symbol can be correctly applied. For example, the following two statements are both true:

10<0and100

In addition, the “or equal to” component of an inclusive inequality allows us to correctly write the following:

1010

The logical use of the word “or” requires that only one of the conditions need be true: the “less than” or the “equal to.”

Example 5

Fill in the blank with <, =, or >: 2___12.

Solution:

Use > because the graph of −2 is to the right of the graph of −12 on a number line. Therefore, −2 > −12, which reads, “negative two is greater than negative twelve.

Answer: −2 > −12

An algebraic inequalityAlgebraic expressions related with the symbols , <, , and >., such as x2, is read, “x is greater than or equal to 2.” Here the letter x is a variable, which can represent any real number. However, the statement x2 imposes a condition on the variable. SolutionsValues that can be used in place of the variable to satisfy the given condition. are the values for x that satisfy the condition. This inequality has infinitely many solutions for x, some of which are 2, 3, 4.1, 5, 20, and 20.001. Since it is impossible to list all of the solutions, a system is needed that allows a clear communication of this infinite set. Common ways of expressing solutions to an inequality are by graphing them on a number line, using interval notation, or using set notation.

To express the solution graphically, draw a number line and shade in all the values that are solutions to the inequality. This is called the graph of the solution setSolutions to an algebraic expression expressed on a number line.. Interval and set notation follow:

"xisgreaterthanorequalto2"x2

Intervalnotation:[2,)Setnotation:{x|x2}

In this example, there is an inclusive inequality, which means that the lower-bound 2 is included in the solution set. Denote this with a closed dot on the number line and a square bracket in interval notation. The symbol ∞ is read as “infinityThe symbol ∞ indicates the interval is unbounded to the right.” and indicates that the set is unbounded to the right on a number line. If using a standard keyboard, use (inf) as a shortened form to denote infinity. Now compare the notation in the previous example to that of the strict, or noninclusive, inequality that follows:

"xislessthan3"x<3

Intervalnotation:(,3)Setnotation:{x|x<3}

Strict inequalities imply that solutions may get very close to the boundary point, in this case 3, but not actually include it. Denote this idea with an open dot on the number line and a round parenthesis in interval notation. The symbol −∞ is read as “negative infinityThe symbol −∞ indicates the interval is unbounded to the left.” and indicates that the set is unbounded to the left on a number line. Infinity is a bound to the real numbers, but is not itself a real number: it cannot be included in the solution set and thus is always enclosed with a parenthesis.

Interval notation is textual and is determined after graphing the solution set on a number line. The numbers in interval notation should be written in the same order as they appear on the number line, with smaller numbers in the set appearing first. Set notation, sometimes called set-builder notation, allows us to describe the set using familiar mathematical notation. For example,

{x|x2}

Here, x describes the type of number. This implies that the variable x represents a real number. The statement x2 is the condition that describes the set using mathematical notation. At this point in our study of algebra, it is assumed that all variables represent real numbers. For this reason, you can omit the “”, and write

{x|x2}

Example 6

Graph the solution set and give the interval and set notation equivalents: x<20.

Solution:

Use an open dot at −20, because of the strict inequality <, and shade all real numbers to the left.

Answer: Interval notation: (,20); set notation: {x|x<20}

A compound inequalityTwo or more inequalities in one statement joined by the word “and” or by the word “or.” is actually two or more inequalities in one statement joined by the word “and” or by the word “or”. Compound inequalities with the logical “or” require that either condition must be satisfied. Therefore, the solution set of this type of compound inequality consists of all the elements of the solution sets of each inequality. When we join these individual solution sets it is called the unionThe set formed by joining the individual solution sets indicated by the logical use of the word “or” and denoted with the symbol ., denoted . For example,

x<3orx6

Intervalnotation:(,3)[6,)Setnotation:{x|x<3orx6}

An inequality such as, 1x<3 reads, “negative one is less than or equal to x and x is less than three.” This is actually a compound inequality because it can be decomposed as follows:

1xandx<3

The logical “and” requires that both conditions must be true. Both inequalities will be satisfied by all the elements in the intersectionThe set formed by the shared values of the individual solution sets that is indicated by the logical use of the word “and,” denoted with the symbol ., denoted , of the solution sets of each.

Example 7

Graph and give the interval notation equivalent: 1x<3.

Solution:

Determine the intersection, or overlap, of the two solution sets to x<3 and x1. The solutions to each inequality are sketched above the number line as a means to determine the intersection, which is graphed on the number line below.

Here, 3 is not a solution because it solves only one of the inequalities. Alternatively, we may interpret 1x<3 as all possible values for x between, or bounded by, −1 and 3 where −1 is included in the solution set.

Answer: Interval notation: [1,3); set notation: {x|1x<3}

In this text, we will often point out the equivalent notation used to express mathematical quantities electronically using the standard symbols available on a keyboard.

×"*"" >= "÷"/"" <= "" != "

Many calculators, computer algebra systems, and programming languages use the notation presented above, in quotes.

Absolute Value

The absolute valueThe absolute value of a number represents the distance from the graph of the number to zero on a number line. of a real number a, denoted |a|, is defined as the distance between zero (the origin) and the graph of that real number on the number line. Since it is a distance, it is always positive. For example,

|4|=4and|4|=4

Both 4 and −4 are four units from the origin, as illustrated below:

Also, it is worth noting that,

|0|=0

The algebraic definition of the absolute value of a real number a follows:

|a|={aifa0aifa<0

This is called a piecewise definitionA definition that changes depending on the value of the variable.. The result depends on the quantity a. If a is nonnegative, as indicated by the inequality a0, then the absolute value will be that number a. If a is negative, as indicated by the inequality a<0, then the absolute value will be the opposite of that number, −a. The results will be the same as the geometric definition. For example, to determine |4| we make note that the value is negative and use the second part of the definition. The absolute value will be the opposite of −4.

|4|=(4)=4

At this point, we can determine what real numbers have certain absolute values.

Example 8

Determine the values represented by x: |x|=6.

Solution:

Think of a real number whose distance to the origin is 6 units. There are two solutions: the distance to the right of the origin and the distance to the left of the origin, namely {±6}. The symbol ± is read “plus or minus” and indicates that there are two answers, one positive and one negative.

|6|=6and|6|=6

Answer: x=±6

Example 9

Determine the values represented by x: |x|=6.

Solution:

Here we wish to find a value where the distance to the origin is negative. Since negative distance is not defined, this equation has no solution. Use the empty set Ø to denote this.

Answer: Ø

The absolute value can be expressed textually using the notation abs(a). We often encounter negative absolute values, such as |3| or −abs(3). Notice that the negative sign is in front of the absolute value symbol. In this case, work the absolute value first and then find the opposite of the result.

|3||3|and=3=3

Try not to confuse this with the double negative property, which states that (3)=+3.

Example 10

Simplify: (|50|).

Solution:

First, find the absolute value of −50 and then apply the double-negative property.

(|50|)=(50)=50

Answer: 50

Key Takeaways

  • Algebra is often described as the generalization of arithmetic. The systematic use of variables, used to represent real numbers, allows us to communicate and solve a wide variety of real-world problems. Therefore, it is important to review the subsets of real numbers and their properties.
  • The number line allows us to visually display real numbers by associating them with unique points on a line.
  • Special notation is used to communicate equality and order relationships between numbers on a number line.
  • The absolute value of a real number is defined geometrically as the distance between zero and the graph of that number on a number line. Alternatively, the absolute value of a real number is defined algebraically in a piecewise manner. If a real number a is nonnegative, then the absolute value will be that number a. If a is negative, then the absolute value will be the opposite of that number, −a.

Topic Exercises

    Part A: Real Numbers

      Use set notation to list the described elements.

    1. Every other positive odd number up to 21.

    2. Every other positive even number up to 22.

    3. The even prime numbers.

    4. Rational numbers that are also irrational.

    5. The set of negative integers.

    6. The set of negative even integers.

    7. Three consecutive odd integers starting with 13.

    8. Three consecutive even integers starting with 22.

      Determine the prime factorization of the given composite number.

    1. 195

    2. 78

    3. 330

    4. 273

    5. 180

    6. 350

      Reduce to lowest terms.

    1. 4230

    2. 10570

    3. 84120

    4. 315420

    5. 6045

    6. 144120

    7. 64128

    8. 72216

    9. 025

    10. 330

    Part B: Number Line and Notation

      Graph the following sets of numbers.

    1. {−5, 5, 10, 15}

    2. {−4, −2, 0, 2, 4}

    3. {32,12,0,1,2}

    4. {34,14,0,12,34}

    5. {−5,−4,−3,−1, 1}

    6. {−40, −30, −20, 10, 30}

      Simplify.

    1. −(−10)

    2. (35)

    3. −(−(−12))

    4. ((53))

    5. (((12)))

    6. ((((34))))

      Fill in the blank with <, =, or >.

    1. −10 _____ −15

    2. −101 _____ −100

    3. −33 _____ 0

    4. 0 _____ −50

    5. −(−(−2)) _____ −(−3)

    6. ((12)) _____ 14

    7. ((23)) _____ (12)

    8. (23) _____ (((23)))

      True or False.

    1. 0=0

    2. 55

    3. 1.032 is irrational.

    4. 0 is a nonnegative number.

    5. Any integer is a rational number.

    6. The constant π is rational.

      Graph the solution set and give the interval notation equivalent.

    1. x<1

    2. x>3

    3. x8

    4. x6

    5. 10x<4

    6. 3<x7

    7. 40<x<0

    8. 12x4

    9. x<5andx0

    10. x10andx40

    11. x7andx<10

    12. x<1andx>3

    13. x<2orx5

    14. x0orx4

    15. x<6orx>2

    16. x<0orx5

      Write an equivalent inequality.

    1. All real numbers less than −15.

    2. All real numbers greater than or equal to −7.

    3. All real numbers less than 6 and greater than zero.

    4. All real numbers less than zero and greater than −5.

    5. All real numbers less than or equal to 5 or greater than 10.

    6. All real numbers between −2 and 2.

      Determine the inequality given the answers expressed in interval notation.

    1. (,12)

    2. [8,)

    3. (,0]

    4. (0,)

    5. (6,14)

    6. (0,12]

    7. [5,25)

    8. [30,10]

    9. (,2)[3,)

    10. (,19][12,)

    11. (,2)(0,)

    12. (,15](5,)

    Part C: Absolute Value

      Simplify.

    1. |9|

    2. |14|

    3. |4|

    4. |8|

    5. |58|

    6. (|72|)

    7. |(7)|

    8. |(10)|

    9. (|2|)

    10. (|10|)

    11. (|(5)|)

    12. ((|20|))

      Determine the values represented by a.

    1. |a|=10

    2. |a|=7

    3. |a|=12

    4. |a|=94

    5. |a|=0

    6. |a|=1

    Part D: Discussion Board

    1. Research and discuss the origins and evolution of algebra.

    2. Research and discuss reasons why algebra is a required subject today.

    3. Solution sets to inequalities can be expressed using a graph, interval notation, or set notation. Discuss the merits and drawbacks of each method. Which do you prefer?

    4. Research and discuss the Fundamental Theorem of Algebra. Illustrate its idea with an example and share your results.

Answers

  1. {1, 5, 9, 13, 17, 21}

  2. {2}

  3. {…,−3, −2, −1}

  4. {13, 15, 17}

  5. 3513

  6. 23511

  7. 22335

  8. 75

  9. 710

  10. 43

  11. 12

  12. 0

  1. 10

  2. −12

  3. 12

  4. >

  5. <

  6. <

  7. <

  8. True

  9. False

  10. True

  11. (−∞, −1);

  12. [8,);

  13. [10,4);

  14. (40,0);

  15. [0,5);

  16. (,7);

  17. (,2)[5,);

  18. (,)=;

  19. x<15

  20. 0<x<6

  21. x5orx>10

  22. x<12

  23. x0

  24. 6<x<14

  25. 5x<25

  26. x<2orx3

  27. x<2orx>0

  1. 9

  2. −4

  3. 58

  4. −7

  5. 2

  6. 5

  7. a=±10

  8. a=±12

  9. a=0

  1. Answer may vary

  2. Answer may vary

1.2 Operations with Real Numbers

Learning Objectives

  1. Review the properties of real numbers.
  2. Simplify expressions involving grouping symbols and exponents.
  3. Simplify using the correct order of operations.

Working with Real Numbers

In this section, we continue to review the properties of real numbers and their operations. The result of adding real numbers is called the sumThe result of adding. and the result of subtracting is called the differenceThe result of subtracting.. Given any real numbers a, b, and c, we have the following properties of addition:

Additive Identity Property:

a+0=0+a=a

Additive Inverse Property:

a+(a)=(a)+a=0

Associative Property:

(a+b)+c=a+(b+c)

Commutative Property:

a+b=b+a

Given any real number a, a+0=0+a=a.

Given any real number a, a+(a)=(a)+a=0.

Given real numbers a, b and c, (a+b)+c=a+(b+c).

Given real numbers a and b, a+b=b+a.

It is important to note that addition is commutative and subtraction is not. In other words, the order in which we add does not matter and will yield the same result. However, this is not true of subtraction.

5+10=10+515=1551010555

We use these properties, along with the double-negative property for real numbers, to perform more involved sequential operations. To simplify things, make it a general rule to first replace all sequential operations with either addition or subtraction and then perform each operation in order from left to right.

Example 1

Simplify: 10(10)+(5).

Solution:

Replace the sequential operations and then perform them from left to right.

10(10)+(5)=10+105  Replace()withaddition(+).                     Replace+()withsubtraction().=05         =5

Answer: −5

Adding or subtracting fractions requires a common denominatorA denominator that is shared by more than one fraction.. Assume the common denominator c is a nonzero integer and we have

ac+bc=a+bcandacbc=abc

Example 2

Simplify: 29115+845.

Solution:

First determine the least common multiple (LCM) of 9, 15, and 45. The least common multiple of all the denominators is called the least common denominatorThe least common multiple of a set of denominators. (LCD). We begin by listing the multiples of each given denominator:

{9,18,27,36,45,54,63,72,81,90,}Multiplesof9{15,30,45,60,75,90,}Multiplesof15{45,90,135}Multiplesof45

Here we see that the LCM(9, 15, 45) = 45. Multiply the numerator and the denominator of each fraction by values that result in equivalent fractions with the determined common denominator.

29115+845=295511533+845=1045345+845

Once we have equivalent fractions, with a common denominator, we can perform the operations on the numerators and write the result over the common denominator.

=103+845=1545

And then reduce if necessary,

=15÷1545÷15=13

Answer: 13

Finding the LCM using lists of multiples, as described in the previous example, is often very cumbersome. For example, try making a list of multiples for 12 and 81. We can streamline the process of finding the LCM by using prime factors.

12=22381=34

The least common multiple is the product of each prime factor raised to the highest power. In this case,

LCM(12,81)=2234=324

Often we will find the need to translate English sentences involving addition and subtraction to mathematical statements. Below are some common translations.

n+2Thesumofanumberand2.2nThedifferenceof2andanumber.n2Here2issubtractedfromanumber.

Example 3

What is 8 subtracted from the sum of 3 and 12?

Solution:

We know that subtraction is not commutative; therefore, we must take care to subtract in the correct order. First, add 3 and 12 and then subtract 8 as follows:

Perform the indicated operations.

(3+12)8=(3122+12)8=(6+12)8=728122=7162=92

Answer: 92

The result of multiplying real numbers is called the productThe result of multiplying. and the result of dividing is called the quotientThe result of dividing.. Given any real numbers a, b, and c, we have the following properties of multiplication:

Zero Factor Property:

a0=0a=0

Multiplicative Identity Property:

a1=1a=a

Associative Property:

(ab)c=a(bc)

Commutative Property:

ab=ba

Given any real number a, a0=0a=0.

Given any real number a, a1=1a=a.

Given any real numbers a, b and c, (ab)c=a(bc).

Given any real numbers a and b, ab=ba.

It is important to note that multiplication is commutative and division is not. In other words, the order in which we multiply does not matter and will yield the same result. However, this is not true of division.

510=10550=505÷1010÷50.52

We will use these properties to perform sequential operations involving multiplication and division. Recall that the product of a positive number and a negative number is negative. Also, the product of two negative numbers is positive.

Example 4

Multiply: 5(3)(2)(4).

Solution:

Multiply two numbers at a time as follows:

Answer: −120

Because multiplication is commutative, the order in which we multiply does not affect the final answer. However, when sequential operations involve multiplication and division, order does matter; hence we must work the operations from left to right to obtain a correct result.

Example 5

Simplify: 10÷(2)(5).

Solution:

Perform the division first; otherwise the result will be incorrect.

Notice that the order in which we multiply and divide does affect the result. Therefore, it is important to perform the operations of multiplication and division as they appear from left to right.

Answer: 25

The product of two fractions is the fraction formed by the product of the numerators and the product of the denominators. In other words, to multiply fractions, multiply the numerators and multiply the denominators:

abcd=acbd

Example 6

Multiply: 452512.

Solution:

Multiply the numerators and multiply the denominators. Reduce by dividing out any common factors.

452512=425512=4125551123=53

Answer: 53

Two real numbers whose product is 1 are called reciprocalsTwo real numbers whose product is 1.. Therefore, ab and ba are reciprocals because abba=abab=1. For example,

2332=66=1

Because their product is 1, 23 and 32 are reciprocals. Some other reciprocals are listed below:

58and857and1745and54

This definition is important because dividing fractions requires that you multiply the dividend by the reciprocal of the divisor.

ab÷cd=abcddcdc=abdc1=abdc

In general,

ab÷cd=abdc=adbc

Example 7

Simplify: 54÷3512.

Solution:

Perform the multiplication and division from left to right.

54÷3512=545312=551432=2524

In algebra, it is often preferable to work with improper fractions. In this case, we leave the answer expressed as an improper fraction.

Answer: 2524

Try this! Simplify: 1234÷18.

Answer: 3

Grouping Symbols and Exponents

In a computation where more than one operation is involved, grouping symbols help tell us which operations to perform first. The grouping symbolsParentheses, brackets, braces, and the fraction bar are the common symbols used to group expressions and mathematical operations within a computation. commonly used in algebra are:

()   Parentheses[]   Brackets{}   Braces    Fractionbar

All of the above grouping symbols, as well as absolute value, have the same order of precedence. Perform operations inside the innermost grouping symbol or absolute value first.

Example 8

Simplify: 2(45215).

Solution:

Perform the operations within the parentheses first.

2(45215)=2(4533215)=2(1215215)=2(1015)=213323=623=43

Answer: 43

Example 9

Simplify: 5|4(3)||3|(57).

Solution:

The fraction bar groups the numerator and denominator. Hence, they should be simplified separately.

5|4(3)||3|(57)=5|4+3||3|(2)=5|7||3|+2=573+2=25=25

Answer: 25

If a number is repeated as a factor numerous times, then we can write the product in a more compact form using exponential notationThe compact notation an used when a factor a is repeated n times.. For example,

5555=54

The baseThe factor a in the exponential notation an. is the factor and the positive integer exponentThe positive integer n in the exponential notation an that indicates the number of times the base is used as a factor. indicates the number of times the base is repeated as a factor. In the above example, the base is 5 and the exponent is 4. Exponents are sometimes indicated with the caret (^) symbol found on the keyboard, 5^4 = 5*5*5*5. In general, if a is the base that is repeated as a factor n times, then

When the exponent is 2 we call the result a squareThe result when the exponent of any real number is 2., and when the exponent is 3 we call the result a cubeThe result when the exponent of any real number is 3.. For example,

52=55=255squared53=555=1255cubed

If the exponent is greater than 3, then the notation an is read, “a raised to the nth power.” The base can be any real number,

(2.5)2=(2.5)(2.5)=6.25(23)3=(23)(23)(23)=827(2)4=(2)(2)(2)(2)=1624=12222=16

Notice that the result of a negative base with an even exponent is positive. The result of a negative base with an odd exponent is negative. These facts are often confused when negative numbers are involved. Study the following four examples carefully:

The base is (−3).

The base is 3.

(3)4=(3)(3)(3)(3)=+81(3)3=(3)(3)(3)=27

34=13333=8133=1333=27

The parentheses indicate that the negative number is to be used as the base.

Example 10

Calculate:

  1. (13)3
  2. (13)4

Solution:

Here 13 is the base for both problems.

  1. Use the base as a factor three times.

    (13)3=(13)(13)(13)=127

  2. Use the base as a factor four times.

    (13)4=(13)(13)(13)(13)=+181

Answers:

  1. 127
  2. 181

Try this! Simplify:

  1. 24
  2. (2)4

Answers:

  1. −16
  2. 16

Order of Operations

When several operations are to be applied within a calculation, we must follow a specific order to ensure a single correct result.

  1. Perform all calculations within the innermost parentheses or grouping symbol first.
  2. Evaluate all exponents.
  3. Apply multiplication and division from left to right.
  4. Perform all remaining addition and subtraction operations last from left to right.

Note that multiplication and division should be worked from left to right. Because of this, it is often reasonable to perform division before multiplication.

Example 11

Simplify: 5324÷612+2.

Solution:

First, evaluate 53 and then perform multiplication and division as they appear from left to right.

5324÷612+2=5324÷612+2=12524÷612+2=125412+2=1252+2=123+2=125

Multiplying first would have led to an incorrect result.

Answer: 125

Example 12

Simplify: 1052+(3)4.

Solution:

Take care to correctly identify the base when squaring.

1052+(3)4=1025+81=35+81=46

Answer: 46

We are less likely to make a mistake if we work one operation at a time. Some problems may involve an absolute value, in which case we assign it the same order of precedence as parentheses.

Example 13

Simplify: 75|22+(3)2|.

Solution:

Begin by performing the operations within the absolute value first.

75|22+(3)2|=75|4+9|=75|5|=755=725=18

Subtracting 75 first will lead to incorrect results.

Answer: −18

Try this! Simplify: 62[15(2)3](2)4.

Answer: −45

Key Takeaways

  • Addition is commutative and subtraction is not. Furthermore, multiplication is commutative and division is not.
  • Adding or subtracting fractions requires a common denominator; multiplying or dividing fractions does not.
  • Grouping symbols indicate which operations to perform first. We usually group mathematical operations with parentheses, brackets, braces, and the fraction bar. We also group operations within absolute values. All groupings have the same order of precedence: the operations within the innermost grouping are performed first.
  • When using exponential notation an, the base a is used as a factor n times. Parentheses indicate that a negative number is to be used as the base. For example, (5)2 is positive and 52 is negative.
  • To ensure a single correct result when applying operations within a calculation, follow the order of operations. First, perform operations in the innermost parentheses or groupings. Next, simplify all exponents. Perform multiplication and division operations from left to right. Finally, perform addition and subtraction operations from left to right.

Topic Exercises

    Part A: Working with Real Numbers

      Perform the operations. Reduce all fractions to lowest terms.

    1. 33(15)+(8)

    2. 109+(6)

    3. 23+(7)(10)

    4. 1(1)1

    5. 12+1316

    6. 15+12110

    7. 23(14)16

    8. 32(29)56

    9. 34(12)58

    10. 1532(710)

    11. Subtract 3 from 10.

    12. Subtract −2 from 16.

    13. Subtract 56 from 4.

    14. Subtract 12 from 32.

    15. Calculate the sum of −10 and 25.

    16. Calculate the sum of −30 and −20.

    17. Find the difference of 10 and 5.

    18. Find the difference of −17 and −3.

      The formula d=|ba| gives the distance between any two points on a number line. Determine the distance between the given numbers on a number line.

    1. 10 and 15

    2. 6 and 22

    3. 0 and 12

    4. −8 and 0

    5. −5 and −25

    6. −12 and −3

      Determine the reciprocal of the following.

    1. 13

    2. 25

    3. 34

    4. −12

    5. a where a0

    6. 1a

    7. ab where a0

    8. 1ab

      Perform the operations.

    1. 4(5)÷2

    2. (15)(3)÷(9)

    3. 22÷(11)(2)

    4. 50÷(25)(4)

    5. 23(910)

    6. 58(1625)

    7. 76(67)

    8. 159(95)

    9. 45(25)÷1625

    10. (92)(32)÷2716

    11. 85÷521540

    12. 316÷5812

    13. Find the product of 12 and 7.

    14. Find the product of 23 and 12.

    15. Find the quotient of −36 and 12.

    16. Find the quotient of 34 and 9.

    17. Subtract 10 from the sum of 8 and −5.

    18. Subtract −2 from the sum of −5 and −3.

    19. Joe earns $18.00 per hour and “time and a half” for every hour he works over 40 hours. What is his pay for 45 hours of work this week?

    20. Billy purchased 12 bottles of water at $0.75 per bottle, 5 pounds of assorted candy at $4.50 per pound, and 15 packages of microwave popcorn costing $0.50 each for his party. What was his total bill?

    21. James and Mary carpooled home from college for the Thanksgiving holiday. They shared the driving, but Mary drove twice as far as James. If Mary drove for 210 miles, then how many miles was the entire trip?

    22. A 634 foot plank is to be cut into 3 pieces of equal length. What will be the length of each piece?

    23. A student earned 72, 78, 84, and 90 points on her first four algebra exams. What was her average test score? (Recall that the average is calculated by adding all the values in a set and dividing that result by the number of elements in the set.)

    24. The coldest temperature on Earth, −129° F, was recorded in 1983 at Vostok Station, Antarctica. The hottest temperature on Earth, 136° F, was recorded in 1922 at Al’ Aziziyah, Libya. Calculate the temperature range on Earth.

    Part B: Grouping Symbols and Exponents

      Perform the operations.

    1. 7{3[6(10)]}

    2. (912)[6(83)]

    3. 12{5(103)}

    4. 23{6+(69)}

    5. 5{2[3(432)]}
    6. 12{6[(1253)]}
    7. 5|5(6)||5||3|
    8. |912|(3)|16|3(4)
    9. |5(7)|(2)|2|+|3|
    10. 1|9(34)||2|+(8(10))

      Perform the operations.

    1. 122

    2. (12)2

    3. 122

    4. (12)2

    5. 54

    6. (5)4

    7. (12)3

    8. (12)3

    9. (34)2

    10. (52)3

    11. (1)22

    12. (1)13

    13. (1)12

    14. (1)5

    15. 102

    16. 104

    Part C: Order of Operations

      Simplify.

    1. 53(432)

    2. 85(332)

    3. (5)2+3(242)

    4. 62(52+47)
    5. 53[3(232)+(3)2]
    6. 105[(25)23]
    7. [5232][2(5+(4)2)]
    8. 72[(27)2(8)2]
    9. 316÷(51212+23)4
    10. 6[(23)2(12)2]÷(2)2
    11. 325+42232
    12. (3+(2)2)4342+1
    13. 52+(3)22382+6(10)
    14. (4)2+(3)392(12+22)*10
    15. 522|5|
    16. 24+6|2452|
    17. (4|7282|)
    18. 3(52|6|)
    19. (3)2|2+(3)3|42
    20. 522|3324|(2)5
    21. 5|5|(2|7|)3
    22. 102+2(|5|363)
    23. 23|12(43)2|
    24. 24|10312÷15|
    25. Calculate the sum of the squares of the first three consecutive positive odd integers.

    26. Calculate the sum of the squares of the first three consecutive positive even integers.

    27. What is 6 subtracted from the sum of the squares of 5 and 8?

    28. What is 5 subtracted from the sum of the cubes of 2 and 3?

    Part D: Discussion Board

    1. What is PEMDAS and what is it missing?

    2. Does 0 have a reciprocal? Explain.

    3. Explain why we need a common denominator in order to add or subtract fractions.

    4. Explain why (10)4 is positive and 104 is negative.

Answers

  1. 40

  2. −20

  3. 23

  4. 34

  5. 58

  6. 7

  7. 296

  8. 15

  9. 5

  10. 5 units

  11. 12 units

  12. 20 units

  13. 3

  14. 43

  15. 1a

  16. ba

  17. 10

  18. −4

  19. 35

  20. −1

  21. 12

  22. 625

  23. 84

  24. −3

  25. −7

  26. $855

  27. 315 miles

  28. 81 points

  1. −12

  2. −1

  3. 75

  4. −3

  5. 0

  6. 144

  7. −144

  8. −625

  9. 18

  10. 916

  11. 1

  12. −1

  13. −100

  1. 20

  2. −17

  3. 41

  4. 35

  5. 97

  6. 35

  7. 52

  8. −35

  9. 11

  10. −36

  11. 150

  12. 1118

  13. 35

  14. 83

  1. Answer may vary

  2. Answer may vary

1.3 Square and Cube Roots of Real Numbers

Learning Objectives

  1. Calculate the exact and approximate value of the square root of a real number.
  2. Calculate the exact and approximate value of the cube root of a real number.
  3. Simplify the square and cube root of a real number.
  4. Apply the Pythagorean theorem.

The Definition of Square and Cube Roots

A square rootThat number that when multiplied by itself yields the original number. of a number is a number that when multiplied by itself yields the original number. For example, 4 is a square root of 16, because 42=16. Since (4)2=16, we can say that −4 is a square root of 16 as well. Every positive real number has two square roots, one positive and one negative. For this reason, we use the radical signThe symbol used to denote a square root. to denote the principal (nonnegative) square rootThe non-negative square root. and a negative sign in front of the radical to denote the negative square root.

16=4           Positivesquarerootof1616=4        Negativesquarerootof16

Zero is the only real number with exactly one square root.

0=0

If the radicandThe number within a radical., the number inside the radical sign, is nonzero and can be factored as the square of another nonzero number, then the square root of the number is apparent. In this case, we have the following property:

a2=a,ifa0

It is important to point out that a is required to be nonnegative. Note that (3)23 because the radical denotes the principal square root. Instead,

(3)2=9=3

This distinction will be carefully considered later in the course.

Example 1

Find the square root:

  1. 121
  2. 0.25
  3. 49

Solution:

  1. 121=112=11
  2. 0.25=0.52=0.5
  3. 49=(23)2=23

Example 2

Find the negative square root:

  1. 64
  2. 1

Solution:

  1. 64=82=8
  2. 1=12=1

The radicand may not always be a perfect square. If a positive integer is not a perfect square, then its square root will be irrational. Consider 5, we can obtain an approximation by bounding it using the perfect squares 4 and 9 as follows:

4<5<92<5<3

With this we conclude that 5 is somewhere between 2 and 3. This number is better approximated on most calculators using the square root button, .

52.236because2.236^25

Next, consider the square root of a negative number. To determine the square root of −9, you must find a number that when squared results in −9,

9=?or(?)2=9

However, any real number squared always results in a positive number,

(3)2=9and(3)2=9

The square root of a negative number is currently left undefined. Try calculating 9 on your calculator; what does it say? For now, we will state that 9 is not a real number. The square root of a negative number is defined later in the course.

A cube rootThe number that when multiplied by itself three times yields the original number, denoted by 3. of a number is a number that when multiplied by itself three times yields the original number. Furthermore, we denote a cube root using the symbol 3, where 3 is called the indexThe positive integer n in the notation n that is used to indicate an nth root.. For example,

83=2,because23=8

The product of three equal factors will be positive if the factor is positive, and negative if the factor is negative. For this reason, any real number will have only one real cube root. Hence the technicalities associated with the principal root do not apply. For example,

83=2,because(2)3=8

In general, given any real number a, we have the following property:

a33=a

When simplifying cube roots, look for factors that are perfect cubes.

Example 3

Find the cube root:

  1. 1253
  2. 03
  3. 8273

Solution:

  1. 1253=533=5
  2. 03=033=0
  3. 8273=(23)33=23

Example 4

Find the cube root:

  1. 273
  2. 13

Solution:

  1. 273=(3)33=3
  2. 13=(1)33=1

It may be the case that the radicand is not a perfect cube. If this is the case, then its cube root will be irrational. For example, 23 is an irrational number, which can be approximated on most calculators using the root button x. Depending on the calculator, we typically type in the index prior to pushing the button and then the radicand as follows:

3yx2=

Therefore, we have

231.260,because1.260^32

We will extend these ideas using any integer as an index later in this course. It is important to point out that a square root has index 2; therefore, the following are equivalent:

a2=a

In other words, if no index is given, it is assumed to be the square root.

Simplifying Square and Cube Roots

It will not always be the case that the radicand is a perfect square. If not, we use the following two properties to simplify the expression. Given real numbers An and Bn where B0,

Product Rule for Radicals:

ABn=AnBn

Quotient Rule for Radicals:

ABn=AnBn

Given real numbers An and Bn, ABn=AnBn.

Given real numbers An and Bn, ABn=AnBn.

A simplified radicalA radical where the radicand does not consist of any factors that can be written as perfect powers of the index. is one where the radicand does not consist of any factors that can be written as perfect powers of the index. Given a square root, the idea is to identify the largest square factor of the radicand and then apply the property shown above. As an example, to simplify 12, notice that 12 is not a perfect square. However, 12 does have a perfect square factor, 12=43. Apply the property as follows: 12=43          Applytheproductruleforradicals.         =43      Simplify.=23            The number 23 is a simplified irrational number. You are often asked to find an approximate answer rounded off to a certain decimal place. In that case, use a calculator to find the decimal approximation using either the original problem or the simplified equivalent. 12=233.46 As a check, calculate 12 and 23 on a calculator and verify that the results are both approximately 3.46.

Example 5

Simplify: 135.

Solution:

Begin by finding the largest perfect square factor of 135.

135=335=3235=915

Therefore,

135=915          Applytheproductruleforradicals.=915      Simplify.=315            

Answer: 315

Example 6

Simplify: 108169.

Solution:

We begin by finding the prime factorizations of both 108 and 169. This will enable us to easily determine the largest perfect square factors.

108=2233=22323169=132

Therefore,

108169=22323132          Applytheproductandquotientruleforradicals.=22323132      Simplify.=23313=6313

Answer: 6313

Example 7

Simplify: 5162.

Solution:

5162=5812=5812=592=452=452

Answer: 452

Try this! Simplify: 4150.

Answer: 206

A cube root is simplified if it does not contain any factors that can be written as perfect cubes. The idea is to identify the largest cube factor of the radicand and then apply the product or quotient rule for radicals. As an example, to simplify 803, notice that 80 is not a perfect cube. However, 80=810 and we can write,

803=8103          Applytheproductruleforradicals.         =83103      Simplify.=2103            

Example 8

Simplify: 1623.

Solution:

Begin by finding the largest perfect cube factor of 162.

162=342=3332=276

Therefore,

1623=2763         Applytheproductruleforradicals.=27363     Simplify.=363           

Answer: 363

Example 9

Simplify: 163433.

Solution:

163433=1823733=138323733=12237=2237

Answer: 2237

Try this! Simplify: 22563.

Answer: 843

Consider the following two calculations,

81=92=981=92=(9)2=(3)2=9

Notice that it does not matter if we apply the exponent first or the square root first. This is true for any positive real number. We have the following,

a2=(a)2=a,ifa0

Example 10

Simplify: (10)2.

Solution:

Apply the fact that (a)2=a if a is nonnegative.

(10)2=10

Pythagorean Theorem

A right triangleA triangle with an angle that measures 90°. is a triangle where one of the angles measures 90°. The side opposite the right angle is the longest side, called the hypotenuseThe longest side of a right triangle; it will always be the side opposite the right angle., and the other two sides are called legsThe sides of a right triangle that are not the hypotenuse.. Numerous real-world applications involve this geometric figure. The Pythagorean theoremThe hypotenuse of any right triangle is equal to the square root of the sum of the squares of the lengths of the triangle’s legs. states that given any right triangle with legs measuring a and b units, the square of the measure of the hypotenuse c is equal to the sum of the squares of the measures of the legs, a2+b2=c2. In other words, the hypotenuse of any right triangle is equal to the square root of the sum of the squares of its legs.

Example 11

Calculate the diagonal of a square with sides measuring 5 units.

Solution:

The diagonal of a square will form an isosceles right triangle where the two equal legs measure 5 units each.

We can use the Pythagorean theorem to determine the length of the hypotenuse.

c=a2+b2=52+52=25+25=50=252=252=52

Answer: 52 units

The Pythagorean theorem actually states that having side lengths satisfying the property a2+b2=c2 is a necessary and sufficient condition of right triangles. In other words, if we can show that the sum of the squares of the lengths of the legs of the triangle is equal to the square of the hypotenuse, then it must be a right triangle.

Example 12

Determine whether or not a triangle with legs a=1 cm and b=2 cm and hypotenuse b=5 cm is a right triangle.

Solution:

If the legs satisfy the condition a2+b2=c2 then the Pythagorean theorem guarantees that the triangle is a right triangle.

a2+b2=c2(1)2+(2)2=?(5)21+4=55=5

Answer: Yes, the described triangle is a right triangle.

Key Takeaways

  • The square root of a number is a number that when squared results in the original number. The principal square root of a positive real number is the positive square root. The square root of a negative number is currently left undefined.
  • When simplifying the square root of a number, look for perfect square factors of the radicand. Apply the product or quotient rule for radicals and then simplify.
  • The cube root of a number is a number that when cubed results in the original number. Every real number has only one real cube root.
  • When simplifying cube roots, look for perfect cube factors of the radicand. Apply the product or quotient rule for radicals and then simplify.
  • The Pythagorean theorem gives us a necessary and sufficient condition of right triangles: a2+b2=c2 if and only if a, b and c represent the lengths of the sides of a right triangle.

Topic Exercises

    Part A: Square and Cube Roots

      Simplify.

    1. 81

    2. 49

    3. 16

    4. 100

    5. 2516
    6. 964
    7. 14
    8. 1100
    9. 1

    10. 25

    11. 0.36

    12. 1.21

    13. (5)2

    14. (6)2

    15. 264

    16. 336

    17. 104

    18. 825

    19. 643

    20. 1253

    21. 273

    22. 13

    23. 03

    24. 0.0083

    25. 0.0643

    26. 83

    27. 10003

    28. (8)33
    29. (15)33
    30. 12163
    31. 27643
    32. 183
    33. 1273
    34. 53433

    35. 45123

    36. 1083

    37. 6643

    38. 883

      Use a calculator to approximate to the nearest hundredth.

    1. 3

    2. 10

    3. 19

    4. 7

    5. 35

    6. 23

    7. 33

    8. 63

    9. 283

    10. 93

    11. 4103

    12. 3123

    13. Determine the set consisting of the squares of the first twelve positive integers.

    14. Determine the set consisting of the cubes of the first twelve positive integers.

    Part B: Simplifying Square Roots and Cube Roots

      Simplify.

    1. 18

    2. 50

    3. 24

    4. 40

    5. 5081
    6. 5425
    7. 472

    8. 327

    9. 580

    10. 6128

    11. 340

    12. 5160

    13. 163

    14. 543

    15. 813

    16. 243

    17. 481253
    18. 135643
    19. 75003

    20. 256863

    21. 21623

    22. 5963

    23. (64)2

    24. (25)2

    25. (2)2

    26. (6)2

    Part C: Pythagorean Theorem

    1. If the two legs of a right triangle measure 3 units and 4 units, then find the length of the hypotenuse.

    2. If the two legs of a right triangle measure 6 units and 8 units, then find the length of the hypotenuse.

    3. If the two equal legs of an isosceles right triangle measure 7 units, then find the length of the hypotenuse.

    4. If the two equal legs of an isosceles right triangle measure 10 units, then find the length of the hypotenuse.

    5. Calculate the diagonal of a square with sides measuring 3 centimeters.

    6. Calculate the diagonal of a square with sides measuring 10 centimeters.

    7. Calculate the diagonal of a square with sides measuring 6 centimeters.

    8. Calculate the diagonal of a square with sides measuring 10 centimeters.

    9. Calculate the length of the diagonal of a rectangle with dimensions 4 centimeters by 8 centimeters.

    10. Calculate the length of the diagonal of a rectangle with dimensions 8 meters by 10 meters.

    11. Calculate the length of the diagonal of a rectangle with dimensions 3 meters by 2 meters.

    12. Calculate the length of the diagonal of a rectangle with dimensions 6 meters by 10 meters.

    13. To ensure that a newly built gate is square, the measured diagonal must match the distance calculated using the Pythagorean theorem. If the gate measures 4 feet by 4 feet, what must the diagonal measure in inches? (Round off to the nearest tenth of an inch.)

    14. If a doorframe measures 3.5 feet by 6.6 feet, what must the diagonal measure to ensure that the frame is a perfect rectangle?

      Determine whether or not the given triangle with legs a and b and hypotenuse c is a right triangle or not.

    1. a=3, b=7, and c=10

    2. a=5, b=12, and c=13

    3. a=8, b=15, and c=17

    4. a=7, b=24, and c=30

    5. a=3, b=2, and c=13

    6. a=7, b=4, and c=11

    7. a=4, b=3, and c=19

    8. a=6, b=15, and c=21

    Part D: Discussion Board

    1. What does your calculator say after taking the square root of a negative number? Share your results on the discussion board and explain why it says that.

    2. Research and discuss the history of the Pythagorean theorem.

    3. Research and discuss the history of the square root.

    4. Discuss the importance of the principal square root. Why is it that the same issue does not come up with cube roots? Provide some examples with your explanation.

Answers

  1. 9

  2. −4

  3. 54

  4. 12

  5. Not a real number.

  6. 0.6

  7. 5

  8. 16

  9. −20

  10. 4

  11. −3

  12. 0

  13. 0.4

  14. −10

  15. −15

  16. 34

  17. 13

  18. 32

  19. 24

  20. 1.73

  21. 4.36

  22. 6.71

  23. 1.44

  24. 3.04

  25. 8.62

  26. {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144}

  1. 32

  2. 26

  3. 529
  4. 242

  5. 205

  6. Not a real number.

  7. 223

  8. 333

  9. 2635
  10. 3543

  11. 663

  12. 64

  13. 2

  1. 5 units

  2. 72 units

  3. 32 centimeters

  4. 23 centimeters

  5. 45 centimeters

  6. 7 meters

  7. The diagonal must measure approximately 67.9 inches.

  8. Not a right triangle.

  9. Right triangle.

  10. Right triangle.

  11. Right triangle.

  1. Answer may vary

  2. Answer may vary

1.4 Algebraic Expressions and Formulas

Learning Objectives

  1. Identify the parts of an algebraic expression.
  2. Apply the distributive property.
  3. Evaluate algebraic expressions.
  4. Use formulas that model common applications.

Algebraic Expressions and the Distributive Property

In algebra, letters called variables are used to represent numbers. Combinations of variables and numbers along with mathematical operations form algebraic expressionsCombinations of variables and numbers along with mathematical operations used to generalize specific arithmetic operations., or just expressions. The following are some examples of expressions with one variable, x:

2x+3

x29

1x+xx+2

3x+x

TermsComponents of an algebraic expression separated by addition operators. in an algebraic expression are separated by addition operators and factorsComponents of a term separated by multiplication operators. are separated by multiplication operators. The numerical factor of a term is called the coefficientThe numerical factor of a term.. For example, the algebraic expression x2y2+6xy3 can be thought of as x2y2+6xy+(3) and has three terms. The first term, x2y2, represents the quantity 1x2y2=1xxyy where 1 is the coefficient and x and y are the variables. All of the variable factors with their exponents form the variable part of a termAll the variable factors with their exponents.. If a term is written without a variable factor, then it is called a constant termA term written without a variable factor.. Consider the components of x2y2+6xy3,

Terms

Coefficient

Variable Part

x2y2

1

x2y2

6xy

6

xy

−3

−3

 

The third term in this expression, −3, is called a constant term because it is written without a variable factor. While a variable represents an unknown quantity and may change, the constant term does not change.

Example 1

List all coefficients and variable parts of each term: 10a25abb2.

Solution:

We want to think of the third term in this example b2 as 1b2.

Terms

Coefficient

Variable Part

10a2

10

a2

5ab

−5

ab

b2

−1

b2

Answer: Coefficients: {5,1,10}; Variable parts: {a2,ab,b2}

In our study of algebra, we will encounter a wide variety of algebraic expressions. Typically, expressions use the two most common variables, x and y. However, expressions may use any letter (or symbol) for a variable, even Greek letters, such as alpha (α) and beta (β). Some letters and symbols are reserved for constants, such as π3.14159 and e2.71828. Since there is only a limited number of letters, you will also use subscripts,x1,x2,x3,x4,, to indicate different variables.

The properties of real numbers are important in our study of algebra because a variable is simply a letter that represents a real number. In particular, the distributive propertyGiven any real numbers a, b, and c, a(b+c)=ab+ac or (b+c)a=ba+ca. states that if given any real numbers a, b and c, then,

a(b+c)=ab+ac

This property is one that we apply often when simplifying algebraic expressions. To demonstrate how it will be used, we simplify 2(53) in two ways, and observe the same correct result.

Working parenthesis first.

Using the distributive property.

2(53)=2(2)=4

2(53)=2523=106=4

Certainly, if the contents of the parentheses can be simplified we should do that first. On the other hand, when the contents of parentheses cannot be simplified any further, we multiply every term within it by the factor outside of it using the distributive property. Applying the distributive property allows us to multiply and remove the parentheses.

Example 2

Simplify: 5(2a+5b)2c.

Solution:

Multiply only the terms grouped within the parentheses for which we are applying the distributive property.

=5(2a)+55b2c=10a+25b2c

Answer: 10a+25b2c

Recall that multiplication is commutative and therefore we can write the distributive property in the following manner, (b+c)a=ba+ca.

Example 3

Simplify: (3x4y+1)3.

Solution:

Multiply all terms within the parenthesis by 3.

(3x4y+1)3=3x34y3+13=9x12y+3

Answer:9x12y+3

Terms whose variable parts have the same variables with the same exponents are called like termsConstant terms or terms whose variable parts have the same variables with the same exponents., or similar termsUsed when referring to like terms.. Furthermore, constant terms are considered to be like terms. If an algebraic expression contains like terms, apply the distributive property as follows:

5x+7x=(5+7)x=12x4x2+5x27x2=(4+57)x2=2x2

In other words, if the variable parts of terms are exactly the same, then we can add or subtract the coefficients to obtain the coefficient of a single term with the same variable part. This process is called combining like termsAdding or subtracting like terms within an algebraic expression to obtain a single term with the same variable part.. For example,

12x2y3+3x2y3=15x2y3

Notice that the variable factors and their exponents do not change. Combining like terms in this manner, so that the expression contains no other similar terms, is called simplifying the expressionThe process of combining like terms until the expression contains no more similar terms.. Use this idea to simplify algebraic expressions with multiple like terms.

Example 4

Simplify: x210x+8+5x26x1.

Solution:

Identify the like terms and add the corresponding coefficients.

1x210x+8+5x26x1Combineliketerms.=6x216x+7

Answer: 6x216x+7

Example 5

Simplify: a2b2ab2(2a2b25ab+1).

Solution:

Distribute −2 and then combine like terms.

a2b2ab2(2a2b25ab+1)=a2b2ab4a2b2+10ab2=3a2b2+9ab2

Answer: 3a2b2+9ab2

Evaluating Algebraic Expressions

An algebraic expression can be thought of as a generalization of particular arithmetic operations. Performing these operations after substituting given values for variables is called evaluatingThe process of performing the operations of an algebraic expression for given values of the variables.. In algebra, a variable represents an unknown value. However, if the problem specifically assigns a value to a variable, then you can replace that letter with the given number and evaluate using the order of operations.

Example 6

Evaluate:

  1. 5x2 where x=23
  2. y2y6 where y=4

Solution:

To avoid common errors, it is a best practice to first replace all variables with parentheses, and then replace, or substituteThe act of replacing a variable with an equivalent quantity., the appropriate given value.

a.

5x2=5()2=5(23)2=1032133=1063=43

b.

y2y6=()2()6=(4)2(4)6=16+46=14

Answer:

  1. 43
  2. 14

Often algebraic expressions will involve more than one variable.

Example 7

Evaluate a38b3 where a=1 and b=12.

Solution:

After substituting in the appropriate values, we must take care to simplify using the correct order of operations.

a38b3=()38()3Replacevariableswithparentheses.=(1)38(12)3Substituteintheappropriatevalues.=18(18)Simplify.=11=2

Answer: −2

Example 8

Evaluate x2y22x1 where x=32 and y=3.

Solution:

x2y22x1=()2()22()1=(32)2(3)22(32)1=94931

At this point we have a complex fraction. Simplify the numerator and then multiply by the reciprocal of the denominator.

=9491444=27441=274(14)=2716

Answer: 2716

The answer to the previous example can be written as a mixed number, 2716=11116. Unless the original problem has mixed numbers in it, or it is an answer to a real-world application, solutions will be expressed as reduced improper fractions.

Example 9

Evaluate b24ac where a=1, b=7, and c=14.

Solution:

Substitute in the appropriate values and then simplify.

b24ac=()24()()=(7)24(1)(14)=49+4(14)=49+1=50=252=52

Answer: 52

Try this! Evaluate 3πVhπh where V=25π and h=3.

Answer: 5

Using Formulas

The main difference between algebra and arithmetic is the organized use of variables. This idea leads to reusable formulasA reusable mathematical model using algebraic expressions to describe a common application., which are mathematical models using algebraic expressions to describe common applications. For example, the volume of a right circular cone depends on its radius r and height h and is modeled by the formula:

V=13πr2h

In this equation, variables and constants are used to describe the relationship between volume and the length of the base and height. If the radius of the base measures 3 meters and the height measures 5 meters, then the volume can be calculated using the formula as follows: V=13πr2h=13π(3m)2(5m)=13π935m3=15πm3 Using π3.14, we can approximate the volume: V15(3.14)=47.1 cubic meters.

 

A list of formulas that describe the area and perimeter of common plane figures follows. The letter P represents perimeter and is measured in linear units. The letter A represents area and is measured in square units.

A list of formulas that describe the surface area and volume of common figures follows. Here SA represents surface area and is measured in square units. The letter V represents volume and is measured in cubic units.

Example 10

The diameter of a spherical balloon is 10 inches. Determine the volume rounded off to the nearest hundredth.

Solution:

The formula for the volume of a sphere is

V=43πr3

This formula gives the volume in terms of the radius, r. Therefore, divide the diameter by 2 and then substitute into the formula. Here, r=102=5 inches and we have

V=43πr3=43π(5in)3=43π125in3=500π3in3523.60in3

Answer: The volume of the balloon is approximately 523.60 cubic inches.

Formulas can be found in a multitude of subjects. For example, uniform motionThe distance D after traveling at an average rate r for some time t can be calculated using the formula D=rt. is modeled by the formula D=rt, which expresses distance D, in terms of the average rate, or speed, r and the time traveled at that rate, t. This formula, D=rt, is used often and is read, “distance equals rate times time.

Example 11

Jim’s road trip took 212 hours at an average speed of 66 miles per hour. How far did he travel?

Solution:

Substitute the appropriate values into the formula and then simplify.

D=rt=(66mihr)(212hr)=66152mi=335mi=165mi

Answer: Jim traveled 165 miles.

Simple interestModeled by the formula I=prt, where p represents the principal amount invested at an annual interest rate r for t years.I is given by the formula I=prt, where p represents the principal amount invested at an annual interest rate r for t years.

Example 12

Calculate the simple interest earned on a 2-year investment of $1,250 at an annual interest rate of 334%.

Solution:

Convert 334% to a decimal number before using it in the formula.

r=334%=3.75%=0.0375

Use this and the fact that p = $1,250 and t = 2 years to calculate the simple interest.

I=prt=(1,250)(0.0375)(2)=93.75

Answer: The simple interest earned is $93.75.

Key Takeaways

  • Think of algebraic expressions as generalizations of common arithmetic operations that are formed by combining numbers, variables, and mathematical operations.
  • The distributive property a(b+c)=ab+ac, is used when multiplying grouped algebraic expressions. Applying the distributive property allows us to remove parentheses.
  • Combine like terms, or terms whose variable parts have the same variables with the same exponents, by adding or subtracting the coefficients to obtain the coefficient of a single term with the same variable part. Remember that the variable factors and their exponents do not change.
  • To avoid common errors when evaluating, it is a best practice to replace all variables with parentheses and then substitute the appropriate values.
  • The use of algebraic expressions allows us to create useful and reusable formulas that model common applications.

Topic Exercises

    Part A: Algebraic Expressions and the Distributive Property

      List all of the coefficients and variable parts of each term.

    1. 5x2+x1

    2. y29y+3

    3. 5x23xy+y2

    4. a2b2+2ab4

    5. x2y+xy23xy+9

    6. x4x3+x2x+2

      Multiply.

    1. 5(3x5)

    2. 3(4x1)

    3. 2(2x25x+1)

    4. 5(6x23x1)

    5. 23(9y2+12y3)
    6. 34(8y2+20y+4)
    7. 12(13a256a+712)
    8. 9(19a253a+1)
    9. 9(a22b2)

    10. 5(3x2y2)

    11. (5a23ab+b2)6

    12. (a2b29ab3)7

    13. (5x2xy+y2)

    14. (x2y26xy1)

      Combine like terms.

    1. 18x5x+3x

    2. 30x50x+10x

    3. 3y4+2y12

    4. 12y+715y6

    5. 2x23x+2+5x26x+1

    6. 9x2+7x510x28x+6

    7. 35a212+13a2+45
    8. 16a2+2343a219
    9. 12y2+23y3+35y2+13y73
    10. 56x2+18x112x2+34x45
    11. a2b2+5ab2+7a2b26ab+12

    12. a212ab+4b26a2+10ab5b2

    13. 3x2y+12xy5xy2+5xy8x2y+2xy2

    14. 10x2y+2xy4xy2+2x2y8xy+5xy2

    15. 7m2n9mn+mn26m2n+mn2mn2

    16. m2n5mn+5mn23m2n+5mn+2mn2

    17. x2n3xn+5+2x2n4xn3

    18. 5y2n3yn+13y2n2yn1

      Simplify.

    1. 52(4x+8)

    2. 86(2x1)

    3. 2(x27x+1)+3x7

    4. 5(x2+4x1)+8x25

    5. 5ab4(ab+5)

    6. 5(7ab)+2ab

    7. 2a2+3(a2+4)

    8. 73y+2(y23y2)

    9. 8x23x5(x2+4x1)

    10. 25y6(y2y+2)

    11. a2b25+3(a2b23ab+2)

    12. a23ab2(a2ab+1)

    13. 10y2+6(3y2+2y+4)

    14. 4m23mn(m23mn+n2)

    15. x2n3xn+5(x2nxn+1)

    16. 3(y2n2yn+1)+4y2n5

    Part B: Evaluating Algebraic Expressions

      Evaluate.

    1. 2x+3 where x=2

    2. 8x5 where x=1

    3. x2x+5 where x=5

    4. 2x28x+1 where x=3

    5. x2x+22x1 where x=12

    6. 9x2+x23x4 where x=23

    7. (3y2)(y+5) where y=23

    8. (3x+2)(5x+1) where x=15

    9. (3x1)(x8) where x=1

    10. (7y+5)(y+1) where y=2

    11. y6y3+2 where y=1

    12. y5+y33 where y=2

    13. a25b2 where a=2 and b=1

    14. a32b3 where a=3 and b=2

    15. (x2y)(x+2y) where x=2 and y=5

    16. (4x3y)(xy) where x=4 and y=3

    17. a2ab+b2 where a=1 and b=2

    18. x2y2xy+2 where x=3 and y=2

    19. a4b4 where a=2 and b=3

    20. a62a3b3b6 where a=2 and b=1

      Evaluate b24ac given the following values.

    1. a=6, b=1 and c=1

    2. a=15, b=4 and c=4

    3. a=34, b=2 and c=4

    4. a=12, b=2 and c=30

    5. a=1, b=2 and c=1

    6. a=1, b=4 and c=50

    7. a=1, b=1 and c=116

    8. a=2, b=13 and c=1

    Part C: Using Formulas

      Convert the following temperatures to degrees Celsius given C=59(F32), where F represents degrees Fahrenheit.

    1. 95°F

    2. 86°F

    3. 32°F

    4. −40°F

       

    5. Calculate the perimeter and area of a rectangle with dimensions 12 feet by 5 feet.

    6. Calculate the perimeter and area of a rectangle with dimensions 5 meters by 1 meter.

    7. Calculate the surface area and volume of a sphere with radius 6 centimeters.

    8. The radius of the base of a right circular cylinder measures 4 inches and the height measures 10 inches. Calculate the surface area and volume.

    9. Calculate the volume of a sphere with a diameter of 18 centimeters.

    10. The diameter of the base of a right circular cone measures 6 inches. If the height is 112 feet, then calculate its volume.

    11. Given that the height of a right circular cylinder is equal to the radius of the base, derive a formula for the surface area in terms of the radius of the base.

    12. Given that the area of the base of a right circular cylinder is 25π square inches, find the volume if the height is 1 foot.

    13. Jose was able to drive from Tucson to Phoenix in 2 hours at an average speed of 58 mph. How far is Phoenix from Tucson?

    14. If a bullet train can average 152 mph, then how far can it travel in 34 of an hour?

    15. Margaret traveled for 134 hour at an average speed of 68 miles per hour. How far did she travel?

    16. The trip from Flagstaff, AZ to the Grand Canyon national park took 112 hours at an average speed of 54 mph. How far is the Grand Canyon national park from Flagstaff?

    17. Calculate the simple interest earned on a 3-year investment of $2,500 at an annual interest rate of 514%.

    18. Calculate the simple interest earned on a 1-year investment of $5,750 at an annual interest rate of 258%.

    19. What is the simple interest earned on a 5-year investment of $20,000 at an annual interest rate of 6%?

    20. What is the simple interest earned on a 1-year investment of $50,000 at an annual interest rate of 4.5%?

    21. The time t in seconds an object is in free fall is given by the formula t=s4, where s represents the distance in feet the object has fallen. How long does it take an object to fall 32 feet? (Give the exact answer and the approximate answer to the nearest hundredth.)

    22. The current I measured in amperes, is given by the formula I=PR, where P is the power usage measured in watts, and R is the resistance measured in ohms. If a light bulb uses 60 watts of power and has 240 ohms of resistance, then how many amperes of current are required?

    Part D: Discussion Board

    1. Find and post a useful mathematical model. Demonstrate its use with some values.

    2. Research and discuss the history of the variable. What can we use if we run out of letters?

    3. Find and post a link to a useful resource describing the Greek alphabet.

    4. Given the algebraic expression 53(9x1), explain why we do not subtract 5 and 3 first.

    5. Do we need a separate distributive property for more than two terms? For example, a(b+c+d)=ab+ac+ad. Explain.

    6. How can we check to see if we have simplified an expression correctly?

Answers

  1. Coefficients: {5,1,1}; variable parts: {x2,x}

  2. Coefficients: {5,3,1}; variable parts: {x2,xy,y2}

  3. Coefficients: {1,3,9}; variable parts: {x2y,xy2,xy}

  4. 15x25

  5. 4x2+10x2

  6. 6y2+8y2

  7. 4a210a+7

  8. 9a218b2

  9. 30a218ab+6b2

  10. 5x2+xyy2

  11. 16x

  12. 5y16

  13. 7x29x+3

  14. 1415a2+310
  15. 1110y2+y163
  16. 8a2b2ab+10

  17. 5x2y+17xy3xy2

  18. m2n8mnmn2

  19. 3x2n7xn+2

  20. 8x11

  21. 2x211x5

  22. ab20

  23. 2a2+14

  24. 3x223x+5

  25. 4a2b29ab+1

  26. 7y22y+2

  27. 6x2n8xn+5

  1. 7

  2. 35

  3. 118

  4. 0

  5. 36

  6. 4

  7. −1

  8. −96

  9. 3

  10. −65

  11. 5

  12. 4

  13. 22

  14. 52

  1. 35°C

  2. 0°C

  3. P = 34 feet; A = 60 square feet

  4. SA = 144π square centimeters; V = 288π cubic centimeters

  5. 972π cubic centimeters

  6. SA=4πr2

  7. 116 miles

  8. 119 miles

  9. $393.75

  10. $6,000

  11. 21.41 seconds

  1. Answer may vary

  2. Answer may vary

  3. Answer may vary

1.5 Rules of Exponents and Scientific Notation

Learning Objectives

  1. Review the rules of exponents.
  2. Review the definition of negative exponents and zero as an exponent.
  3. Work with numbers using scientific notation.

Review of the Rules of Exponents

In this section, we review the rules of exponents. Recall that if a factor is repeated multiple times, then the product can be written in exponential form xn. The positive integer exponent n indicates the number of times the base x is repeated as a factor.

Consider the product of x4 and x6,

Expanding the expression using the definition produces multiple factors of the base which is quite cumbersome, particularly when n is large. For this reason, we have useful rules to help us simplify expressions with exponents. In this example, notice that we could obtain the same result by adding the exponents.

x4x6=x4+6=x10Productruleforexponents

In general, this describes the product rule for exponentsxmxn=xm+n; the product of two expressions with the same base can be simplified by adding the exponents.. In other words, when multiplying two expressions with the same base we add the exponents. Compare this to raising a factor involving an exponent to a power, such as (x6)4.

Here we have 4 factors of x6, which is equivalent to multiplying the exponents.

(x6)4=x64=x24Powerruleforexponents

This describes the power rule for exponents(xm)n=xmn; a power raised to a power can be simplified by multiplying the exponents.. Now we consider raising grouped products to a power. For example,

(x2y3)4=x2y3x2y3x2y3x2y3=x2x2x2x2y3y3y3y3      Commutativeproperty=x2+2+2+2y3+3+3+3=x8y12

After expanding, we are left with four factors of the product x2y3. This is equivalent to raising each of the original grouped factors to the fourth power and applying the power rule.

(x2y3)4=(x2)4(y3)4=x8y12

In general, this describes the use of the power rule for a product as well as the power rule for exponents. In summary, the rules of exponents streamline the process of working with algebraic expressions and will be used extensively as we move through our study of algebra. Given any positive integers m and n where x,y0 we have

Product rule for exponents:

xmxn=xm+n

Quotient rule for exponents:

xmxn=xmn

Power rule for exponents:

(xm)n=xmn

Power rule for a product:

(xy)n=xnyn

Power rule for a quotient:

(xy)n=xnyn

(xy)n=xnyn; if a product is raised to a power, then apply that power to each factor in the product.

(xy)n=xnyn; if a quotient is raised to a power, then apply that power to the numerator and the denominator.

These rules allow us to efficiently perform operations with exponents.

Example 1

Simplify: 1041012103.

Solution:

1041012103=1016103          Productrule=10163        Quotientrule=1013

Answer: 1013

In the previous example, notice that we did not multiply the base 10 times itself. When applying the product rule, add the exponents and leave the base unchanged.

Example 2

Simplify: (x5x4x)2.

Solution:

Recall that the variable x is assumed to have an exponent of one, x=x1.

(x5x4x)2=(x5+4+1)2=(x10)2=x102=x20

Answer: x20

The base could in fact be any algebraic expression.

Example 3

Simplify: (x+y)9(x+y)13.

Solution:

Treat the expression (x+y) as the base.

(x+y)9(x+y)13=(x+y)9+13=(x+y)22

Answer: (x+y)22

The commutative property of multiplication allows us to use the product rule for exponents to simplify factors of an algebraic expression.

Example 4

Simplify: 8x5y3x7y3.

Solution:

Multiply the coefficients and add the exponents of variable factors with the same base.

8x5y3x7y3=83x5x7y1y3      Commutativeproperty=24x5+7y1+3              Powerruleforexponents=24x12y4

Answer: 24x12y4

Division involves the quotient rule for exponents.

Example 5

Simplify: 33x7y5(xy)1011x6y(xy)3.

Solution:

33x7y5(xy)1011x6y(xy)3=3311x76y51(xy)103=3x1y4(xy)7

Answer: 3xy4(xy)7

The power rule for a quotient allows us to apply that exponent to the numerator and denominator. This rule requires that the denominator is nonzero and so we will make this assumption for the remainder of the section.

Example 6

Simplify: (4a2bc4)3.

Solution:

First apply the power rule for a quotient and then the power rule for a product.

(4a2bc4)3=(4a2b)3(c4)3            Powerruleforaquotient=(4)3(a2)3(b)3(c4)3           Powerruleforaproduct=64a6b3c12

Answer: 64a6b3c12

Using the quotient rule for exponents, we can define what it means to have zero as an exponent. Consider the following calculation:

1=2525=5252=522=50

Twenty-five divided by twenty-five is clearly equal to one, and when the quotient rule for exponents is applied, we see that a zero exponent results. In general, given any nonzero real number x and integer n,

1=xnxn=xnn=x0

This leads us to the definition of zero as an exponentx0=1; any nonzero base raised to the 0 power is defined to be 1.,

x0=1x0

It is important to note that 00 is indeterminate. If the base is negative, then the result is still positive one. In other words, any nonzero base raised to the zero power is defined to be equal to one. In the following examples assume all variables are nonzero.

Example 7

Simplify:

  1. (2x)0
  2. 2x0

Solution:

  1. Any nonzero quantity raised to the zero power is equal to 1.

    (2x)0=1

  2. In the example, 2x0, the base is x, not −2x.

    2x0=2x0=21=2

Noting that 20=1 we can write,

123=2023=203=23

In general, given any nonzero real number x and integer n,

1xn=x0xn=x0n=xnx0

This leads us to the definition of negative exponentsxn=1xn, given any integer n, where x is nonzero.:

xn=1xnx0

An expression is completely simplified if it does not contain any negative exponents.

Example 8

Simplify: (4x2y)2.

Solution:

Rewrite the entire quantity in the denominator with an exponent of 2 and then simplify further.

(4x2y)2=1(4x2y)2=1(4)2(x2)2(y)2=116x4y2

Answer: 116x4y2

Sometimes negative exponents appear in the denominator.

Example 9

Simplify: x3y4.

Solution:

x3y4=1x31y4=1x3y41=y4x3

Answer: y4x3

The previous example suggests a property of quotients with negative exponentsxnym=ymxn, given any integers m and n, where x0 and y0.. Given any integers m and n where x0 and y0, then

xnym=1xn1ym=1xnym1=ymxn

This leads us to the property

xnym=ymxn

In other words, negative exponents in the numerator can be written as positive exponents in the denominator and negative exponents in the denominator can be written as positive exponents in the numerator.

Example 10

Simplify: 5x3y3z4.

Solution:

Take care with the coefficient −5, recognize that this is the base and that the exponent is actually positive one: 5=(5)1. Hence, the rules of negative exponents do not apply to this coefficient; leave it in the numerator.

5x3y3z4=5x3y3z4=5y3z4x3

Answer: 5y3z4x3

In summary, given integers m and n where x,y0 we have

Zero exponent:

x0=1

Negative exponent:

xn=1xn

Quotients with negative exponents:

xnym=ymxn

Furthermore, all of the rules of exponents defined so far extend to any integer exponents. We will expand the scope of these properties to include any real number exponents later in the course.

Try this! Simplify: (2x2y3z)4.

Answer: x8z416y12

Scientific Notation

Real numbers expressed using scientific notationReal numbers expressed the form a×10n, where n is an integer and 1a<10. have the form, a×10n where n is an integer and 1a<10. This form is particularly useful when the numbers are very large or very small. For example,

9,460,000,000,000,000m=9.46×1015m      Onelightyear0.000000000025m=2.5×1011mRadiusofahydrogenatom

It is cumbersome to write all the zeros in both of these cases. Scientific notation is an alternative, compact representation of these numbers. The factor 10n indicates the power of ten to multiply the coefficient by to convert back to decimal form:

This is equivalent to moving the decimal in the coefficient fifteen places to the right.

A negative exponent indicates that the number is very small:

This is equivalent to moving the decimal in the coefficient eleven places to the left.

 

Converting a decimal number to scientific notation involves moving the decimal as well. Consider all of the equivalent forms of 0.00563 with factors of 10 that follow:

0.00563=0.0563×101=0.563×102=5.63×103=56.3×104=563×105

While all of these are equal, 5.63×103 is the only form expressed in correct scientific notation. This is because the coefficient 5.63 is between 1 and 10 as required by the definition. Notice that we can convert 5.63×103 back to decimal form, as a check, by moving the decimal three places to the left.

xmxn=xmn; the quotient of two expressions with the same base can be simplified by subtracting the exponents.

Example 11

Write 1,075,000,000,000 using scientific notation.

Solution:

Here we count twelve decimal places to the left of the decimal point to obtain the number 1.075.

1,075,000,000,000=1.075×1012

Answer: 1.075×1012

Example 12

Write 0.000003045 using scientific notation.

Solution:

Here we count six decimal places to the right to obtain 3.045.

0.000003045=3.045×106

Answer: 3.045×106

Often we will need to perform operations when using numbers in scientific notation. All the rules of exponents developed so far also apply to numbers in scientific notation.

Example 13

Multiply: (4.36×105)(5.3×1012).

Solution:

Use the fact that multiplication is commutative, and apply the product rule for exponents.

(4.36×105)(5.30×1012)=(4.365.30)×(1051012)=23.108×105+12=2.3108×101×107=2.3108×101+7=2.3108×108

Answer: 2.3108×108

Example 14

Divide: (3.24×108)÷(9.0×103).

Solution:

(3.24×108)(9.0×103)=(3.249.0)×(108103)=0.36×108(3)=0.36×108+3=3.6×101×1011=3.6×101+11=3.6×1010

Answer: 3.6×1010

Example 15

The speed of light is approximately 6.7×108 miles per hour. Express this speed in miles per second.

Solution:

A unit analysis indicates that we must divide the number by 3,600.

6.7×108milesperhour=6.7×108miles1hour(1hour60minutes)(1minutes60seconds)=6.7×108miles3600seconds=(6.73600)×1080.0019×108         roundedtotwosignificantdigits=1.9×103×108=1.9×103+8=1.9×105

Answer: The speed of light is approximately 1.9×105 miles per second.

Example 16

The Sun moves around the center of the galaxy in a nearly circular orbit. The distance from the center of our galaxy to the Sun is approximately 26,000 light-years. What is the circumference of the orbit of the Sun around the galaxy in meters?

Solution:

One light-year measures 9.46×1015 meters. Therefore, multiply this by 26,000 or 2.60×104 to find the length of 26,000 light years in meters.

(9.46×1015)(2.60×104)=9.462.60×101510424.6×1019=2.46×1011019=2.46×1020

The radius r of this very large circle is approximately 2.46×1020 meters. Use the formula C=2πr to calculate the circumference of the orbit.

C=2πr2(3.14)(2.46×1020)=15.4×1020=1.54×1011020=1.54×1021

Answer: The circumference of the Sun’s orbit is approximately 1.54×1021 meters.

Try this! Divide: (3.15×105)÷(12×1013).

Answer: 2.625×107

Key Takeaways

  • When multiplying two quantities with the same base, add exponents: xmxn=xm+n.
  • When dividing two quantities with the same base, subtract exponents: xmxn=xmn.
  • When raising powers to powers, multiply exponents: (xm)n=xmn.
  • When a grouped quantity involving multiplication and division is raised to a power, apply that power to all of the factors in the numerator and the denominator: (xy)n=xnyn and (xy)n=xnyn.
  • Any nonzero quantity raised to the 0 power is defined to be equal to 1: x0=1.
  • Expressions with negative exponents in the numerator can be rewritten as expressions with positive exponents in the denominator: xn=1xn.
  • Expressions with negative exponents in the denominator can be rewritten as expressions with positive exponents in the numerator: 1xm=xm.
  • Take care to distinguish negative coefficients from negative exponents.
  • Scientific notation is particularly useful when working with numbers that are very large or very small.

Topic Exercises

    Part A: Rules of Exponents

      Simplify. (Assume all variables represent nonzero numbers.)

    1. 104107

    2. 7372

    3. 102104105
    4. 757972
    5. x3x2

    6. y5y3

    7. a8a6a5
    8. b4b10b8
    9. x2nx3nxn
    10. xnx8nx3n
    11. (x5)3

    12. (y4)3

    13. (x4y5)3

    14. (x7y)5

    15. (x2y3z4)4

    16. (xy2z3)2

    17. (5x2yz3)2

    18. (2xy3z4)5

    19. (x2yz5)n

    20. (xy2z3)2n

    21. (xx3x2)3

    22. (y2y5y)2

    23. a2(a4)2a3
    24. aa3a2(a2)3
    25. (2x+3)4(2x+3)9

    26. (3y1)7(3y1)2

    27. (a+b)3(a+b)5

    28. (x2y)7(x2y)3

    29. 5x2y3xy2

    30. 10x3y22xy

    31. 6x2yz33xyz4

    32. 2xyz2(4x2y2z)

    33. 3xny2n5x2y

    34. 8x5nyn2x2ny

    35. 40x5y3z4x2y2z
    36. 8x2y5z316x2yz
    37. 24a8b3(a5b)108a5b3(a5b)2
    38. 175m9n5(m+n)725m8n(m+n)3
    39. (2x4y2z)6
    40. (3xy4z7)5
    41. (3ab22c3)3
    42. (10a3b3c2)2
    43. (2xy4z3)4
    44. (7x9yz4)3
    45. (xy2z3)n
    46. (2x2y3z)n
    47. (5x)0

    48. (3x2y)0

    49. 5x0

    50. 3x2y0

    51. (2a2b0c3)5

    52. (3a4b2c0)4

    53. (9x3y2z0)23xy2
    54. (5x0y5z)325y2z0
    55. 2x3

    56. (2x)2

    57. a4a5a2

    58. b8b3b4

    59. a8a3a6
    60. b10b4b2
    61. 10x3y2

    62. 3x5y2

    63. 3x2y2z1

    64. 5x4y2z2

    65. 25x3y25x1y3
    66. 9x1y3z53x2y2z1
    67. (5x3y2z)3
    68. (7x2y5z2)2
    69. (2x3zy2)5
    70. (5x5z22y3)3
    71. (12x3y2z2x7yz8)3
    72. (150xy8z290x7y2z)2
    73. (9a3b4c23a3b5c7)4
    74. (15a7b5c83a6b2c3)3

      The value in dollars of a new mobile phone can be estimated by using the formula V=210(2t+1)1, where t is the number of years after purchase.

    1. How much was the phone worth new?

    2. How much will the phone be worth in 1 year?

    3. How much will the phone be worth in 3 years?

    4. How much will the phone be worth in 10 years?

    5. How much will the phone be worth in 100 years?

    6. According to the formula, will the phone ever be worthless? Explain.

    7. The height of a particular right circular cone is equal to the square of the radius of the base, h=r2. Find a formula for the volume in terms of r.

    8. A sphere has a radius r=3x2. Find the volume in terms of x.

    Part B: Scientific Notation

      Convert to a decimal number.

    1. 5.2×108

    2. 6.02×109

    3. 1.02×106

    4. 7.44×105

      Rewrite using scientific notation.

    1. 7,050,000

    2. 430,000,000,000

    3. 0.00005001

    4. 0.000000231

      Perform the operations.

    1. (1.2×109)(3×105)

    2. (4.8×105)(1.6×1020)

    3. (9.1×1023)(3×1010)

    4. (5.5×1012)(7×1025)

    5. 9.6×10161.2×104
    6. 4.8×10142.4×106
    7. 4×1088×1010
    8. 2.3×10239.2×103
    9. 987,000,000,000,000×23,000,000

    10. 0.00000000024×0.00000004

    11. 0.000000000522÷0.0000009

    12. 81,000,000,000÷0.0000648

    13. The population density of Earth refers to the number of people per square mile of land area. If the total land area on Earth is 5.751×107 square miles and the population in 2007 was estimated to be 6.67×109 people, then calculate the population density of Earth at that time.

    14. In 2008 the population of New York City was estimated to be 8.364 million people. The total land area is 305 square miles. Calculate the population density of New York City.

    15. The mass of Earth is 5.97×1024 kilograms and the mass of the Moon is 7.35×1022 kilograms. By what factor is the mass of Earth greater than the mass of the Moon?

    16. The mass of the Sun is 1.99×1030 kilograms and the mass of Earth is 5.97×1024 kilograms. By what factor is the mass of the Sun greater than the mass of Earth? Express your answer in scientific notation.

    17. The radius of the Sun is 4.322×105 miles and the average distance from Earth to the Moon is 2.392×105 miles. By what factor is the radius of the Sun larger than the average distance from Earth to the Moon?

    18. One light year, 9.461×1015 meters, is the distance that light travels in a vacuum in one year. If the distance from our Sun to the nearest star, Proxima Centauri, is estimated to be 3.991×1016 meters, then calculate the number of years it would take light to travel that distance.

    19. It is estimated that there are about 1 million ants per person on the planet. If the world population was estimated to be 6.67 billion people in 2007, then estimate the world ant population at that time.

    20. The radius of the earth is 6.3×106 meters and the radius of the sun is 7.0×108 meters. By what factor is the radius of the Sun larger than the radius of the Earth?

    21. A gigabyte is 1×109 bytes and a megabyte is 1×106 bytes. If the average song in the MP3 format consumes about 4.5 megabytes of storage, then how many songs will fit on a 4-gigabyte memory card?

    22. Water weighs approximately 18 grams per mole. If one mole is about 6×1023 molecules, then approximate the weight of each molecule of water.

    Part C: Discussion Board

    1. Use numbers to show that (x+y)nxn+yn.

    2. Why is 00 indeterminate?

    3. Explain to a beginning algebra student why 222345.

    4. René Descartes (1637) established the usage of exponential form: a2, a3, and so on. Before this, how were exponents denoted?

Answers

  1. 1011

  2. 10

  3. x5

  4. a9

  5. x4n

  6. x15

  7. x12y15

  8. x8y12z16

  9. 25x4y2z6

  10. x2nynz5n

  11. x18

  12. a7

  13. (2x+3)13

  14. (a+b)8

  15. 15x3y3

  16. 18x3y2z7

  17. 15xn+2y2n+1

  18. 10x3y

  19. 3a3(a5b)8

  20. 64x24y12z6

  21. 27a3b68c9
  22. 16x4y16z12
  23. xny2nz3n
  24. 1

  25. −5

  26. 32a10c15

  27. 27x5y2

  28. 2x3
  29. a

  30. a11

  31. 10y2x3
  32. 3y2x2z
  33. 5y5x2
  34. x9125y6z3
  35. x15y1032z5
  36. 216y3x12z21
  37. a24b481c20
  38. $210

  39. $30

  40. $1.04

  41. V=13πr4

  1. 520,000,000

  2. 0.00000102

  3. 7.05×106

  4. 5.001×105

  5. 3.6×1014

  6. 2.73×1034

  7. 8×1020

  8. 5×1019

  9. 2.2701×1022

  10. 5.8×104

  11. About 116 people per square mile

  12. 81.2

  13. 1.807

  14. 6.67×1015 ants

  15. Approximately 889 songs

  1. Answer may vary

  2. Answer may vary

1.6 Polynomials and Their Operations

Learning Objectives

  1. Identify a polynomial and determine its degree.
  2. Add and subtract polynomials.
  3. Multiply and divide polynomials.

Definitions

A polynomialAn algebraic expression consisting of terms with real number coefficients and variables with whole number exponents. is a special algebraic expression with terms that consist of real number coefficients and variable factors with whole number exponents. Some examples of polynomials follow:

3x2

7xy+5

32x3+3x212x+1

6x2y4xy3+7

The degree of a termThe exponent of the variable. If there is more than one variable in the term, the degree of the term is the sum their exponents. in a polynomial is defined to be the exponent of the variable, or if there is more than one variable in the term, the degree is the sum of their exponents. Recall that x0=1; any constant term can be written as a product of x0 and itself. Hence the degree of a constant term is 0.

Term

Degree

3x2

2

6x2y

2+1=3

7a2b3

2+3=5

8

0, since 8=8x0

2x

1, since 2x=2x1

The degree of a polynomialThe largest degree of all of its terms. is the largest degree of all of its terms.

Polynomial

Degree

4x53x3+2x1

5

6x2y5xy3+7

4, because 5xy3 has degree 4.

12x+54

1, because 12x=12x1

Of particular interest are polynomials with one variableA polynomial where each term has the form anxn, where an is any real number and n is any whole number., where each term is of the form anxn. Here an is any real number and n is any whole number. Such polynomials have the standard form:

anxn+an1xn1++a1x+a0

Typically, we arrange terms of polynomials in descending order based on the degree of each term. The leading coefficientThe coefficient of the term with the largest degree. is the coefficient of the variable with the highest power, in this case, an.

Example 1

Write in standard form: 3x4x2+5x3+72x4.

Solution:

Since terms are defined to be separated by addition, we write the following:

3x4x2+5x3+72x4=3x+(4)x2+5x3+7+(2)x4

In this form, we can see that the subtraction in the original corresponds to negative coefficients. Because addition is commutative, we can write the terms in descending order based on the degree as follows:

=(2)x4+5x3+(4)x2+3x+7=2x4+5x34x2+3x+7

Answer: 2x4+5x34x2+3x+7

We classify polynomials by the number of terms and the degree:

Expression

Classification

Degree

5x7

Monomial (one term)

7

8x61

Binomial (two terms)

6

3x2+x1

Trinomial (three terms)

2

5x32x2+3x6

Polynomial (many terms)

3

Polynomial with one term.

Polynomial with two terms.

Polynomial with three terms.

We can further classify polynomials with one variable by their degree:

Polynomial

Name

5

Constant (degree 0)

2x+1

Linear (degree 1)

3x2+5x3

Quadratic (degree 2)

x3+x2+x+1

Cubic (degree 3)

7x4+3x37x+8

Fourth-degree polynomial

A polynomial with degree 0.

A polynomial with degree 1.

A polynomial with degree 2.

A polynomial with degree 3.

In this text, we call any polynomial of degree n4 an nth-degree polynomial. In other words, if the degree is 4, we call the polynomial a fourth-degree polynomial. If the degree is 5, we call it a fifth-degree polynomial, and so on.

Example 2

State whether the following polynomial is linear or quadratic and give the leading coefficient: 25+4xx2.

Solution:

The highest power is 2; therefore, it is a quadratic polynomial. Rewriting in standard form we have

x2+4x+25

Here x2=1x2 and thus the leading coefficient is −1.

Answer: Quadratic; leading coefficient: −1

Adding and Subtracting Polynomials

We begin by simplifying algebraic expressions that look like +(a+b) or (a+b). Here, the coefficients are actually implied to be +1 and −1 respectively and therefore the distributive property applies. Multiply each term within the parentheses by these factors as follows:

+(a+b)=+1(a+b)=(+1)a+(+1)b=a+b(a+b)=1(a+b)=(1)a+(1)b=ab

Use this idea as a means to eliminate parentheses when adding and subtracting polynomials.

Example 3

Add: 9x2+(x25).

Solution:

The property +(a+b)=a+b allows us to eliminate the parentheses, after which we can then combine like terms.

9x2+(x25)=9x2+x25=10x25

Answer: 10x25

Example 4

Add: (3x2y24xy+9)+(2x2y26xy7).

Solution:

Remember that the variable parts have to be exactly the same before we can add the coefficients.

(3x2y24xy+9)+(2x2y26xy7)=3x2y24xy+9+2x2y26xy7=5x2y210xy+2

Answer: 5x2y210xy+2

When subtracting polynomials, the parentheses become very important.

Example 5

Subtract: 4x2(3x2+5x).

Solution:

The property (a+b)=ab allows us to remove the parentheses after subtracting each term.

4x2(3x2+5x)=4x23x25x=x25x

Answer: x25x

Subtracting a quantity is equivalent to multiplying it by −1.

Example 6

Subtract: (3x22xy+y2)(2x2xy+3y2).

Solution:

Distribute the −1, remove the parentheses, and then combine like terms. Multiplying the terms of a polynomial by −1 changes all the signs.

=3x22xy+y22x2+xy3y2=x2xy2y2

Answer: x2xy2y2

Try this! Subtract: (7a22ab+b2)(a22ab+5b2).

Answer: 6a24b2

Multiplying Polynomials

Use the product rule for exponents, xmxn=xm+n, to multiply a monomial times a polynomial. In other words, when multiplying two expressions with the same base, add the exponents. To find the product of monomials, multiply the coefficients and add the exponents of variable factors with the same base. For example,

7x48x3=78x4x3Commutativeproperty=56x4+3    Productruleforexponents=56x7

To multiply a polynomial by a monomial, apply the distributive property, and then simplify each term.

Example 7

Multiply: 5xy2(2x2y2xy+1).

Solution:

Apply the distributive property and then simplify.

=5xy22x2y25xy2xy+5xy21=10x3y45x2y3+5xy2

Answer: 10x3y45x2y3+5xy2

To summarize, multiplying a polynomial by a monomial involves the distributive property and the product rule for exponents. Multiply all of the terms of the polynomial by the monomial. For each term, multiply the coefficients and add exponents of variables where the bases are the same.

In the same manner that we used the distributive property to distribute a monomial, we use it to distribute a binomial. (a+b)(c+d)=(a+b)c+(a+b)d=ac+bc+ad+bd=ac+ad+bc+bd Here we apply the distributive property multiple times to produce the final result. This same result is obtained in one step if we apply the distributive property to a and b separately as follows:

This is often called the FOIL method. Multiply the first, outer, inner, and then last terms.

Example 8

Multiply: (6x1)(3x5).

Solution:

Distribute 6x and −1 and then combine like terms.

(6x1)(3x5)=6x3x6x5+(1)3x(1)5=18x230x3x+5=18x233x+5

Answer: 18x233x+5

Consider the following two calculations:

(a+b)2=(a+b)(a+b)=a2+ab+ba+b2=a2+ab+ab+b2=a2+2ab+b2

(ab)2=(ab)(ab)=a2abba+b2=a2abab+b2=a22ab+b2

This leads us to two formulas that describe perfect square trinomialsThe trinomials obtained by squaring the binomials (a+b)2=a2+2ab+b2 and (ab)2=a22ab+b2.:

(a+b)2=a2+2ab+b2(ab)2=a22ab+b2

We can use these formulas to quickly square a binomial.

Example 9

Multiply: (3x+5)2.

Solution:

Here a=3x and b=5. Apply the formula:

Answer: 9x2+30x+25

This process should become routine enough to be performed mentally. Our third special product follows: (a+b)(ab)=a2ab+bab2=a2ab+abb2=a2b2 This product is called difference of squaresThe special product obtained by multiplying conjugate binomials (a+b)(ab)=a2b2.:

(a+b)(ab)=a2b2

The binomials (a+b) and (ab) are called conjugate binomialsThe binomials (a+b) and (ab).. When multiplying conjugate binomials the middle terms are opposites and their sum is zero; the product is itself a binomial.

Example 10

Multiply: (3xy+1)(3xy1).

Solution:

(3xy+1)(3xy1)=(3xy)23xy+3xy12=9x2y21

Answer: 9x2y21

Try this! Multiply: (x2+5y2)(x25y2).

Answer: (x425y4)

Example 11

Multiply: (5x2)3.

Solution:

Here we perform one product at a time.

Answer: 125x2150x2+60x8

Dividing Polynomials

Use the quotient rule for exponents, xmxn=xmn, to divide a polynomial by a monomial. In other words, when dividing two expressions with the same base, subtract the exponents. In this section, we will assume that all variables in the denominator are nonzero.

Example 12

Divide: 24x7y58x3y2.

Solution:

Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

24x7y58x3y2=248x73y52=3x4y3

Answer: 3x4y3

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property: a+bc=ac+bc Applying this property will result in terms that can be treated as quotients of monomials.

Example 13

Divide: 5x4+25x315x25x2.

Solution:

Break up the fraction by dividing each term in the numerator by the monomial in the denominator, and then simplify each term.

5x4+25x315x25x2=5x45x2+25x35x215x25x2=55x42+255x32155x22=1x2+5x13x0=x2+5x31

Answer: x2+5x3

We can check our division by multiplying our answer, the quotient, by the monomial in the denominator, the divisor, to see if we obtain the original numerator, the dividend.

DividendDivisor=Quotient

5x4+25x315x25x2=x2+5x3

or

or

Dividend=DivisorQuotient

5x4+25x315x2=5x2(x2+5x3)

The same technique outlined for dividing by a monomial does not work for polynomials with two or more terms in the denominator. In this section, we will outline a process called polynomial long divisionThe process of dividing two polynomials using the division algorithm., which is based on the division algorithm for real numbers. For the sake of clarity, we will assume that all expressions in the denominator are nonzero.

Example 14

Divide: x3+3x28x4x2.

Solution:

Here x2 is the divisor and x3+3x28x4 is the dividend. To determine the first term of the quotient, divide the leading term of the dividend by the leading term of the divisor.

Multiply the first term of the quotient by the divisor, remembering to distribute, and line up like terms with the dividend.

Subtract the resulting quantity from the dividend. Take care to subtract both terms.

Bring down the remaining terms and repeat the process.

Notice that the leading term is eliminated and that the result has a degree that is one less. The complete process is illustrated below:

Polynomial long division ends when the degree of the remainder is less than the degree of the divisor. Here, the remainder is 0. Therefore, the binomial divides the polynomial evenly and the answer is the quotient shown above the division bar.

x3+3x28x4x2=x2+5x+2

To check the answer, multiply the divisor by the quotient to see if you obtain the dividend as illustrated below:

x3+3x28x4=(x2)(x2+5x+2)

This is left to the reader as an exercise.

Answer: x2+5x+2

Next, we demonstrate the case where there is a nonzero remainder.

Just as with real numbers, the final answer adds to the quotient the fraction where the remainder is the numerator and the divisor is the denominator. In general, when dividing we have: DividendDivisor=Quotient+RemainderDivisor If we multiply both sides by the divisor we obtain, Dividend=Quotient×Divisor+Remainder

Example 15

Divide: 6x25x+32x1.

Solution:

Since the denominator is a binomial, begin by setting up polynomial long division.

To start, determine what monomial times 2x1 results in a leading term 6x2. This is the quotient of the given leading terms: (6x2)÷(2x)=3x. Multiply 3x times the divisor 2x1, and line up the result with like terms of the dividend.

Subtract the result from the dividend and bring down the constant term +3.

Subtracting eliminates the leading term. Multiply 2x1 by −1 and line up the result.

Subtract again and notice that we are left with a remainder.

The constant term 2 has degree 0 and thus the division ends. Therefore,

6x25x+32x1=3x1+22x1

To check that this result is correct, we multiply as follows:

quotient×divisor+remainder=(3x1)(2x1)+2=6x23x2x+1+2=6x25x+2=dividend

Answer: 3x1+22x1

Occasionally, some of the powers of the variables appear to be missing within a polynomial. This can lead to errors when lining up like terms. Therefore, when first learning how to divide polynomials using long division, fill in the missing terms with zero coefficients, called placeholdersTerms with zero coefficients used to fill in all missing exponents within a polynomial..

Example 16

Divide: 27x3+643x+4.

Solution:

Notice that the binomial in the numerator does not have terms with degree 2 or 1. The division is simplified if we rewrite the expression with placeholders:

27x3+64=27x3+0x2+0x+64

Set up polynomial long division:

We begin with 27x3÷3x=9x2 and work the rest of the division algorithm.

Answer: 9x212x+16

Example 17

Divide: 3x42x3+6x2+23x7x22x+5.

Solution:

Begin the process by dividing the leading terms to determine the leading term of the quotient 3x4÷x2=3x2. Take care to distribute and line up the like terms. Continue the process until the remainder has a degree less than 2.

The remainder is x2. Write the answer with the remainder:

3x42x3+6x2+23x7x22x+5=3x2+4x1+x2x22x+5

Answer: 3x2+4x1+x2x22x+5

Polynomial long division takes time and practice to master. Work lots of problems and remember that you may check your answers by multiplying the quotient by the divisor (and adding the remainder if present) to obtain the dividend.

Try this! Divide: 6x413x3+9x214x+63x2.

Answer: 2x33x2+x423x2

Key Takeaways

  • Polynomials are special algebraic expressions where the terms are the products of real numbers and variables with whole number exponents.
  • The degree of a polynomial with one variable is the largest exponent of the variable found in any term. In addition, the terms of a polynomial are typically arranged in descending order based on the degree of each term.
  • When adding polynomials, remove the associated parentheses and then combine like terms. When subtracting polynomials, distribute the −1, remove the parentheses, and then combine like terms.
  • To multiply polynomials apply the distributive property; multiply each term in the first polynomial with each term in the second polynomial. Then combine like terms.
  • When dividing by a monomial, divide all terms in the numerator by the monomial and then simplify each term.
  • When dividing a polynomial by another polynomial, apply the division algorithm.

Topic Exercises

    Part A: Definitions

      Write the given polynomials in standard form.

    1. 1xx2

    2. y5+y2

    3. y3y2+5y3

    4. 812a2+a3a

    5. 2x2+6x5x3+x4

    6. a35+a2+2a4a5+6a

      Classify the given polynomial as a monomial, binomial, or trinomial and state the degree.

    1. x2x+2

    2. 510x3

    3. x2y2+5xy6

    4. 2x3y2

    5. x41

    6. 5

      State whether the polynomial is linear or quadratic and give the leading coefficient.

    1. 19x2

    2. 10x2

    3. 2x3

    4. 100x

    5. 5x2+3x1

    6. x1

    7. x62x2

    8. 15x

    Part B: Adding and Subtracting Polynomials

      Simplify.

    1. (5x23x2)+(2x26x+7)

    2. (x2+7x12)+(2x2x+3)

    3. (x2+5x+10)+(x210)

    4. (x21)+(4x+2)

    5. (10x2+3x2)(x26x+1)

    6. (x23x8)(2x23x8)

    7. (23x2+34x1)(16x2+52x12)

    8. (45x258x+106)(310x223x+35)

    9. (x2y2+7xy5)(2x2y2+5xy4)

    10. (x2y2)(x2+6xy+y2)

    11. (a2b2+5ab2)+(7ab2)(4a2b2)

    12. (a2+9ab6b2)(a2b2)+7ab

    13. (10x2y8xy+5xy2)(x2y4xy)+(xy2+4xy)

    14. (2m2n6mn+9mn2)(m2n+10mn)m2n

    15. (8x2y25xy+2)(x2y2+5)+(2xy3)

    16. (x2y2)(5x22xyy2)(x27xy)

    17. (16a22ab+34b2)(53a2+45b2)+118ab

    18. (52x22y2)(75x212xy+73y2)12xy

    19. (x2n+5xn2)+(2x2n3xn1)

    20. (7x2nxn+5)(6x2nxn8)

    21. Subtract 4y3 from y2+7y10.

    22. Subtract x2+3x2 from 2x2+4x1.

    23. A right circular cylinder has a height that is equal to the radius of the base, h=r. Find a formula for the surface area in terms of h.

    24. A rectangular solid has a width that is twice the height and a length that is 3 times that of the height. Find a formula for the surface area in terms of the height.

    Part C: Multiplying Polynomials

      Multiply.

    1. 8x22x

    2. 10x2y5x3y2

    3. 2x(5x1)

    4. 4x(3x5)

    5. 7x2(2x6)

    6. 3x2(x2x+3)

    7. 5y4(y22y+3)

    8. 52a3(24a26a+4)

    9. 2xy(x27xy+y2)

    10. 2a2b(a23ab+5b2)
    11. xn(x2+x+1)

    12. xn(x2nxn1)

    13. (x+4)(x5)

    14. (x7)(x6)

    15. (2x3)(3x1)

    16. (9x+1)(3x+2)

    17. (3x2y2)(x25y2)

    18. (5y2x2)(2y23x2)

    19. (3x+5)(3x5)

    20. (x+6)(x6)

    21. (a2b2)(a2+b2)

    22. (ab+7)(ab7)

    23. (4x5y2)(3x2y)

    24. (xy+5)(xy)

    25. (x5)(x23x+8)

    26. (2x7)(3x2x+1)

    27. (x2+7x1)(2x23x1)

    28. (4x2x+6)(5x24x3)

    29. (x+8)2

    30. (x3)2

    31. (2x5)2

    32. (3x+1)2

    33. (a3b)2

    34. (7ab)2

    35. (x2+2y2)2
    36. (x26y)2
    37. (a2a+5)2
    38. (x23x1)2
    39. (x3)3

    40. (x+2)3

    41. (3x+1)3

    42. (2x3)3

    43. (x+2)4

    44. (x3)4

    45. (2x1)4

    46. (3x1)4

    47. (x2n+5)(x2n5)
    48. (xn1)(x2n+4xn3)
    49. (x2n1)2
    50. (x3n+1)2
    51. Find the product of 3x2 and x25x2.

    52. Find the product of x2+4 and x31.

    53. Each side of a square measures 3x3 units. Determine the area in terms of x.

    54. Each edge of a cube measures 2x2 units. Determine the volume in terms of x.

    Part D: Dividing Polynomials

      Divide.

    1. 125x5y225x4y2
    2. 256x2y3z564x2yz2
    3. 20x312x2+4x4x
    4. 15x475x3+18x23x2
    5. 12a2b+28ab24ab4ab
    6. 2a4b3+16a2b2+8ab32ab2
    7. x3+x23x+9x+3
    8. x34x29x+20x5
    9. 6x311x2+7x62x3
    10. 9x39x2x+13x1
    11. 16x3+8x239x+174x3
    12. 12x356x2+55x+302x5
    13. 6x4+13x39x2x+63x+2
    14. 25x410x3+11x27x+15x1
    15. 20x4+12x3+9x2+10x+52x+1
    16. 25x445x326x2+36x115x2
    17. 3x4+x21x2
    18. x4+x3x+3
    19. x310x2
    20. x3+15x+3
    21. y5+1y+1
    22. y6+1y+1
    23. x44x3+6x27x1x2x+2
    24. 6x4+x32x2+2x+43x2x+1
    25. 2x37x2+8x3x22x+1
    26. 2x4+3x36x24x+3x2+x3
    27. x4+4x32x24x+1x21
    28. x4+x1x2+1