How To Solve Factoring Problems


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1 05W4801AM1.qxd 8/19/08 8:45 PM Page 241 Factoring, Solving Equations, and Problem Solving Factoring by Using the Distributive Property 5.2 Factoring the Difference of Two Squares 5.3 Factoring Trinomials of the Form x 2 bx c 5.4 Factoring Trinomials of the Form ax 2 bx c 5.5 Factoring, Solving Equations, and Problem Solving Algebraic equations can be used to solve a large variety of problems involving geometric relationships. Photodisc/Getty Images A flower garden is in the shape of a right triangle with one leg 7 meters longer than the other leg and the hypotenuse 1 meter longer than the longer leg. Find the lengths of all three sides of the right triangle. A popular geometric formula, called the Pythagorean theorem, serves as a guideline for setting up an equation to solve this problem. We can use the equation x 2 1x x 82 2 to determine that the sides of the right triangle are 5 meters, 12 meters, and 13 meters long. The distributive property has allowed us to combine similar terms and multiply polynomials. In this chapter, we will see yet another use of the distributive property as we learn how to factor polynomials. Factoring polynomials will allow us to solve other kinds of equations, which will, in turn, help us to solve a greater variety of word problems. Exciting videos of all objective concepts are available in a variety of delivery models. 241
2 05W4801AM1.qxd 8/19/08 8:45 PM Page 242 INTERNET PROJECT Pythagoras is widely known for the Pythagorean theorem pertaining to right triangles. Do an Internet search to determine at least two other fields where Pythagoras made significant contributions. Pythagoras also founded a school. While conducting your search, find the name given to the students attending Pythagoras' school and some of the school rules for students. Can you think of any modernday schools that might have the same requirements? 5.1 Factoring by Using the Distributive Property OBJECTIVES 1 Find the Greatest Common Factor 2 Factor Out the Greatest Common Factor 3 Factor by Grouping 4 Solve Equations by Factoring 5 Solve Word Problems Using Factoring 1 Find the Greatest Common Factor In Chapter 1, we found the greatest common factor of two or more whole numbers by inspection or by using the prime factored form of the numbers. For example, by inspection we see that the greatest common factor of 8 and 12 is 4. This means that 4 is the largest whole number that is a factor of both 8 and 12. If it is difficult to determine the greatest common factor by inspection, then we can use the prime factorization technique as follows: 42 2 # 3 # # 5 # 7 We see that 2 # 7 14 is the greatest common factor of 42 and 70. It is meaningful to extend the concept of greatest common factor to monomials. Consider the next example. EXAMPLE 1 Find the greatest common factor of 8x 2 and 12x 3. 8x 2 2 # 2 # 2 # x # x 12x 3 2 # 2 # 3 # x # x # x 2 # 2 # x # x 4x 2 Therefore, the greatest common factor is. Find the greatest common factor of 14a 2 and 7a 5. 7a 2 By the greatest common factor of two or more monomials we mean the monomial with the largest numerical coefficient and highest power of the variables that is a factor of the given monomials. 242
3 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring by Using the Distributive Property 243 EXAMPLE 2 Find the greatest common factor of 16x 2 y, 24x 3 y 2, and 32xy. 16x 2 y 2 # 2 # 2 # 2 # x # x # y 24x 3 y 2 2 # 2 # 2 # 3 # x # x # x # y # y 32xy 2 # 2 # 2 # 2 # 2 # x # y Therefore, the greatest common factor is. 2 # 2 # 2 # x # y 8xy Find the greatest common factor of 18m 2 n 4, 4m 3 n 5, and 10m 4 n 3. 2m 2 n 3 2 Factor Out the Greatest Common Factor We have used the distributive property to multiply a polynomial by a monomial; for example, 3x1x 22 3x 2 6x Suppose we start with 3x 2 6x and want to express it in factored form. We use the distributive property in the form ab ac a(b c). 3x 2 6x 3x1x2 3x122 3x1x 22 3x is the greatest common factor of 3x 2 and 6x Use the distributive property The next four examples further illustrate this process of factoring out the greatest common monomial factor. EXAMPLE 3 Factor 12x 3 8x 2. 12x 3 8x 2 4x 2 13x2 4x x 2 13x 22 ab ac a(b c) Factor 15a 2 21a 6. 3a 2 (5 7a 4 ) EXAMPLE 4 Factor 12x 2 y 18xy 2. 12x 2 y 18xy 2 6xy12x2 6xy13y2 6xy12x 3y2 Factor 8m 3 n 2 2m 6 n. 2m 3 n(4n m 3 )
4 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 5 Factor 24x 3 30x 4 42x 5. 24x 3 30x 4 42x 5 6x x 3 15x2 6x 3 17x 2 2 6x x 7x 2 2 Factor 48y 8 16y 6 24y 4. 8y 4 (6y 4 2y 2 3) EXAMPLE 6 Factor 9x 2 9x. 9x 2 9x 9x1x2 9x112 9x1x 12 Factor 8b 3 8b 2. 8b 2 (b 1) We want to emphasize the point made just before Example 3. It is important to realize that we are factoring out the greatest common monomial factor. We could factor an expression such as 9x 2 9x in Example 6 as 91x 2 x2, 313x 2 3x2, 1 3x13x 32, or even, but it is the form 9x1x 12 that we want. We 2 118x2 18x2 can accomplish this by factoring out the greatest common monomial factor; we sometimes refer to this process as factoring completely. A polynomial with integral coefficients is in completely factored form if these conditions are met: 1. It is expressed as a product of polynomials with integral coefficients. 2. No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients. Thus 91x 2 x2, 313x 2 3x2, and 3x13x 22 are not completely factored because they violate condition 2. The form violates both conditions x2 18x2 1 and 2. 3 Factor by Grouping Sometimes there may be a common binomial factor rather than a common monomial factor. For example, each of the two terms of x(y 2) z(y 2) has a common binomial factor of (y 2). Thus we can factor (y 2) from each term and get x(y 2) z(y 2) (y 2)(x z) Consider a few more examples involving a common binomial factor: a1b c2 d1b c2 1b c21a d2 x1x 22 31x 22 1x 221x 32 x1x 52 41x 52 1x 521x 42
5 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring by Using the Distributive Property 245 It may be that the original polynomial exhibits no apparent common monomial or binomial factor, which is the case with ab 3a bc 3c However, by factoring a from the first two terms and c from the last two terms, we see that ab 3a bc 3c a(b 3) c(b 3) Now a common binomial factor of (b 3) is obvious, and we can proceed as before: a(b 3) c(b 3) (b 3)(a c) This factoring process is called factoring by grouping. Let s consider two more examples of factoring by grouping. EXAMPLE 7 Factor each polynomial completely. (a) x 2 x 5x 5 (b) 6x 2 4x 3x 2 (a) x 2 x 5x 5 x1x 12 51x 12 1x 121x 52 (b) 6x 2 4x 3x 2 2x13x x 22 13x 2212x 12 Factor x from first two terms and 5 from last two terms Factor common binomial factor of (x 1) from both terms Factor 2x from first two terms and 1 from last two terms Factor common binomial factor of (3x 2) from both terms Factor each polynomial completely. (a) ab 5a 3b 15 (b) xy 2x 4y 8 (a) (a 3)(b 5) (b) (x 4)(y 2) 4 Solve Equations by Factoring Suppose we are told that the product of two numbers is 0. What do we know about the numbers? Do you agree we can conclude that at least one of the numbers must be 0? The next property formalizes this idea. Property 5.1 For all real numbers a and b, ab 0 if and only if a 0 or b 0 Property 5.1 provides us with another technique for solving equations.
6 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 8 Solve x 2 6x 0. To solve equations by applying Property 5.1, one side of the equation must be a product, and the other side of the equation must be zero. This equation already has zero on the righthand side of the equation, but the lefthand side of this equation is a sum. We will factor the lefthand side, x 2 6x, to change the sum into a product. x 2 6x 0 x1x 62 0 x 0 or x 6 0 x 0 or ab 0 if and only if a 0 or b 0 The solution set is { 6, 0}. (Be sure to check both values in the original equation.) x 6 Solve y 2 7y 0. {0, 7} EXAMPLE 9 Solve x 2 12x. In order to solve this equation by Property 5.1, we will first get zero on the righthand side of the equation by adding 12x to each side. Then we factor the expression on the lefthand side of the equation. x 2 12x x 2 12x 0 Added 12x to both sides x1x x 0 or x 12 0 ab 0 if and only if a 0 or b 0 x 0 or x 12 The solution set is {0, 12}. Solve a 2 15a. {0,15} Remark: Note in Example 9 that we did not divide both sides of the original equation by x. Doing so would cause us to lose the solution of 0. EXAMPLE 10 Solve 4x 2 3x 0. 4x 2 3x 0 x14x 32 0 x 0 or 4x 3 0 x 0 or 4x 3 ab 0 if and only if a 0 or b 0 x 0 or x 3 4 The solution set is e 0, 3. 4 f
7 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring by Using the Distributive Property 247 Solve 5y 2 2y 0. e 2 5, 0 f EXAMPLE 11 Solve x1x 22 31x In order to solve this equation by Property 5.1, we will factor the lefthand side of the equation. The greatest common factor of the terms is (x 2). x1x 22 31x x 221x 32 0 x 2 0 or x 3 0 x 2 or x 3 The solution set is 5 3, 26. ab 0 if and only if a 0 or b 0 Solve m(m 2) 5(m 2) 0. { 5, 2} 5 Solve Word Problems Using Factoring Each time we expand our equationsolving capabilities, we gain more techniques for solving word problems. Let s solve a geometric problem with the ideas we learned in this section. EXAMPLE 12 s Apply Your Skill The area of a square is numerically equal to twice its perimeter. Find the length of a side of the square. Sketch a square and let s represent the length of each side (see Figure 5.1). Then the area is represented by and the perimeter by 4s. Thus s 2 s s Figure 5.1 s s 2 214s2 s 2 8s s 2 8s 0 s1s 82 0 s 0 or s 8 0 s 0 or s 8 Because 0 is not a reasonable answer to the problem, the solution is 8. (Be sure to check this solution in the original statement of the example!) The area of a square is numerically equal to three times its perimeter. Find the length of a side of the square. The length is 12
8 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving CONCEPT QUIZ For Problems 1 10, answer true or false. 1. The greatest common factor of 6x 2 y 3 12x 3 y 2 18x 4 y is 2x 2 y. 2. If the factored form of a polynomial can be factored further, then it has not met the conditions for being considered factored completely. 3. Common factors are always monomials. 4. If the product of x and y is zero, then x is zero or y is zero. 5. The factored form 3a(2a 2 4) is factored completely. 6. The solutions for the equation x(x 2) 7 are 7 and The solution set for x 2 7x is {7}. 8. The solution set for x(x 2) 3(x 2) 0 is {2, 3}. 9. The solution set for 3x x 2 is { 3, 0}. 10. The solution set for x(x 6) 2(x 6) is { 6}. Problem Set Find the Greatest Common Factor x 2 y 40xy 55y 5y16x 2 8x 112 For Problems 1 10, find the greatest common factor of the given expressions y and 30xy 6y 2. 32x and 40xy 8x 3. 60x 2 y and 84xy 2 12xy ab 3 and 70a 2 b 2 14ab x 3, 8x, and 24x 2 2x 8. 72xy, 36x 2 y, and 84xy 2 12xy 9. 16a 2 b 2, 40a 2 b 3, and 56a 3 b 4 8a 2 b a 3 b 3, 42a 2 b 4, and 49ab 5 7ab 3 72x 3 and 63x 2 48a 2 b 2 and 96ab 4 9x 2 48ab x 3 3x 2 4x 34. x 4 x 3 x 2 x12x 2 3x 42 x21x2 x y 5 24y 3 20y 2 4y 2 111y 3 6y a 18a 3 26a 5 14a 2 b 3 35ab 2 49a 3 b 24a 3 b 2 36a 2 b 4 60a 4 b 3 x1y 12 z1y 12 a1c d2 21c d2 a1b 42 c1b 42 2a17 9a 2 13a 4 2 7ab12ab 2 5b 7a a 2 b 2 12a 3b 2 5a 2 b2 1y 121x z2 1c d21a 22 1b 421a c2 2 Factor Out the Greatest Common Factor 42. x1y 62 31y 62 1y 621x 32 For Problems 11 46, factor each polynomial completely. 43. x1x 32 61x 32 1x 321x x 12y 412x 3y x 24y 613x 4y2 44. x1x 72 91x 72 1x 721x xy 21y 7y12x x 40xy 8x13 5y x1x 12 31x 12 1x 1212x x 2 45x 9x12x x 28x 3 4x13 7x x1x 82 51x 82 1x 8214x xy 2 30x 2 y x 2 y 2 49x 2 y 6xy12y 5x2 7x 2 y14y a 2 b 60a 3 b ab 3 45a 2 b 2 12a 2 b13 5ab 3 2 5ab 2 113b 9a xy 3 25x 2 y x 2 y 2 29x 2 y xy 2 116y 25x2 x 2 y112y ab 72cd xy 72zw 818ab 9cd2 915xy 8zw a 2 b 4 27a 2 b 26. 7a 3 b 5 42a 2 b 6 9a 2 b1b a 2 b 5 1a 6b x 4 y 2 60x 6 y x 5 y 3 42x 8 y 2 4x 4 y(13y 15x 2 ) 14x 5 y 2 15y 3x x 2 y 2 8x 2 y x 2 y 3 12xy 3 8x 2 y15y 12 12xy 3 17x x 15xy 21x 2 3x14 5y 7x2 3 Factor by Grouping For Problems 47 60, use the process of factoring by grouping to factor each polynomial x 5y bx by 1x y215 b x 7y zx zy bx by cx cy 2x 2y ax ay 1x y217 z2 1x y21b c2 1x y212 a2 51. ac bc a b 1a b21c 12 Blue problem numbers indicate Enhanced WebAssign Problems.
9 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring by Using the Distributive Property x y ax ay 1x y211 a2 x 2 5x 12x 60 1x 521x 122 x 2 3x 7x 21 1x 321x 72 x 2 2x 8x 16 1x 221x 82 x 2 4x 9x 36 1x 421x 92 2x 2 x 10x 5 12x 121x 52 3x 2 2x 18x 12 13x 221x 62 6n 2 3n 8n 4 12n 1213n x13x x x 62 x1x 62 0 {4, 6} 84. x1x 92 21x 92 { 9, 2} e 2 3, 7 f 5 Solve Word Problems Using Factoring For Problems 85 91, set up an equation and solve each problem. 85. The square of a number equals nine times that number. Find the number. 0 or n 2 8n 15n 6 15n 2214n Suppose that four times the square of a number equals 20 times that number. What is the number? 0 or 5 4 Solve Equations by Factoring For Problems 61 84, solve each equation. 61. x 2 8x 0 {0, 8} 62. x 2 12x 0 {0, 12} 63. x 2 x 0 5 1, n 2 5n 50, e 0, 3 2y2 3y f 69. e 3 7x2 3x 70. 7, 0 f 71. 3n 2 15n 0 5 5, e 0, 3 4x2 6x f 75. 7x x , x x 2 50, x e 5 2x , 0 f 81. x1x 52 41x , 46 x 2 7x 0 n 2 2n 4y2 7y 0 5x2 2x 6n 2 24n 0 12x2 8x 9x x x x 2 7x 5x2 THOUGHTS INTO WORDS 5 7, , 06 e 2 5, 0 f e 0, 2 3 f e 0, 7 4 f 50, 46 50, , 06 e 7 5, 0 f 92. Suppose that your friend factors 24x 2 y 36xy like this: 24x 2 y 36xy 4xy16x 92 14xy213212x 32 12xy12x 32 Is this correct? Would you suggest any changes? 87. The area of a square is numerically equal to five times its perimeter. Find the length of a side of the square. 20 units 88. The area of a square is 14 times as large as the area of a triangle. One side of the triangle is 7 inches long, and the altitude to that side is the same length as a side of the square. Find the length of a side of the square. Also find the areas of both figures, and be sure that your answer 343 checks. 49 inches, 2401 square inches, square inches Suppose that the area of a circle is numerically equal to the perimeter of a square whose length of a side is the same as the length of a radius of the circle. Find the length of a side of the square. Express your answer in terms of p. 4 p 90. One side of a parallelogram, an altitude to that side, and one side of a rectangle all have the same measure. If an adjacent side of the rectangle is 20 centimeters long, and the area of the rectangle is twice the area of the parallelogram, find the areas of both figures. See answer below 91. The area of a rectangle is twice the area of a square. If the rectangle is 6 inches long, and the width of the rectangle is the same as the length of a side of the square, find the dimensions of both the rectangle and the square. The square is 3 inches by 3 inches and the rectangle is 3 inches by 6 inches 93. The following solution is given for the equation x1x x1x x 2 10x 0 x1x x 0 or x 10 0 x 0 or x 10 The solution set is {0, 10}. Is this solution correct? Would you suggest any changes? 90. The area of the parallelogram is 10(10) 100 square centimeters and the area of the rectangle is 10(20) 200 square centimeters
10 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving FURTHER INVESTIGATIONS 94. The total surface area of a right circular cylinder is given by the formula A 2pr 2 2prh, where r represents the radius of a base, and h represents the height of the cylinder. For computational purposes, it may be more convenient to change the form of the right side of the formula by factoring it. A 2pr 2 2prh 2pr1r h2 Use A 2pr1r h2 to find the total surface area of 22 each of the following cylinders. Use as an approximation for p. 7 a. r 7 centimeters and h 12 centimeters 836 square centimeters b. r 14 meters and h 20 meters 2992 square meters c. r 3 feet and h 4 feet 132 square feet d. r 5 yards and h 9 yards 440 square yards 95. The formula A P Prt yields the total amount of money accumulated (A) when P dollars are invested at r percent simple interest for t years. For computational purposes it may be convenient to change the right side of the formula by factoring. A P Prt P11 rt2 Use A P(1 rt) to find the total amount of money accumulated for each of the following investments. a. $100 at 8% for 2 years $116 b. $200 at 9% for 3 years $254 c. $500 at 10% for 5 years $750 d. $1000 at 10% for 10 years $2000 For Problems 96 99, solve each equation for the indicated variable. 96. ax bx c for x x c a b 97. b 2 x 2 cx 0 for x x 0 or x c 98. 5ay 2 by for y y 0 or y b b 2 5a c 99. y ay by c 0 for y y 1 a b Answers to the Example Practice Skills 1. 7a m 2 n a 2 (5 7a 4 ) 4. 2m 3 n(4n m 3 ) 5. 8y 4 (6y 4 2y 2 3) 6. 8b 2 (b 1) 7. (a) (a 3)(b 5) (b) (x 4)(y 2) 8. {0, 7} 9. {0, 15} 10. e { 5, 2} 5, 0 f 12. The length is 12. Answers to the Concept Quiz 1. False 2. True 3. False 4. True 5. False 6. False 7. False 8. True 9. True 10. False 5.2 Factoring the Difference of Two Squares OBJECTIVES 1 Factor the Difference of Two Squares 2 Solve Equations by Factoring the Difference of Two Squares 3 Solve Word Problems Using Factoring 1 Factor the Difference of Two Squares In Section 4.3, we noted some special multiplication patterns. One of these patterns was 1a b21a b2 a 2 b 2
11 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring the Difference of Two Squares 251 Here is another version of that pattern: Difference of Two Squares a 2 b 2 (a b)(a b) To apply the differenceoftwosquares pattern is a fairly simple process, as these next examples illustrate. The steps inside the box are often performed mentally. x x x 2 16y 2 64 y 2 1x x x2 2 14y y2 2 1x 621x 62 12x 5212x 52 13x 4y213x 4y2 18 y218 y2 Because multiplication is commutative, the order of writing the factors is not important. For example, (x 6)(x 6) can also be written as (x 6)(x 6). Remark: You must be careful not to assume an analogous factoring pattern for the sum of two squares; it does not exist. For example, x 2 4 (x 2)(x 2) because (x 2)(x 2) x 2 4x 4. We say that the sum of two squares is not factorable using integers. The phrase using integers is necessary because x 2 4 could be 1 written as, but such factoring is of no help. Furthermore, we do not 2 12x2 82 consider 1121x 2 42 as factoring x 2 4. It is possible that both the technique of factoring out a common monomial factor and the difference of two squares pattern can be applied to the same polynomial. In general, it is best to look for a common monomial factor first. EXAMPLE 1 Factor 2x x x x 521x 52 Common factor of 2 Difference of squares Factor 3y (y 2)(y 2) In Example 1, by expressing 2x 2 50 as 2(x 5)(x 5), we say that it has been factored completely. That means the factors 2, x 5, and x 5 cannot be factored any further using integers. EXAMPLE 2 Factor completely 18y 3 8y. 18y 3 8y 2y19y y13y 2213y 22 Common factor of 2y Difference of squares Factor completely 32x 3 2x. 2x(4x 1)(4x 1) Sometimes it is possible to apply the differenceoftwosquares pattern more than once. Consider the next example.
12 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 3 Factor completely x x x 2 421x x 2 421x 221x 22 Factor completely a (a 2 9)(a 3)(a 3) The following examples should help you to summarize the factoring ideas presented thus far. 5x x y 2 15 y215 y2 3 3x x x211 x2 36x 2 49y 2 16x 7y216x 7y2 a 2 9 9x 17y is not factorable using integers is not factorable using integers 2 Solve Equations by Factoring the Difference of Two Squares Each time we learn a new factoring technique, we also develop more power for solving equations. Let s consider how we can use the differenceofsquares factoring pattern to help solve certain kinds of equations. EXAMPLE 4 Solve x x 2 25 x x 521x 52 0 x 5 0 or x 5 0 x 5 or x 5 The solution set is { 5, 5}. Check these answers! Remember: ab 0 if and only if a 0 or b 0 Solve x { 8, 8} EXAMPLE 5 Solve 9x x x
13 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring the Difference of Two Squares x 5213x x 5 0 or 3x 5 0 3x 5 or 3x 5 x 5 3 or x 5 3 The solution set is e 5. 3, 5 3 f Solve 4x 2 9. e 3 2, 3 2 f EXAMPLE 6 Solve 5y y y y 2 4 y y 221y 22 0 y 2 0 or y 2 0 y 2 or y 2 The solution set is { 2, 2}. Check it! Divide both sides by 5 Solve 3x { 5, 5} EXAMPLE 7 Solve x 3 9x 0. x 3 9x 0 x1x x1x 321x 32 0 x 0 or x 3 0 or x 3 0 x 0 or x 3 or x 3 The solution set is { 3, 0, 3}. Solve x 3 49x 0. { 7, 0, 7} The more we know about solving equations, the more easily we can solve word problems.
14 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving 3 Solve Word Problems Using Factoring EXAMPLE 8 Apply Your Skill The combined area of two squares is 20 square centimeters. Each side of one square is twice as long as a side of the other square. Find the lengths of the sides of each square. We can sketch two squares and label the sides of the smaller square s (see Figure 5.2). Then the sides of the larger square are 2s. The sum of the areas of the two squares is 20 square centimeters, so we set up and solve the following equation: s 2 12s s 2 4s s 2 20 s 2 4 s s 221s 22 0 s 2 0 or s 2 0 s 2 or s 2 Because s represents the length of a side of a square, we must disregard the solution 2. Thus one square has sides of length 2 centimeters, and the other square has sides of length 2(2) 4 centimeters. The combined area of two squares is 250 square feet. Each side of one square is three times as long as a side of the other square. Find the lengths of the sides of each square. 5 feet and 15 feet s s Figure 5.2 2s 2s CONCEPT QUIZ For Problems 1 8, answer true or false. 1. A binomial that has two perfect square terms that are subtracted is called the difference of two squares. 2. The sum of two squares is factorable using integers. 3. When factoring it is usually best to look for a common factor first. 4. The polynomial 4x 2 y 2 factors into (2x y)(2x y). 5. The completely factored form of y 4 81 is (y 2 9)(y 2 9). 6. The solution set for x 2 16 is { 4}. 7. The solution set for 5x 3 5x 0 is { 1, 0, 1}. 8. The solution set for x 4 9x 2 0 is { 3, 0, 3}. Problem Set Factor the Difference of Two Squares For Problems 1 12, use the differenceofsquares pattern to factor each polynomial. 1. x 2 1 1x 121x x x 1021x 102 x 2 25 x x 1121x 112 1x 521x x 2 4y 2 6. x 2 36y 2 1x 2y21x 2y2 1x 6y21x 6y2 7. 9x 2 y y 2 64x 2 13x y213x y2 17y 8x217y 8x a 2 25b a 2 81b 2 16a 5b216a 5b2 12a 9b212a 9b n n n211 2n2 12 3n212 3n2 Blue problem numbers indicate Enhanced WebAssign Problems.
15 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring the Difference of Two Squares 255 For Problems 13 44, factor each polynomial completely. Indicate any that are not factorable using integers. Don t forget to look for a common monomial factor first x x 221x x x x 2 18y x 3y21x 3y2 19. x 3 25x x1x 521x x 2 9y 2 Not factorable x 2 36xy 9x15x 4y x x213 x a 4 16a 2 4a 2 1a x 2 25y 2 Not factorable 75 3x x215 x2 x214 x x x 4 1x 321x 321x x x 4 x 2 x 2 1x x 5 2x 3 x 3 1x x 3 48x 3x1x x 20x x11 2x211 2x x x 421x x x x 3 24x x 3 y 12xy x 3 y 18xy 3 3xy15x 2y215x 2y2 2xy14x 3y214x 3y x 4 81y x xy 3y212x 3y214x 2 9y 2 2 1x 121x 121x x x 4 16y 4 13 x213 x219 x x 2y213x 2y219x 2 4y Solve Equations by Factoring the Difference of Two Squares For Problems 45 68, solve each equation. 9a 4 81a 2 71x 121x x x 2 32y 2 81x 2y21x 2y2 2x 3 2x 2x1x 121x 12 18x 42y 613x 7y2 9a 2 1a x1x x 36x 3 4x11 3x211 3x2 9x x 121x e x2 0 3, 1 45x2 0 3 f 63. 4x 3 400x { 10, 0, 10} x2 81 e 9 8, 9 8 f e 68. 2, 0, 1 36x3 9x 2 f 2x 3 98x 0 5 7, 0, 76 e 5 9, 5 81x f 64x3 4x 3 Solve Word Problems Using Factoring 1 e 2, 1 2 f 1 e 4, 0, 1 4 f For Problems set up an equation and solve the problem. 69. Fortynine less than the square of a number equals zero. Find the number. 7 or The cube of a number equals nine times the number. Find the number. 3, 0, or Suppose that five times the cube of a number equals 80 times the number. Find the number. 4, 0, or Ten times the square of a number equals 40. Find the number. 2 or The sum of the areas of two squares is 234 square inches. Each side of the larger square is five times the length of a side of the smaller square. Find the length of a side of each square. 3 inches and 15 inches 74. The difference of the areas of two squares is 75 square feet. Each side of the larger square is twice the length of a side of the smaller square. Find the length of a side of each square. 5 feet and 10 feet 75. Suppose that the length of a certain rectangle is times its width, and the area of that same rectangle is 160 square centimeters. Find the length and width of the rectangle. 20 centimeters and 8 centimeters 76. Suppose that the width of a certain rectangle is threefourths of its length and that the area of this same rectangle is 108 square meters. Find the length and width of the rectangle. 12 meters and 9 meters 45. x , n 2 5 2, x , n , The sum of the areas of two circles is 80p square meters. Find the length of a radius of each circle if one of them is twice as long as the other. 4 meters and 8 meters 49. e e 3 2, 3 4x2 9 3, 4 9x f 2 f 51. n , n , e 54. e 6 7, 6 49x2 36 5, 2 25x2 4 5 f 7 f 55. 3x , x , x 3 48x 0 5 4, 0, x 3 x 0 5 1, 0, n 3 16n 5 4, 0, n 3 8n 5 2, 0, The area of a triangle is 98 square feet. If one side of the triangle and the altitude to that side are of equal length, find the length. 14 feet 79. The total surface area of a right circular cylinder is 100p square centimeters. If a radius of the base and the altitude of the cylinder are the same length, find the length of a radius. 5 centimeters 80. The total surface area of a right circular cone is 192p square feet. If the slant height of the cone is equal in length to a diameter of the base, find the length of a radius. 8 feet
16 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving THOUGHTS INTO WORDS 81. How do we know that the equation x has no solutions in the set of real numbers? 82. Why is the following factoring process incomplete? 16x x 8214x 82 How should the factoring be done? 83. Consider the following solution: 4x x x 321x or x 3 0 or x or x 3 or x 3 The solution set is { 3, 3}. Is this a correct solution? Do you have any suggestion to offer the person who worked on this problem? FURTHER INVESTIGATIONS The following patterns can be used to factor the sum of two cubes and the difference of two cubes, respectively. a 3 b 3 1a b21a 2 ab b 2 2 a 3 b 3 1a b21a 2 ab b 2 2 Consider these examples: x 3 8 1x x 221x 2 2x 42 x 3 1 1x x 121x 2 x 12 Use the sumoftwocubes and the differenceoftwocubes patterns to factor each polynomial. 84. x x 121x 2 x 12 x 3 8 1x 221x 2 2x n n n 321n 2 3n 92 1n 421n 2 4n x 3 27y a 3 64b 3 12x 3y214x 2 6xy 9y a 4b219a 2 12ab 16b x a x211 2x 4x a211 3a 9a x 3 8y x 3 y 3 1x 2y21x 2 2xy 4y x y214x 2 2xy y a 3 b x 3 8y 3 1ab 121a 2 b 2 ab 12 13x 2y219x 2 6xy 4y n x 3 8y 3 12 n214 2n n x 2y2125x 2 10xy 4y n x 3 13n 5219n 2 15n x2116 4x x 2 2 Answers to the Example Practice Skills 1. 3(y 2)(y 2) 2. 2x(4x 1)(4x 1) 3. (a 2 9)(a 3)(a 3) 4. { 8, 8} 5. e 3 6. { 5, 5} 2, 3 2 f 7. { 7, 0, 7} 8. 5 feet and 15 feet Answers to the Concept Quiz 1. True 2. False 3. True 4. False 5. False 6. False 7. True 8. True 5.3 Factoring Trinomials of the Form x 2 bx c OBJECTIVES 1 Factor Trinomials of the Form x 2 bx c 2 Use Factoring of Trinomials to Solve Equations 3 Solve Word Problems Including Consecutive Number Problems 4 Use the Pythagorean Theorem to Solve Problems 1 Factor Trinomials of the Form x 2 bx c One of the most common types of factoring used in algebra is to express a trinomial as the product of two binomials. In this section, we will consider trinomials where the coefficient of the squared term is 1, that is, trinomials of the form x 2 bx c.
17 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring Trinomials of the Form x 2 bx c 257 Again, to develop a factoring technique we first look at some multiplication ideas. Consider the product (x r)(x s), and use the distributive property to show how each term of the resulting trinomial is formed. 1x r21x s2 x1x2 x1s2 r1x2 r1s x 2 (s r)x rs Note that the coefficient of the middle term is the sum of r and s and that the last term is the product of r and s. These two relationships are used in the next examples. EXAMPLE 1 Factor x 2 7x 12. We need to fill in the blanks with two numbers whose sum is 7 and whose product is 12. x 2 7x 12 (x )(x ) This can be done by setting up a table showing possible numbers. Product Sum 1(12) (6) (4) The bottom line contains the numbers that we need. Thus x 2 7x 12 1x 321x 42 Factor y 2 9y 20. (y 5)(y 4) EXAMPLE 2 Factor x 2 11x 24. We need two numbers whose product is 24 and whose sum is 11. Product Sum The third line contains the numbers that we want. Thus x 2 11x 24 1x 321x 82 Factor m 2 8m 15. (m 3)(m 5)
18 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 3 Factor x 2 3x 10. We need two numbers whose product is 10 and whose sum is 3. Product Sum The bottom line is the key line. Thus x 2 3x 10 1x 521x 22 Factor y 2 5y 24. (y 8)(y 3) EXAMPLE 4 Factor x 2 2x 8. We need two numbers whose product is 8 and whose sum is 2. Product Sum The third line has the information we want. x 2 2x 8 1x 421x 22 Factor y 2 4y 12. (y 2)(y 6) The tables in the last four examples illustrate one way of organizing your thoughts for such problems. We showed complete tables; that is, for Example 4, we included the last line even though the desired numbers were obtained in the third line. If you use such tables, keep in mind that as soon as you get the desired numbers, the table need not be continued beyond that point. Furthermore, there will be times that you will be able to find the numbers without using a table. The key ideas are the product and sum relationships. EXAMPLE 5 Factor x 2 13x 12. Product ( 1)( 12) 12 Sum ( 1) ( 12) 13
19 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring Trinomials of the Form x 2 bx c 259 We need not complete the table. x 2 13x 12 1x 121x 122 Factor y 2 7y 6. (y 1)(y 6) In the next example, we refer to the concept of absolute value. Recall that the absolute value is the number without regard for the sign. For example, and EXAMPLE 6 Factor x 2 x 56. Note that the coefficient of the middle term is 1. Therefore, we are looking for two numbers whose product is 56; their sum is 1, so the absolute value of the negative number must be 1 larger than the absolute value of the positive number. The numbers are 8 and 7, and we have x 2 x 56 1x 821x 72 Factor a 2 a 12. (a 4)(a 3) EXAMPLE 7 Factor x 2 10x 12. Product Sum Because the table is complete and no two factors of 12 produce a sum of 10, we conclude that x 2 10x 12 is not factorable using integers. Factor y 2 7y 18. Not factorable In a problem such as Example 7, we need to be sure that we have tried all possibilities before we conclude that the trinomial is not factorable. 2 Use Factoring of Trinomials to Solve Equations The property ab 0 if and only if a 0 or b 0 continues to play an important role as we solve equations that involve the factoring ideas of this section. Consider the following examples.
20 05W4801AM1.qxd 8/19/08 8:45 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 8 Solve x 2 8x x 2 8x x 321x 52 0 Factor the left side x 3 0 or x 5 0 Use ab 0 if and only if a 0 or b 0 x 3 or x 5 The solution set is { 5, 3}. Solve y 2 14y {2, 12} EXAMPLE 9 Solve x 2 5x 6 0. x 2 5x 6 0 1x 621x 12 0 x 6 0 or x 1 0 x 6 or x 1 The solution set is { 6, 1}. Solve a 2 3a { 7, 4} EXAMPLE 10 Solve y 2 4y 45. y 2 4y 45 y 2 4y y 921y 52 0 y 9 0 or y 5 0 y 9 or y 5 The solution set is { 5, 9}. Solve x 2 18x 40. { 2, 20} Don t forget that we can always check to be absolutely sure of our solutions. Let s check the solutions for Example 10. If y 9, then y 2 4y 45 becomes
21 05W4801AM1.qxd 8/19/08 8:45 PM Page Factoring Trinomials of the Form x 2 bx c 261 If y 5, then y 2 4y 45 becomes Solve Word Problems Including Consecutive Number Problems The more we know about factoring and solving equations, the more easily we can solve word problems. EXAMPLE 11 Apply Your Skill Find two consecutive integers whose product is 72. Let n represent one integer. Then n 1 represents the next integer. n1n n 2 n 72 n 2 n n 921n 82 0 n 9 0 or n 8 0 n 9 or n 8 The product of the two integers is 72 If n 9, then n If n 8, then n Thus the consecutive integers are 9 and 8 or 8 and 9. Find two consecutive integers whose product is , 10 or 10, 11 EXAMPLE 12 Apply Your Skill A rectangular plot is 6 meters longer than it is wide. The area of the plot is 16 square meters. Find the length and width of the plot. We let w represent the width of the plot, and then w 6 represents the length (see Figure 5.3). w Figure 5.3 w + 6
22 05W4801AM1.qxd 8/19/08 8:46 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving Using the area formula A lw, we obtain w1w w 2 6w 16 w 2 6w w 821w 22 0 w 8 0 or w 2 0 w 8 or w 2 The solution 8 is not possible for the width of a rectangle, so the plot is 2 meters wide, and its length (w 6) is 8 meters. A rectangular plot is 2 yards longer than it is wide. The area of the plot is 80 square yards. Find the width and length of the plot. 8 yards, 10 yards 4 Use the Pythagorean Theorem to Solve Problems The Pythagorean theorem, an important theorem pertaining to right triangles, can also serve as a guideline for solving certain types of problems. The Pythagorean theorem states that in any right triangle, the square of the longest side (called the hypotenuse) is equal to the sum of the squares of the other two sides (called legs); see Figure 5.4. We can use this theorem to help solve a problem. a 2 + b 2 = c 2 c b a Figure 5.4 EXAMPLE 13 Apply Your Skill Suppose that the lengths of the three sides of a right triangle are consecutive whole numbers. Find the lengths of the three sides. Let s represent the length of the shortest leg. Then s 1 represents the length of the other leg, and s 2 represents the length of the hypotenuse. Using the Pythagorean theorem as a guideline, we obtain the following equation: Sum of squares of two legs Square of hypotenuse s 2 1s s 22 2 Solving this equation yields s 2 s 2 2s 1 s 2 4s 4 2s 2 2s 1 s 2 4s 4
23 05W4801AM1.qxd 8/19/08 8:46 PM Page Factoring Trinomials of the Form x 2 bx c 263 s 2 2s 1 4s 4 s 2 2s 1 4 s 2 2s 3 0 1s 321s 12 0 s 3 0 or s 1 0 s 3 or s 1 Add s 2 to both sides Add 4s to both sides Add 4 to both sides The solution of 1 is not possible for the length of a side, so the shortest side (s) is of length 3. The other two sides (s 1 and s 2) have lengths of 4 and 5. The length of one leg of a right triangle is 7 inches more then the length of the other leg. The length of the hypotenuse is 13 inches. Find the length of the two legs. 5 inches and 12 inches CONCEPT QUIZ For Problems 1 10, answer true or false. 1. Any trinomial of the form x 2 bx c can be factored (using integers) into the product of two binomials. 2. To factor x 2 4x 60 we look for two numbers whose product is 60 and whose sum is A trinomial of the form x 2 bx c will never have a common factor other than If n represents an odd integer, then n 1 represents the next consecutive odd integer. 5. The Pythagorean theorem only applies to right triangles. 6. In a right triangle the longest side is called the hypotenuse. 7. The polynomial x 2 25x 72 is not factorable. 8. The polynomial x 2 27x 72 is not factorable. 9. The solution set of the equation x 2 2x 63 0 is { 9, 7}. 10. The solution set of the equation x 2 5x 66 0 is { 11, 6}. Problem Set Factor Trinomials of the Form x 2 bx c For Problems 1 30, factor each trinomial completely. Indicate any that are not factorable using integers. 1. x 2 10x x 2 9x 14 1x 421x 62 1x 221x x 2 13x x 2 11x 24 1x 521x 82 1x 321x x 2 11x x 2 5x 4 1x 221x 92 1x 121x n 2 11n n 2 7n 10 1n 721n 42 1n 221n n 2 6n n 2 3n 18 1n 921n 32 1n 321n n 2 6n n 2 4n 45 1n 1021n 42 1n 521n t 2 12t t 2 20t 96 Not factorable 1t 821t x 2 18x x 2 14x 32 1x 621x 122 Not factorable 17. x 2 5x x 2 11x 42 1x 1121x 62 1x 321x y 2 y y 2 y 30 1y 921y 82 1y 521y x 2 21x x 2 21x 90 1x 521x 162 1x 621x x 2 6x x 2 8x 36 1x 1221x 62 Not factorable 25. x 2 10x x2 12x x 2 3xy 10y x2 4xy 12y2 Not factorable 1x 421x 162 1x 2y21x 5y2 1x 2y21x 6y2 29. a 2 4ab 32b a 2 3ab 54b 2 1a 8b21a 4b2 1a 6b21a 9b2 2 Use Factoring of Trinomials to Solve Equations For Problems 31 50, solve each equation. 31. x 2 10x x 2 9x , , x 2 9x x 2 9x , 66 51, 86 Blue problem numbers indicate Enhanced WebAssign Problems.
24 05W4801AM1.qxd 8/19/08 8:46 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving 35. x 2 3x , n 2 5n , n 2 6n , t 2 t , x 2 16x 28 0 {2, 14} 44. x 2 18x 45 0 {3, 15} 45. x 2 11x , x1x , x 2 2x 24 0 { 6, 4} 50. x 2 6x 16 0 { 2, 8} x 2 x , 46 n 2 3n , 36 n 2 8n , 126 t 2 t , 86 x 2 8x 20 x1x , , Suppose that the width of a certain rectangle is 3 inches less than its length. The area is numerically 6 less than twice the perimeter. Find the length and width of the rectangle. 9 inches by 6 inches 60. The sum of the areas of a square and a rectangle is 64 square centimeters. The length of the rectangle is 4 centimeters more than a side of the square, and the width of the rectangle is 2 centimeters more than a side of the square. Find the dimensions of the rectangle and the square. 8 centimeters by 6 centimeters and 4 centimeters by 4 centimeters 61. The perimeter of a rectangle is 30 centimeters, and the area is 54 square centimeters. Find the width and length of the rectangle. [Hint: Let w represent the width; then 15 w represents the length.] 6 centimeters by 9 centimeters 62. The perimeter of a rectangle is 44 inches, and its area is 120 square inches. Find the width and length of the rectangle. 10 inches by 12 inches 3 Solve Word Problems Including Consecutive Number Problems For Problems 51 64, set up an equation and solve each problem. 51. Find two consecutive integers whose product is and 7 or 7 and Find two consecutive odd whole numbers whose product is and Find two consecutive even whole numbers whose product is and One number is 2 larger than another number. The sum of their squares is 100. Find the numbers. 8 and 6, or 6 and Find four consecutive integers such that the product of the two larger integers is 22 less than twice the product of the two smaller integers. 4, 3, 2, 1 or 7, 8, 9, Find three consecutive integers such that the product of the two smaller integers is 2 more than 10 times the largest integer. 11, 12, and 13 or 2, 1, and One number is 3 smaller than another number. The square of the larger number is 9 larger than 10 times the smaller number. Find the numbers. 4 and 7 or 0 and The area of the floor of a rectangular room is 84 square feet. The length of the room is 5 feet more than its width. Find the width and length of the room. 7 feet and 12 feet 63. An apple orchard contains 84 trees. The number of trees per row is 5 more than the number of rows. Find the number of rows. 7 rows 64. A room contains 54 chairs. The number of rows is 3 less than the number of chairs per row. Find the number of rows. 6 rows of chairs 4 Use the Pythagorean Theorem to Solve Problems For Problems 65 68, set up an equation and solve for each problem. 65. Suppose that one leg of a right triangle is 7 feet shorter than the other leg. The hypotenuse is 2 feet longer than the longer leg. Find the lengths of all three sides of the right triangle. 8 feet, 15 feet, and 17 feet 66. Suppose that one leg of a right triangle is 7 meters longer than the other leg. The hypotenuse is 1 meter longer than the longer leg. Find the lengths of all three sides of the right triangle. 5 meters, 12 meters, and 13 meters 67. Suppose that the length of one leg of a right triangle is 2 inches less than the length of the other leg. If the length of the hypotenuse is 10 inches, find the length of each leg. 6 inches and 8 inches 68. The length of one leg of a right triangle is 3 centimeters more than the length of the other leg. The length of the hypotenuse is 15 centimeters. Find the lengths of the two legs. 9 centimeters and 12 centimeters THOUGHTS INTO WORDS 69. What does the expression not factorable using integers mean to you? 70. Discuss the role that factoring plays in solving equations. 71. Explain how you would solve the equation (x 3) (x 4) 0, and also how you would solve (x 3) (x 4) 8.
25 05W4801AM1.qxd 8/19/08 8:46 PM Page Factoring Trinomials of the Form ax 2 bx c 265 FURTHER INVESTIGATIONS For Problems 72 75, factor each trinomial and assume that all variables appearing as exponents represent positive integers. 72. x 2a 10x a x 2a 13x a 40 1x a 421x a 62 1x a 821x a x 2a 2x a x 2a 6x a 27 1x a 221x a 42 1x a 921x a Suppose that we want to factor n 2 26n 168 so that we can solve the equation n 2 26n We need to find two positive integers whose product is 168 and whose sum is 26. Because the constant term, 168, is rather large, let s look at it in prime factored form: # 2 # 2 # 3 # 7 Now we can mentally form two numbers by using all of these factors in different combinations. Using two 2s and the 3 in one number and the other 2 and the 7 in another number produces 2 # 2 # 3 12 and 2 # Therefore, we can solve the given equation as follows: n 2 26n n 1221n n 12 0 or n 14 0 n 12 or n 14 The solution set is { 14, 12}. Solve each of the following equations. a. n 2 30n { 18, 12} b. n 2 35n { 21, 14} c. n 2 40n {16, 24} d. n 2 40n {15, 25} e. n 2 6n { 24, 18} f. n 2 16n { 16, 32} Answers to the Example Practice Skills 1. (y 5)(y 4) 2. (m 3)(m 5) 3. (y 8)(y 3) 4. (y 2)(y 6) 5. (y 1)(y 6) 6. (a 4)(a 3) 7. Not factorable 8. {2, 12} 9. { 7, 4} 10. { 2, 20} , 10 or 10, yards, 10 yards inches and 12 inches Answers to the Concept Quiz 1. False 2. True 3. True 4. False 5. True 6. True 7. True 8. False 9. True 10. False 5.4 Factoring Trinomials of the Form ax 2 bx c OBJECTIVES 1 Factor Trinomials Where the Leading Coefficient Is Not 1 2 Solve Equations That Involve Factoring 1 Factor Trinomials Where the Leading Coefficient Is Not 1 Now let s consider factoring trinomials where the coefficient of the squared term is not 1. We first illustrate an informal trialanderror technique that works well for certain types of trinomials. This technique relies on our knowledge of multiplication of binomials. EXAMPLE 1 Factor 2x 2 7x 3. By looking at the first term, 2x 2, and the positive signs of the other two terms, we know that the binomials are of the form (2x )(x )
26 05W4801AM1.qxd 8/19/08 8:46 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving Because the factors of the constant term 3 are 1 and 3, we have only two possibilities to try: (2x 3)(x 1) or (2x 1)(x 3) By checking the middle term of both of these products, we find that the second one yields the correct middle term of 7x. Therefore, 2x 2 7x 3 12x 121x 32 Factor 3y 2 16y 5. (3y 1)(y 5) EXAMPLE 2 Factor 6x 2 17x 5. First, we note that 6x 2 can be written as 2x # 3x or 6x # x. Second, because the middle term of the trinomial is negative, and the last term is positive, we know that the binomials are of the form (2x )(3x ) or (6x )(x ) The factors of the constant term 5 are 1 and 5, so we have the following possibilities: 12x 5213x 12 12x 1213x 52 16x 521x 12 16x 121x 52 By checking the middle term for each of these products, we find that the product (2x 5)(3x 1) produces the desired term of 17x. Therefore, 6x 2 17x 5 12x 5213x 12 Factor 4a 2 8a 3. (2a 1)(2a 3) EXAMPLE 3 Factor 8x 2 8x 30. First, we note that the polynomial 8x 2 8x 30 has a common factor of 2. Factoring out the common factor gives us 2(4x 2 4x 15). Now we need to factor 4x 2 4x 15. Now, we note that 4x 2 can be written as 4x # x or 2x # 2x. Also, the last term, 15, can be written as (1)( 15), ( 1)(15), (3)( 5), or ( 3)(5). Thus we can generate the possibilities for the binomial factors as follows: Using 1 and 15 Using 1 and 15 (4x 15)(x 1) (4x 1)(x 15) (4x 1)(x 15) (4x 15)(x 1) (2x 1)(2x 15) (2x 1)(2x 15)
27 05W4801AM1.qxd 8/19/08 8:46 PM Page Factoring Trinomials of the Form ax 2 bx c 267 Using 3 and 5 Using 3 and 5 (4x 3)(x 5) (4x 3)(x 5) (4x 5)(x 3) (4x 5)(x 3) (2x 5)(2x 3) (2x 5)(2x 3) By checking the middle term of each of these products, we find that the product indicated with a check mark produces the desired middle term of 4x. Therefore, 8x 2 8x x 5212x 32 Factor 6y 2 27y 30. 3(y 2)(2y 5) Let s pause for a moment and look back over Examples 1, 2, and 3. Example 3 clearly created the most difficulty because we had to consider so many possibilities. We have suggested one possible format for considering the possibilities, but as you practice such problems, you may develop a format of your own that works better for you. Whatever format you use, the key idea is to organize your work so that you consider all possibilities. Let s look at another example. EXAMPLE 4 Factor 4x 2 6x 9. First, we note that 4x 2 can be written as 4x # x or 2x # 2x. Second, because the middle term is positive and the last term is positive, we know that the binomials are of the form (4x )(x ) or (2x )(2x ) Because 9 can be written as 9 # 1 or 3 # 3, we have only the following five possibilities to try: 14x 921x 12 14x 321x 32 12x 3212x 32 14x 121x 92 12x 1212x 92 When we try all of these possibilities, we find that none of them yields a middle term of 6x. Therefore, 4x 2 6x 9 is not factorable using integers. Factor 6a 2 3a 7. Not factorable Remark: Example 4 illustrates the importance of organizing your work so that you try all possibilities before you conclude that a particular trinomial is not factorable. 2 Solve Equations That Involve Factoring The ability to factor certain trinomials of the form ax 2 bx c provides us with greater equationsolving capabilities. Consider the next examples.
28 05W4801AM1.qxd 8/19/08 8:46 PM Page Chapter 5 Factoring, Solving Equations, and Problem Solving EXAMPLE 5 Solve 3x 2 17x x 2 17x x 5213x 22 0 x 5 0 or 3x 2 0 x 5 or 3x 2 Factoring 3x 2 17x 10 as (x 5)(3x 2) may require some extra work on scratch paper ab 0 if and only if a 0 or b 0 x 5 or x 2 3 The solution set is e 5, 2. Check it! 3 f Solve 2y 2 7y 6 0. e 2, 3 2 f EXAMPLE 6 Solve 24x 2 2x x 2 2x x 3216x x 3 0 or 6x 5 0 4x 3 or 6x 5 x 3 4 or x 5 6 The solution set is e 5. 6, 3 4 f Solve 2a 2 5a e 3 2, 4 f CONCEPT QUIZ For Problems 1 8, answer true or false. 1. Any trinomial of the form ax 2 bx c can be factored (using integers) into the product of two binomials. 2. To factor 2x 2 x 3, we look for two numbers whose product is 3 and whose sum is A trinomial of the form ax 2 bx c will never have a common factor other than The factored form (x 3)(2x 4) is factored completely. 5. The differenceofsquares polynomial 9x 2 25 could be written as the trinomial 9x 2 0x The polynomial 12x 2 11x 12 is not factorable. 7. The solution set of the equation 6x 2 13x 5 0 is e 1. 3, 2 5 f 8. The solution set of the equation 18x 2 39x 20 0 is e 5. 6, 4 3 f
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