# Polynomials. Solving Equations by Using the Zero Product Rule

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1 mil23264_ch05_ :21:05 06:16 PM Page 303 Polynomials 5.1 Addition and Subtraction of Polynomials and Polynomial Functions 5.2 Multiplication of Polynomials 5.3 Division of Polynomials 5.4 Greatest Common Factor and Factoring by Grouping 5.5 Factoring Trinomials 5.6 Factoring Binomials 5.7 Solving Equations by Using the Zero Product Rule 5 In this chapter, we study addition, subtraction, multiplication, and division of polynomials, along with an important operation called factoring. A polynomial function is a function defined by a sum of terms of the form axn, where a is a real number and n is a whole number. In Exercise 68 in Section 5.1, the function P(t) 0.022t t 102 is used to estimate the population P (in millions) of Mexico, t years after

2 304 Chapter 5 Polynomials chapter 5 preview The exercises in this chapter preview contain concepts that have not yet been presented. These exercises are provided for students who want to compare their levels of understanding before and after studying the chapter. Alternatively, you may prefer to work these exercises when the chapter is completed and before taking the exam. Section Given the polynomial: 7x 5 2x 3 4x 6 2 a. Write the polynomial in descending order x 2 5x 6 2x Use synthetic division to divide the polynomials. 5y 2 6y 3 y 2 Section Factor out the greatest common factor. 14x 3 y 4 20x 2 y 3 18xy b. Determine the degree of the polynomial. 14. Factor by grouping. ab ac 2b 2c c. Determine the leading coefficient. For Exercises 2 3, perform the indicated operations x 2 y 3xy 2xy x 2 y 3xy 6xy 2 2 Section 5.5 For Exercises 15 18, factor the polynomials completely. 15. a 2 a y 2 16y t 2 2t 32 12t 2 4t x 3 14x 2 10x 18. 9c 2 12cd 4d 2 4. Given the polynomial function P1x2 3x 2 2x 1, find P1 22. Section 5.2 For Exercises 5 9, multiply the polynomials. 5. 3x 2 y 3 14x 2 y 2 2xy 3 10x 3 y x 5212x z 2214z 2 2z 52 Section 5.6 For Exercises 19 20, factor completely y z 3 8 Section 5.7 For Exercises 21 23, solve the equations. 21. x 2 7x t d c214d c a 6212a 521a Section 5.3 For Exercises 10 11, divide the polynomials p 4 24p 3 8p p c1c c 42 c The length of a rectangle is 1 ft more than twice the width. The area is 36 ft 2. Find the length and width.

5 Section 5.1 Addition and Subtraction of Polynomials and Polynomial Functions 307 example 2 Finding the Opposite of a Polynomial Find the opposite of the polynomials. a. 4x b. 5a 2b c c. 5.5y 4 2.4y 3 1.1y 3 a. The opposite of 4x is (4x), or 4x. Find the opposite of the polynomials. 8. 7z 9. 2p 3q r 1 b. The opposite of 5a 2b c is 15a 2b c2 or equivalently 5a 2b c. c. The opposite of 5.5y 4 2.4y 3 1.1y 3 is 15.5y 4 2.4y 3 1.1y 32 or equivalently 5.5y 4 2.4y 3 1.1y 3. Tip: Notice that the sign of each term is changed when finding the opposite of a polynomial. Definition of Subtraction of Polynomials If A and B are polynomials, then A B A 1 B2. example 3 Subtract the polynomials. a. Subtracting Polynomials a. 13x 2 2x 52 14x 2 7x 22 b. 16x 2 y 2xy 52 1x 2 y 32 13x 2 2x 52 14x 2 7x 22 13x 2 2x x 2 7x 22 3x 2 1 4x 2 2 2x 7x x 2 9x 7 Add the opposite of the second polynomial. Group like terms. Combine like terms. Subtract the polynomials a 2 b 2ab2 1 3a 2 b 2ab a 1 3 p p 2 pb a 1 2 p p pb b. 16x 2 y 2xy 52 1x 2 y 32 16x 2 y 2xy 52 1 x 2 y 32 6x 2 y 1 x 2 y2 1 2xy x 2 y 2xy 8 Add the opposite of the second polynomial. Group like terms. Combine like terms. Tip: Subtraction of polynomials can be performed vertically by vertically aligning like terms. Then add the opposite of the second polynomial. Placeholders (shown in bold) may be used to help line up like terms. 16x 2 y 2xy 52 1x 2 y 32 6x 2 y 2xy 5 Add the 6x 2 y 2xy 5 1x 2 y 0xy 32 opposite. x 2 y 0xy 3 5x 2 y 2xy 8 Answers 8. 7z 9. 2p 3q r a 2 b 4ab p p2 3 2 p

6 308 Chapter 5 Polynomials 12. Subtract 18t 2 4t 32 from 1 6t 2 t 22. example 4 Subtract Subtracting Polynomials 1 2 x4 3 4 x2 1 5 from 3 2 x4 1 2 x2 4x In general, to subtract a from b, we write b a.therefore, to subtract we have 1 2 x4 3 4 x2 1 5 from a 3 2 x4 1 2 x2 4xb a 1 2 x4 3 4 x2 1 5 b 3 2 x4 1 2 x2 4x 3 2 x4 1 2 x2 4x 1 2 x4 3 4 x2 1 5 Subtract the polynomials. 3 4 x 1 Group like terms. 2 x4 1 2 x4 1 2 x2 3 4 x x4 1 2 x4 2 4 x2 3 4 x2 4x x4 5 4 x2 4x 1 5 Write like terms with a common denominator. Combine like terms. x x2 4x 1 5 Simplify. 4. Polynomial Functions A polynomial function is a function defined by a finite sum of terms of the form ax n,where a is a real number and n is a whole number. For example, the functions defined here are polynomial functions: f1x2 3x 8 g1x2 4x 5 2x 3 5x 3 h1x2 1 2 x4 3 5 x3 4x x 1 k1x x 0 which is of the form ax n, where n 0 is a whole number2 The following functions are not polynomial functions: m1x2 1 x 8 q1x2 0x 0 a 1 x x 1, the exponent 1 is not a whole numberb 10x 0 is not of the form ax n 2 Answer t 2 5t 5

7 Section 5.1 Addition and Subtraction of Polynomials and Polynomial Functions 309 example 5 Evaluating a Polynomial Function Given P1x2 x 3 2x 2 x 2, find the function values. a. P1 32 b. P1 12 c. P102 d. P122 a. P1x2 x 3 2x 2 x 2 P Given: P1x2 2x 3 4x Find P(0). 14. Find P( 2). b. P c. P d. P The function values can be confirmed from the graph of y P(x) (Figure 5 1). p(x) (2, 12) 8 ( 1, 0) 4 x ( 3, 8) (0, 2) P(x) x 3 2x 2 x 2 Figure 5-1 example 6 Using Polynomial Functions in Applications The length of a rectangle is 4 m less than 3 times the width. Let x represent the width. Write a polynomial function P that represents the perimeter of the rectangle and simplify the result. Let x represent the width. Then 3x 4 is the length. The perimeter of a rectangle is given by P 2L 2W. Thus P1x2 213x 42 21x2 6x 8 2x 8x 8 3x 4 x 15. The longest side of a triangle is 2 ft less than 4 times the shortest side. The middle side is 3 ft more than twice the shortest side. Let x represent the shortest side. Find a polynomial function P that represents the perimeter of the triangle, and simplify the result. Answers 13. P P P1x2 x 14x 22 12x 32 7x 1

8 310 Chapter 5 Polynomials The yearly cost of tuition at public two-year colleges from 1992 to 2006 can be approximated by T1x2 0.08x 2 61x 1135 for 0 x 14, where x represents the number of years since Find T(13) and interpret the result. 17. Use the function T to approximate the cost of tuition in the year Answers 16. T In the year 2005, tuition for public two-year colleges averaged approximately \$ \$1438 example 7 Applying a Polynomial Function The percent of females between the ages of 18 and 24 who engaged in cigarette smoking in the United States can be approximated by F1x x x 29.1, where x is the number of years since 1997 and F1x2 is measured as a percent (Figure 5-2). a. Evaluate F122 to 1 decimal place, and interpret the meaning in the context of this problem. b. What percent of females between the ages of 18 and 24 smoked in the year 2005? Round to the nearest tenth of a percent. a. F Substitute x 2 into the function In the year x 2 years since 1997), approximately 28.9% of females between the ages of 18 and 24 smoked. b. The year 2005 is 8 years since Substitute x 8 into the function. 22.4% Percent F Percent of Females Who Smoked, United States, F(x) 0.125x x Year (x 0 corresponds to 1997) Source: Center for Disease Control. Figure 5-2 Substitute x 8 into the function. Approximately 22.4% of females in the age group smoked in section 5.1 Practice Exercises Boost your GRADE at mathzone.com! Practice Problems e-professors Study Skills Exercises Self-Tests NetTutor Videos 1. After you do a section of homework, check the answers to the odd-numbered exercises in the back of the text. Choose a method to identify the exercises that you got wrong or had trouble with (i.e., circle the number or put a star by the number). List some reasons why it is important to label these problems. 2. Define the key terms. a. Polynomial b. Coefficient c. Degree of the term d. Monomial e. Binomial f. Trinomial g. Leading term h. Leading coefficient i. Degree of a polynomial j. Like terms k. Polynomial function

9 Section 5.1 Addition and Subtraction of Polynomials and Polynomial Function 311 Objective 1: Polynomials: Basic Definitions For Exercises 3 8, write the polynomial in descending order. Then identify the leading coefficient and the degree. 3. a 2 6a 3 a 4. 2b b 4 5b x 2 x 3x y y 5 y t s 2 For Exercises 9 14, write a polynomial in one variable that is described by the following. (Answers may vary.) 9. A monomial of degree A monomial of degree A trinomial of degree A trinomial of degree A binomial of degree A binomial of degree 2 Objective 2: Addition of Polynomials For Exercises 15 22, add the polynomials and simplify. (See Example 1.) m 2 4m2 15m 2 6m n 3 5n2 12n 3 2n x 4 x 3 x x 3 7x 2 2x x 3 2x x 2 3x a w3 2 9 w2 1.8wb a 3 2 w3 1 9 w2 2.7wb a2.9t t 5 3 b a 8.1t4 1 8 t 1 3 b 21. Add 19x 2 5x 12 to 18x 2 x Add 1 x 3 5x2 to 110x 3 x Objective 3: Subtraction of Polynomials For Exercises 23 28, write the opposite of the given polynomial. (See Example 2.) y x p 3 2p t 2 4t ab 2 a 2 b rs 4r 9s For Exercises 29 38, subtract the polynomials and simplify. (See Examples 3 4.) z 5 z z 5 5z w 4 3w w 4 w x 3 3x 2 x 62 1 x 3 x 2 x x 3 6x x 3 2x a 1 5 a2 1 2 ab 1 10 b2 3b a 3 10 a2 2 5 ab 1 2 b2 5b

10 312 Chapter 5 Polynomials 34. a 4 7 a2 1 7 ab 1 14 b2 7b a 1 2 a2 2 7 ab 9 14 b2 1b 35. Subtract 19x 2 5x 12 from 18x 2 x Subtract 1 x 3 5x2 from 110x 3 x Find the difference of 13x 5 2x 3 42 and 1x 4 2x Find the difference of 17x 10 2x 4 3x2 and 1 4x 3 5x 4 x 52. Mixed Exercises For Exercises 39 50, add or subtract as indicated. Write the answers in descending order y 2 4y y 2 8y y y r 6r r 4 9r s 9 7s s 9 s xy 13x 2 3y2 14x 2 8y p 2 q 2q2 1 2p 2 q ab 23b ab 19b x 2 y 92 18x 2 y p 13p 52 14p q q m 2 14m n 3 1n n 3 Objective 4: Polynomial Functions For Exercises 51 58, determine whether the given function is a polynomial function. If it is a polynomial function, state the degree. If not, state the reason why. 51. g1x g1x2 4x 53. h1x2 2 3 x k1x2 7x 55. p1x2 8x 3 2x x x x q1x2 x 2 4x M1x2 0x 0 5x 58. N1x2 x 2 0x Given P1x2 x 4 2x 5, find the function values. (See Example 5.) a. P(2) b. P( 1) c. P(0) d. P(1) 60. Given N1x2 x 2 5x, find the function values. a. N(1) b. N( 1) c. N(2) d. N(0)

11 Section 5.1 Addition and Subtraction of Polynomials and Polynomial Functions Given H1x2 1 2x 3 x 1 4, find the function values. a. H(0) b. H(2) c. H( 2) d. H( 1) 62. Given K1x2 2 3x 2 1 9, find the function values. a. K(0) b. K(3) c. K( 3) d. K( 1) 63. A rectangular garden is designed to be 3 ft longer than it is wide. Let x represent the width of the garden. Find a function P that represents the perimeter in terms of x. (See Example 6.) 64. A flowerbed is in the shape of a triangle with the larger side 3 times the middle side and the smallest side 2 ft shorter than the middle side. Let x represent the length of the middle side. Find a function P that represents the perimeter in terms of x. 65. The cost in dollars of producing x toy cars is C1x2 2.2x 1. The revenue received is R1x2 5.98x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x. b. Find the profit of producing 50 toy cars. 66. The cost in dollars of producing x lawn chairs is C1x2 2.5x The revenue for selling x chairs is R1x2 6.99x. To calculate profit, subtract the cost from the revenue. a. Write and simplify a function P that represents profit in terms of x. b. Find the profit of producing 100 lawn chairs. 67. The function defined by D1x2 8x 2 86x 2766 approximates the yearly dormitory charges for private four-year colleges between the years 1995 and D(x) is measured in dollars, and x 0 corresponds to the year Find the function values and interpret their meaning in the context of this problem. (See Example 7.) a. D(0) b. D(2) Cost (\$) Yearly Dormitory Cost for Private Four-Year Colleges, D(x) 8x 2 86x 2766 c. D(4) d. D(6) Year (x 0 corresponds to 1995) Source: U.S. National Center for Education Statistics.

12 314 Chapter 5 Polynomials 68. The population of Mexico can be modeled by P1t t t 102, where t is the number of years since 2000 and P is the number of people in millions. a. Evaluate P(0) and P(6), and interpret their meaning in the context of this problem. Round to 1 decimal place if necessary. b. If this trend continues, what will the population of Mexico be in the year 2010? Round to 1 decimal place if necessary. Population (millions) P(t) Population Model for Mexico P(t) 0.022t t Year (t 0 corresponds to 2000) t 69. The number of women, w, who were due child support in the United States between 1995 and 2005 can be approximated by W1t2 143t 5865 where t is the number of years after 1995 and W(t) is measured in thousands. (Source: U.S. Bureau of the Census.) a. Evaluate W(0), W(5), and W(10). b. Interpret the meaning of the function value of W(10). 70. The total amount of child support due (in billions of dollars) in the United States between 1995 and 2005 can be approximated by where t is the number of years after 1995 and D(t) is the amount due (in billions of dollars). a. Evaluate D(0), D(4), and D(8). D1t t b. Interpret the meaning of the function value of D(8). Expanding Your Skills 71. A toy rocket is shot from ground level at an angle of 60 from the horizontal. See the figure. The x- and y-positions of the rocket (measured in feet) vary with time t according to x1t2 25t y1t2 16t t a. Evaluate x(0) and y(0), and write the values as an ordered pair. Interpret the meaning of these function values in the context of this problem. Match the ordered pair with a point on the graph. b. Evaluate x(1) and y(1) and write the values as an ordered pair. Interpret the meaning of these function values in the context of this problem. Match the ordered pair with a point on the graph. c. Evaluate x(2) and y(2), and write the values as an ordered pair. Match the ordered pair with a point on the graph. Vertical Distance (ft) y Path of Rocket x 70 Horizontal Distance (ft)

13 Section 5.2 Multiplication of Polynomials 315 section Multiplying Polynomials Multiplication of Polynomials The properties of exponents covered in Section 1.8 can be used to simplify many algebraic expressions including the multiplication of monomials. To multiply monomials, first use the associative and commutative properties of multiplication to group coefficients and like bases. Then simplify the result by using the properties of exponents. example 1 Multiplying Monomials Multiply the monomials. a. 13x 2 y 7 215x 3 y2 b. 1 3x 4 y x 6 yz 8 2 Objectives 1. Multiplying Polynomials 2. Special Case Products: Difference of Squares and Perfect Square Trinomials 3. Translations Involving Polynomials 4. Applications Involving a Product of Polynomials Multiply the polynomials r 3 s21 4r 4 s 4 2 a. 13x 2 y 7 215x 3 y x 2 x 3 21y 7 y2 15x 5 y 8 Group coefficients and like bases. Add exponents and simplify. b. 1 3x 4 y x 6 yz x 4 x 6 21y 3 y21z 8 2 6x 10 y 4 z 8 Group coefficients and like bases. Add exponents and simplify. The distributive property is used to multiply polynomials: a1b c2 ab ac. example 2 Multiplying a Polynomial by a Monomial Multiply the polynomials. a. 5y 3 12y 2 7y 62 b. 4a 3 b 7 c a2ab 2 c a5 bb Multiply the polynomials. 2. 6b 2 12b 2 3b t 3 a 1 2 t t 2 b a. 5y 3 12y 2 7y 62 15y 3 212y y y2 15y y 5 35y 4 30y 3 Apply the distributive property. Simplify each term. b. 4a 3 b 7 c a2ab 2 c a5 bb 1 4a 3 b 7 c212ab 2 c a 3 b 7 c2a 1 2 a5 bb 8a 4 b 9 c 5 2a 8 b 8 c Apply the distributive property. Simplify each term. Answers 1. 32r 7 s b 4 18b 3 48b t 6 2t 5

14 e 316 Chapter 5 Polynomials Thus far, we have illustrated polynomial multiplication involving monomials. Next, the distributive property will be used to multiply polynomials with more than one term. For example: Note: 1x 321x 52 1x 32x 1x 325 1x 32x 1x 325 x x 3 x x x 2 3x 5x 15 x 2 8x 15 Apply the distributive property. Apply the distributive property again. Combine like terms. Using the distributive property results in multiplying each term of the first polynomial by each term of the second polynomial: 1x 321x 52 x x x 5 3 x 3 5 x 2 5x 3x 15 x 2 8x 15 Multiply the polynomials y 1213y 2 2y 12 14t 5212t 32 13a 4 2a 3 a2 12a 2 a 52 example 3 Multiply the polynomials. Multiplying Polynomials a. 12x x 2 x 52 b. 13y 2217y 62 a. 12x x 2 x 52 Multiply each term in the first polynomial by each term in the second. 12x 2 213x x 2 21 x2 12x x x Apply the distributive property. 6x 4 2x 3 10x 2 12x 2 4x 20 6x 4 2x 3 22x 2 4x 20 Simplify each term. Combine like terms. Tip: Multiplication of polynomials can be performed vertically by a process similar to column multiplication of real numbers. 12x x 2 x 52 3x 2 x 5 2x x 2 4x 20 6x 4 2x 3 10x 2 6x 4 2x 3 22x 2 4x 20 Answers 4. 6y 3 7y t 2 22t a 6 a 5 17a 4 8a 3 a 2 5a Note: When multiplying by the column method, it is important to align like terms vertically before adding terms.

15 Section 5.2 Multiplication of Polynomials 317 b. 13y 2217y 62 13y217y2 13y y Multiply each term in the first polynomial by each term in the second. Apply the distributive property. 21y 2 18y 14y 12 21y 2 4y 12 Simplify each term. Combine like terms. Tip: The acronym, FOIL (first outer inner last) can be used as a memory device to multiply the two binomials. Outer terms First Outer Inner Last First terms 13y 2217y 62 13y217y2 13y y Inner terms 21y 2 18y 14y 12 Last terms 21y 2 4y 12 Note: It is important to realize that the acronym FOIL may only be used when finding the product of two binomials. 2. Special Case Products: Difference of Squares and Perfect Square Trinomials In some cases the product of two binomials takes on a special pattern. I. The first special case occurs when multiplying the sum and difference of the same two terms. For example: Note: II. 12x 3212x 32 4x 2 6x 6x 9 4x 2 9 w Notice that the middle terms are opposites. This leaves only the difference between the square of the first term and the square of the second term. For this reason, the product is called a difference of squares. The sum and difference of the same two terms are called conjugates. For example, we call 2x 3 the conjugate of 2x 3 and vice versa. In general, a b and a b are conjugates of each other. The second special case involves the square of a binomial. For example: 13x 72 2 When squaring a binomial, the product will be a trinomial called a perfect 13x 7213x 72 square trinomial.the first and third 9x 2 21x 21x 49 terms are formed by squaring the terms w of the binomial. The middle term is twice 9x 2 42x 49 the product of the terms in the binomial. 13x x

16 318 Chapter 5 Polynomials Note: The expression 13x 72 2 also results in a perfect square trinomial, but the middle term is negative. 13x 7213x 72 9x 2 21x 21x 49 9x 2 42x 49 The following table summarizes these special case products. Special Case Product Formulas 1. 1a b21a b2 a 2 b 2 The product is called a difference of squares. 2. 1a b2 2 a 2 2ab b 2 1a b2 2 a 2 2ab b 2 } The product is called a perfect square trinomial. It is advantageous for you to become familiar with these special case products because they will be presented again when we factor polynomials. Multiply the polynomials. 7. 1c x 4215x s 2 2t2 2 example 4 Finding Special Products Use the special product formulas to multiply the polynomials. a. 15x 22 2 b. 16c 7d216c 7d2 c. 14x 3 3y a 5x, b 2 a. 15x x x x 2 20x 4 Apply the formula a 2 2ab b 2. Simplify each term. b. 16c 7d216c 7d2 16c2 2 17d2 2 36c 2 49d 2 a 6c, b 7d Apply the formula a 2 b 2. Simplify each term. c. 14x 3 3y a 4x3, b 3y2 14x x 3 213y y Apply the formula 2ab b2. 16x 6 24x 3 y 2 9y 4 Simplify each term. The special case products can be used to simplify more complicated algebraic expressions. example 5 Using Special Products Multiply the following expressions. a. 1x y2 3 b. 3x 1y z243x 1y z24 Answers 7. c 2 6c x s 4 28s 2 t 4t 2

17 Section 5.2 Multiplication of Polynomials 319 a. 1x y2 3 1x y2 2 1x y2 1x 2 2xy y 2 21x y2 1x 2 21x2 1x 2 21y2 12xy21x2 12xy21y2 1y 2 21x2 1y 2 21y2 Rewrite as the square of a binomial and another factor. Expand 1x y2 2 by using the special case product formula. Apply the distributive property. Multiply the polynomials b a 1b 3243a 1b 324 x 3 x 2 y 2x 2 y 2xy 2 xy 2 y 3 x 3 3x 2 y 3xy 2 y 3 Simplify each term. Combine like terms. b. 3x 1y z243x 1y z24 1x2 2 1y z2 2 1x2 2 1y 2 2yz z 2 2 x 2 y 2 2yz z 2 This product is in the form 1a b21a b2, where a x and b 1y z2. Apply the formula a 2 b 2. Expand 1y z2 2 by using the special case product formula. Apply the distributive property. 3. Translations Involving Polynomials example 6 Complete the table. Translating Between English Form and Algebraic Form English Form The square of the sum of x and y The square of the product of 3 and x Algebraic Form x 2 y 2 Translate to algebraic form: 12. The square of the difference of a and b 13. The difference of the square of a and the square of b 14. Translate to English form: a b 2. English Form Algebraic Form Notes The square of the sum of x and y The sum of the squares of x and y The square of the product of 3 and x 1x y2 2 x 2 y 2 13x2 2 The sum is squared, not the individual terms. The individual terms x and y are squared first. Then the sum is taken. The product of 3 and x is taken. Then the result is squared. Answers b 3 6b 2 12b 8 a 2 b 2 6b a b a 2 b The difference of a and the square of b

18 320 Chapter 5 Polynomials 4. Applications Involving a Product of Polynomials A rectangular photograph is mounted on a square piece of cardboard whose sides have length x. The border that surrounds the photo is 3 in. on each side and 4 in. on both top and bottom. x example 7 Applying a Product of Polynomials A box is created from a sheet of cardboard 20 in. on a side by cutting a square from each corner and folding up the sides (Figures 5-3 and 5-4). Let x represent the length of the sides of the squares removed from each corner. a. Find an expression for the volume of the box in terms of x. b. Find the volume if a 4-in. square is removed. 4 in. x x x 3 in. 3 in. x x x 4 in. x 15. Write an expression for the area of the photograph and multiply. 16. Determine the area of the photograph if x is x 20 2x x x 20 2x 20 Figure 5-3 Figure 5-4 a. The volume of a rectangular box is given by the formula V lwh. The length and width can both be expressed as 20 2x. The height of the box is x. Hence the volume is given by V l w h 120 2x2120 2x2x 120 2x2 2 x x 4x 2 2x 400x 80x 2 4x 3 4x 3 80x 2 400x b. If a 4-in. square is removed from the corners of the box, we have x 4 in. Thus the volume is V The volume is 576 in. 3 Answers 15. A 1x 821x 62; A x 2 14x in. 2

19 Section 5.2 Multiplication of Polynomials 321 section 5.2 Study Skills Exercises Practice Exercises Boost your GRADE at mathzone.com! Practice Problems e-professors Self-Tests NetTutor Videos 1. Do you need complete silence, or do you listen to music while you do your homework? Try something different today so that you can compare and choose the best situation for you. 2. Define the key terms. a. Difference of squares b. Conjugates c. Perfect square trinomial Review Exercises 3. Given f 1x2 4x 3 5, find the function values. 4. Given g1x2 x 4 x 2 3, find the function values. a. f 132 b. f 102 c. f 1 22 a. g1 12 b. g122 c. g102 For Exercises 5 6, perform the indicated operations x 2 7x 22 1 x 2 3x x 2 7x 22 1 x 2 3x Can the terms 2x 3 and 3x 2 be combined by addition? Can they be combined by multiplication? Explain. 8. Write the distributive property of multiplication over addition. Give an example of the distributive property. (Answers may vary.) Objective 1: Multiplying Polynomials For Exercises 9 38, multiply the polynomials by using the distributive property and the special product formulas. (See Examples 1 3.) 9. 3ab 1a b a 13 a a b m 3 n 2 1m 2 n 3 3mn 2 4n p 2 q 1p 3 q 3 pq 2 4p x y21x 2y a 521a x x x21x a 9212a b 521b 52

20 322 Chapter 5 Polynomials 21. 1y y p p s 3t215s 2t a 3b214a b n n m m a 4b212.5a 7b x 3.5y214.7x 2y x y213x 2 2xy y h 5k21h 2 2hk 3k x 721x 2 7x x 321x 2 3x a b21a 3 4a 2 b ab 2 b m 2n21m 3 2m 2 n mn 2 2n a 1 a 2b cb 1a 6b c x y 2z215x y z x 2 2x 1213x a 1 2 a2 2ab b 2 b12a b2 Objective 2: Special Case Products: Difference of Squares and Perfect Square Trinomials For Exercises 39 54, multiply by using the special case products. (See Example 4.) 39. 1a 821a b 221b p 1213p q 3215q ax a h k213h k x 1 3 b a1 2 x 1 3 b ax 1 3 b 3 b 1x 7y21x 7y h k x 7y t u 3v a 4b ah kb 54. 1w 92 2 a 2 5 x 1b Multiply the expressions. Explain their similarities. a. 1A B21A B2 56. Multiply the expressions. Explain their similarities. a. 1A B21A B2 b. 31x y2 B431x y2 B4 b. 3A 13h k243a 13h k24 For Exercises 57 62, multiply the expressions. (See Example 5.) w v2 2431w v x y2 6431x y x y2432 1x y a 1b 1243a 1b a 42 b4313a 42 b p 72 q4315p 72 q4

21 Section 5.2 Multiplication of Polynomials Explain how to multiply 1x y Explain how to multiply 1a b2 3. For Exercises 65 68, multiply the expressions. (See Example 5.) x y x 5y a b a 4b Explain how you would multiply the 70. Explain how you would multiply the binomials binomials 1x 221x 6212x 12 1a b21a b212a b212a b2 For Exercises 71 74, multiply the expressions containing more than two factors. (Hint: First multiply two of the factors, then multiply that product by the third factor.) 71. 2a 2 1a 5213a y12y 321y x 321x 321x t 221t 321t 12 Objective 3: Translations Involving Polynomials For Exercises 75 78, translate from English form to algebraic form. (See Example 6.) 75. The square of a plus the cube of b 76. The square of the sum of r and t 77. The difference of x squared and y cubed 78. The square of the product of 3 and a For Exercises 79 82, translate from algebraic form to English form. (See Example 6.) 79. p 3 q a 3 b xy c d2 3 Objective 4: Applications Involving a Product of Polynomials 83. A rectangular garden has a walk around it of width x.the garden is 20 ft by 15 ft. Find a function f in terms of x that represents the combined area of the garden and walk. Simplify the result. (See Example 7.) 84. An 8-in. by 10-in. photograph is in a frame of width x. Find a function g in terms of x that represents the area of the frame. Simplify the result. x 20 ft 15 ft 8 in. x 10 in.

23 Section 5.3 Division of Polynomials 325 section Division by a Monomial Division of Polynomials Division of polynomials is presented in this section as two separate cases. The first case illustrates division by a monomial divisor. The second case illustrates division by a polynomial with two or more terms. To divide a polynomial by a monomial, divide each individual term in the polynomial by the divisor and simplify the result. Objectives 1. Division by a Monomial 2. Long Division 3. Synthetic Division To Divide a Polynomial by a Monomial If a, b, and c are polynomials such that c 0, then a b c a c b c Similarly, a b c a c b c example 1 Divide the polynomials. a. 3x 4 6x 3 9x 3x 3x4 3x 6x3 3x x 3 2x 2 3 Dividing a Polynomial by a Monomial 3x 4 6x 3 9x a. b. 110c 3 d 15c 2 d 2 2cd c 2 d 2 2 3x 9x 3x Divide each term in the numerator by 3x. Simplify each term, using the properties of exponents. Divide. 18y 3 6y 2 12y 1. 6y a 3 b 2 16a 2 b 3 8ab2 1 8ab2 b. 110c 3 d 15c 2 d 2 2cd c 2 d c3 d 15c 2 d 2 2cd 3 5c 2 d 2 10c3 d 5c 2 d 15c2 d 2 2 5c 2 d 2cd3 2 5c 2 d 2 2c d 3 2d 5c Divide each term in the numerator by 5c 2 d 2. Simplify each term. 2. Long Division If the divisor has two or more terms, a long division process similar to the division of real numbers is used. Answers 1. 3y 2 y a 2 b 2ab 2 1

24 326 Chapter 5 Polynomials Divide x 2 6x 82 1x 32 12p 3 2p 2 5p p 1 Concept Connections 5. When is long division required for dividing polynomials? example 2 Using Long Division to Divide Polynomials Divide the polynomials by using long division. x 2 3x 2 14x 10 3x x 2 3x 2 14x 10 3x 2 6x 3x x 2 3x 2 14x 10 3x 2 6x 8x 13x 2 14x 102 1x 22 Divide the leading term in the dividend by the leading term in the divisor. 13x 2 2 x 3x. This is the first term in the quotient. Multiply 3x by the divisor and record the result: 3x1x 22 3x 2 6x. Next, subtract the quantity 3x 2 6x. To do this, add its opposite. 3x 8 x 2 3x 2 14x 10 3x 2 6x 8x 10 8x 16 Bring down next column and repeat the process. Divide the leading term by x: 8x x 8 Multiply the divisor by 8 and record the result: 81x 22 8x 16. 3x 8 x 2 3x 2 14x 10 3x 2 6x 8x 10 8x Subtract the quantity 1 8x 162 by adding its opposite. The remainder is 26. We do not continue because the degree of the remainder is less than the degree of the divisor. Answers 3. 4x 6 10 x p 2 2p 1 4 3p 1 5. Long division is required when the divisor has two or more terms. Summary: The quotient is The remainder is The divisor is The dividend is 3x 8 26 x 2 3x 2 14x 10 The solution to a long division problem is often written in the form: Quotient remainder divisor. Hence This answer can also be written as 13x 2 14x 102 1x 22 3x 8 26 x 2 3x 8 26 x 2

25 Section 5.3 Division of Polynomials 327 The division of polynomials can be checked in the same fashion as the division of real numbers. To check, we know that Dividend (divisor)(quotient) remainder 3x 2 14x 10 1x 22 13x x 2 8x 6x x 2 14x 10 example 3 Using Long Division to Divide Polynomials Divide the polynomials by using long division: 1 2x 3 10x x 42 First note that the dividend has a missing power of x and can be written as 2x 3 10x 2 0x 56.The term 0x is a placeholder for the missing term. It is helpful to use the placeholder to keep the powers of x lined up. x 2 2x 4 2x 3 10x 2 0x 56 2x 3 4x 2 x 2 7x 2x 4 2x 3 10x 2 0x 56 2x 3 4x 2 14x 2 0x 14x 2 28x x 2 7x 14 2x 4 2x 3 10x 2 0x 56 2x 3 4x 2 14x 2 0x 14x 2 28x 28x 56 28x 56 x 2 7x 14 2x 4 2x 3 10x 2 0x 56 2x 3 4x 2 14x 2 0x 14x 2 28x 28x 56 28x 56 Leave space for the missing power of x. Divide 2x 3 2x x 2 to get the first term of the quotient. Subtract by adding the opposite. Bring down the next column. Divide ( 14x 2 ) (2x) 7x to get the next term in the quotient. Subtract by adding the opposite. Bring down the next column. Divide ( 28x) (2x) 14 to get the next term in the quotient. Subtract by adding the opposite. 0 The remainder is 0. The quotient is x 2 7x 14 and the remainder is 0. Divide. 4y 3 2y y 2 Tip: Both the divisor and dividend must be written in descending order before you do polynomial division. Answer 6. 2y 2 2y 1 5 2y 2

26 328 Chapter 5 Polynomials Because the remainder is zero, 2x 4 divides evenly into 2x 3 10x For this reason, the divisor and quotient are factors of 2x 3 10x To check, we have Dividend (divisor) (quotient) remainder 2x 3 10x x 421 x 2 7x x 3 14x 2 28x 4x 2 28x 56 2x 3 10x 2 56 Divide. 7. 1x 3 2x x 2 12 example 4 Divide. Using Long Division to Divide Polynomials 16x 4 15x 3 5x x 2 42 The dividend has a missing power of x and can be written as 6x 4 15x 3 5x 2 0x 4. 2x 2 Leave space for missing The divisor has a missing power of x and can be written as 0x 4. 3x 2 0x 4 6x 4 15x 3 5x 2 0x 4 powers of x. 6x 4 0x 3 8x 2 Divide 6x 4 3x 2 2x 2 to get the first term of the quotient. Answer x 1 7. x 2 x 2 1 2x 2 5x 3x 2 0x 4 6x 2 15x 3 5x 2 0x 4 6x 4 0x 3 8x 2 15x 3 3x 2 0x 15x 3 0x 2 20x 2x 2 5x 1 3x 2 0x 4 6x 4 15x 3 5x 2 0x 4 6x 4 0x 3 8x 2 15x 3 3x 2 0x 15x 3 0x 2 20x 2x 2 5x 1 3x 2 0x 4 6x 4 15x 3 5x 2 0x 4 6x 4 0x 3 8x 2 3x 2 20x 4 3x 2 0x 4 15x 3 3x 2 0x 15x 3 0x 2 20x 3x 2 20x 4 3x 2 0x 4 20x Subtract by adding the opposite. Bring down the next column. Subtract by adding the opposite. Bring down the next column. Subtract by adding the opposite. The remainder is 20x. The degree of 20x is less than the degree of 3x 2 4.

27 Section 5.3 Division of Polynomials 329 Therefore, 16x 4 15x 3 5x x x 2 5x 1 20x 3x Synthetic Division In this section we introduced the process of long division to divide two polynomials. Next, we will learn another technique, called synthetic division, to divide two polynomials. Synthetic division may be used when dividing a polynomial by a first-degree divisor of the form x r, where r is a constant. Synthetic division is considered a shortcut because it uses the coefficients of the divisor and dividend without writing the variables. Consider dividing the polynomials 13x 2 14x 102 1x 22. First note that the divisor x 2 is in the form x r, where r 2. Hence synthetic division can also be used to find the quotient and remainder. Step 1: Write the value of r in a box. 3 Step 3: Skip a line and draw a horizontal line below the list of coefficients. 3x 8 x 2 3x 2 14x 10 13x 2 6x2 8x x Step 5: Multiply the value 6 of r by the number 3 8 below the line Write the result in the next column above the line. Step 2: Step 4: Step 6: Write the coefficients of the dividend to the right of the box. Bring down the leading coefficient from the dividend and write it below the line. Add the numbers in the column above the line , and write the result below the line. Repeat steps 5 and 6 until all columns have been completed. Step 7: To get the final A box is usuresult, we use the 6 16 ally drawn numbers below around the the line. The remainder. number in the last column is the Quotient: 3x 8, remainder 26 remainder. The other numbers are the coefficients of the quotient. Concept Connections Determine whether synthetic division can be used to divide the polynomials. Explain your answer. 8. 1x 3 6x 2 2x 12 12x y 2 5y 82 1y a 3 5a 2 a 22 1a 52 Answers 8. No. The leading coefficient of the divisor is not equal to No. The divisor is not firstdegree. 10. Yes. The divisor is in the form 1x r2, where r 5.

28 330 Chapter 5 Polynomials The degree of the quotient will always be 1 less than that of the dividend. Because the dividend is a second-degree polynomial, the quotient will be a first-degree polynomial. In this case, the quotient is 3x 8 and the remainder is 26. Divide the polynomials by using synthetic division. Identify the quotient and the remainder y 2 4y 2y y 32 example 5 Divide the polynomials division. Using Synthetic Division to Divide Polynomials 15x 4x 3 6 x 4 2 1x 32 by using synthetic As with long division, the terms of the dividend and divisor should be written in descending order. Furthermore, missing powers must be accounted for by using placeholders (shown here in bold). Hence, 5x 4x 3 6 x 4 x 4 4x 3 0x 2 5x 6 To use synthetic division, the divisor must be in the form (x r). The divisor x 3 can be written as x ( 3). Hence, r 3. Step 1: Write the value Step 2: Write the of r in a box. coefficients 1 of the dividend to the right of the box. Answer 11. Quotient: 2y 2 y 1; remainder: 2 Step 3: Skip a line and Step 4: Bring down draw a horizontal the leadline below the ing colist of coefficients. efficient from the dividend and write it below the line. Step 5: Multiply the Step 6: Add the value of r by the 3 numbers number below 1 1 in the the line ( 3 1 3). column Write the result above in the next column the line: above the line. 4 ( 3) 1. Repeat steps 5 and 6: The quotient is x 3 x 2 3x 14. The remainder is 48. remainder constant x-term coefficient x 2 -term coefficient x 3 -term coefficient

29 Section 5.3 Division of Polynomials 331 Tip: It is interesting to compare the long division process to the synthetic division process. For Example 5, long division is shown on the left, and synthetic division is shown on the right. Notice that the same pattern of coefficients used in long division appears in the synthetic division process. x 3 x 2 3x 14 x 3 x 4 4x 3 0x 2 5x 6 1x 4 3x 3 2 x 3 0x 2 1x 3 3x 2 2 3x 2 5x 1 3x 2 9x2 14x 6 114x x 3 x 2 x constant remainder Quotient: x 3 x 2 3x 14 Remainder: 48 example 6 Using Synthetic Division to Divide Polynomials Divide the polynomials by using synthetic division. Identify the quotient and remainder. a. 12m 7 3m 5 4m 4 m 82 1m 22 b. 1p p 32 a. Insert placeholders (bold) for missing powers of m. b. 12m 7 3m 5 4m 4 m 82 1m 22 12m 7 0m 6 3m 5 4m 4 0m 3 0m 2 m 82 1m 22 Because m +2 can be written as m 1 22, r Quotient: 2m 6 4m 5 5m 4 6m 3 12m 2 24m 47 Remainder: 86 1p p 32 1p 4 0p 3 0p 2 0p 812 1p Quotient: p 3 3p 2 9p 27 Remainder: 0ction 4.5 The quotient is 1 degree less than dividend. Insert placeholders (bold) for missing powers of p. Divide the polynomials by using synthetic division. Identify the quotient and the remainder c 4 3c 2 6c 32 1c x x 12 Answers 12. Quotient: 4c 3 8c 2 13c 20; remainder: Quotient: x 2 x 1; remainder: 0

30 332 Chapter 5 Polynomials section 5.3 Practice Exercises Boost your GRADE at mathzone.com! Practice Problems e-professors Self-Tests NetTutor Videos Study Skills Exercises 1. a. How often do you do your math homework? Every day? Three times a week? Once a week? b. Math should be studied every day. List the days and times when you plan to work on math this week. Mon. Tue. Wed. Thurs. Fri. Sat. Sun. 2. Define the key term synthetic division. Review Exercises 3. a. Add 13x 12 12x a. Subtract 1a 10b2 15a b2. b. Multiply 13x 1212x 52. b. Multiply 1a 10b215a b2. 5. a. Subtract 12y y 2 5y a. Add 1x 2 x2 16x 2 x 22. b. Multiply 12y 2 121y 2 5y 12. b. Multiply 1x 2 x216x 2 x 22. For Exercises 7 11, answers may vary. 7. Write an example of a product of a monomial 8. Write an example of a product of two binomials and a binomial and simplify. and simplify. 9. Write an example of a sum of a binomial and a 10. Write an example of the square of a binomial trinomial and simplify. and simplify. 11. Write an example of the product of conjugates and simplify. Objective 1: Division by a Monomial For Exercises 12 25, divide the polynomials. Check your answer by multiplication. (See Example 1.) y 24y 2 6y y p 2 18p 4 30p p x 3 y 12x 2 y 2 4xy xy m 5 n 10m 4 n m 3 n2 15m 3 n2

31 Section 5.3 Division of Polynomials y 4 12y 3 32y y y 5 8y 6 16y 4 10y y p 4 6p 3 2p 2 p2 1 6p q 3 8q 2 q2 1 12q a 3 5a 2 a 52 1a m 5 3m 4 m 3 m 2 9m2 1m s 3 t 5 8s 2 t 4 10st st r 4 w 2 4r 3 w 2w r 3 w p 4 q 7 9p 5 q 6 11p 3 q 42 1p 2 q a 5 b 5 20a 3 b 2 5a 2 b 62 1a 2 b2 Objective 2: Long Division 26. a. Divide 12x 3 7x 2 5x 12 1x 22, and identify the divisor, quotient, and remainder. b. Explain how to check by using multiplication. 27. a. Divide 1x 3 4x 2 7x 32 1x 32, and identify the divisor, quotient, and remainder. b. Explain how to check by using multiplication. For Exercises 28 43, divide the polynomials by using long division. Check your answer by multiplication. (See Examples 2 4.) 28. 1x 2 11x 192 1x x 3 7x 2 13x 32 1x y 3 7y 2 4y 32 1y z 3 2z 2 2z 52 1z a 2 77a a x 2 29x 62 14x y 2 9y y y 2 2y 12 1 y a a x x x 4 x 3 x 2 4x 22 1x 2 x a 5 7a 4 11a 3 22a 2 29a a 2 5a x 4 3x x y 4 25y y n n m m 32

32 334 Chapter 5 Polynomials Objective 3: Synthetic Division 44. Explain the conditions under which you may use synthetic division to divide polynomials. 45. Can synthetic division be used to divide 14x 4 3x 3 7x 92 by 12x 52? Explain why or why not. 46. Can synthetic division be used to divide 16y 5 3y 2 2y 142 by 1y 2 32? Explain why or why not. 47. Can synthetic division be used to divide 13y 4 y 12 by 1y 52? Explain why or why not. 48. The following table represents the result 49. The following table represents the result of a synthetic division. of a synthetic division Use x as the variable Use x as the variable. a. Identify the divisor. a. Identify the divisor. b. Identify the quotient. b. Identify the quotient. c. Identify the remainder. c. Identify the remainder. For Exercises 50 61, divide by using synthetic division. Check your answer by multiplication. (See Examples 5 6.) 50. 1x 2 2x 482 1x x 2 4x 122 1x t 2 3t 42 1t h 2 7h 122 1h y 2 5y 12 1y w 2 w 52 1w y 3 7y 2 4y 32 1y z 3 2z 2 2z 52 1z x 3 3x x y4 25y y w 4 w 2 6w 32 aw 1 2 b y 4 5y 3 y 2 y 32 ay 3 4 b Mixed Exercises For Exercises 62 73, divide the polynomials by using an appropriate method x 3 8x 2 3x 22 1x xy 2 9x 2 y 6x 2 y 2 2 1x 2 y x 2 11x x m 3 4m 2 5m 332 1m 32

33 Midchapter Review y 3 17y 2 30y y 2 2y h 12 63h 9 45h 8 36h h x 4 6x 3 3x 12 12x y 4 3y 3 5y 2 2y 52 1y k 11 32k 10 8k 8 40k k m 3 18m 2 22m m 2 4m x 3 9x 2 10x2 15x k 4 3k 3 4k k 2 12 chapter 5 midchapter review 1. Give the leading coefficient and the degree of the polynomial 5x 6 2x 5 7x 3 5x. 2. Given P1x2 2x 3 3x 2 x, find P y 3 2y 2 y 22 13y 3 4y 32 4x 2 6x 1 2x 1 For Exercises 3 14, perform the indicated operations t 2 6t 22 13t 2 7t 32 5x 2 13x 2 x x a 3 8a 2 16a 8a a 2212a 2 3a 12 1t 3 4t 2 t 92 1t t 2 6t2 12b 3 3b 102 1b 22 1p 521p 52 12p z 5216z k k 92

34 336 Chapter 5 Polynomials Objectives 1. Factoring Out the Greatest Common Factor 2. Factoring Out a Negative Factor 3. Factoring Out a Binomial Factor 4. Factoring by Grouping section 5.4 Greatest Common Factor and Factoring by Grouping 1. Factoring Out the Greatest Common Factor The next three sections of this chapter are devoted to a mathematical operation called factoring. To factor an integer means to write the integer as a product of two or more integers. To factor a polynomial means to express the polynomial as a product of two or more polynomials. In the product , for example, 5 and 7 are factors of 35. In the product 12x 121x 62 2x 2 11x 6, the quantities 12x 12 and 1x 62 are factors of 2x 2 11x 6. The greatest common factor (GCF) of a polynomial is the greatest factor that divides each term of the polynomial evenly. For example, the greatest common factor of 9x 4 18x 3 6x 2 is 3x 2. In other words, 3x 2 is the greatest factor that divides evenly into each term. To factor out the greatest common factor from a polynomial, follow these steps: Steps to Remove the Greatest Common Factor 1. Identify the greatest common factor of all terms of the polynomial. 2. Write each term as the product of the GCF and another factor. 3. Use the distributive property to factor out the greatest common factor. Note: To check the factorization, multiply the polynomials. Factor out the greatest common factor y 5 15y 2 30y 2. 16a 2 b 5 12a 3 b 3 4a 3 b 2 example 1 Factoring Out the Greatest Common Factor Factor out the greatest common factor. a. 12x 3 30x 2 b. 12c 2 d 3 30c 3 d 2 3cd a. 12x 3 30x 2 6x 2 12x2 6x x 2 12x 52 The GCF is 6x 2. Write each term as the product of the GCF and another factor. Factor out 6x 2 by using the distributive property. Avoiding Mistakes: In Example 1(b), the GCF of 3cd is equal to one of the terms of the polynomial. In such a case, you must leave a 1 in place of that term after the GCF is factored out. 3cd 14cd 2 10c 2 d 12 Answers 1. 15y 13y 4 y a 2 b 2 14b 3 3ab a2 Tip: Any factoring problem can be checked by multiplying the factors: Check: 6x 2 12x 52 12x 3 30x 2 b. 12c 2 d 3 30c 3 d 2 3cd 3cd 14cd 2 2 3cd 110c 2 d2 3cd 112 3cd14cd 2 10c 2 d 12 The GCF is 3cd. Write each term as the product of the GCF and another factor. Factor out 3cd by using the distributive property. Check: 3cd14cd 2 10c 2 d 12 12c 2 d 3 30c 3 d 2 3cd

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