6.5 Factoring Special Forms

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1 440 CHAPTER 6. FACTORING 6.5 Factoring Special Forms In this section we revisit two special product forms that we learned in Chapter 5, the first of which was squaring a binomial. Squaring a binomial. Here are two earlier rules for squaring a binomial. 1. (a+b) 2 = a 2 +2ab+b 2 2. (a b) 2 = a 2 2ab+b 2 Perfect Square Trinomials To square a binomial such as (a+b) 2, proceed as follows: 1. Square the first term: a 2 2. Multiply the first and second term, then double: 2ab 3. Square the last term: b 2 EXAMPLE 1. Expand: (2x+3y) 2 Expand: (5a+2b) 2 Solution: Using the pattern (a+b) 2 = a 2 +2ab+b 2, we can expand (2x+3y) 2 as follows: (2x+3y) 2 = (2x) 2 +2(2x)(3y)+(3y) 2 = 4x 2 +6xy +9y 2 Answer: 25a 2 +20ab+4b 2 Note how we square the first and second terms, then produce the middle term of our answer by multiplying the first and second terms and doubling. EXAMPLE 2. Expand: (3u 2 5v 2 ) 2 Expand: (2s 3 7t) 2 Solution: Usingthepattern(a b) 2 = a 2 2ab+b 2, wecanexpand(3u 2 5v 2 ) 2 as follows: (3u 2 5v 2 ) 2 = (3u 2 ) 2 2(3u 2 )(5v 2 )+(5v 2 ) 2 = 9u 4 30u 2 v 2 +25v 4

2 Answer: 4s 6 28s 3 t+49t FACTORING SPECIAL FORMS 441 Note that the sign of the middle term is negative this time. The first and last terms are still positive because we are squaring. Once you ve squared a few binomials, it s time to do all of the work in your head. (i) Square the first term; (ii) multiply the first and second term and double the result; and (iii) square the second term. EXAMPLE 3. Expand each of the following: Expand: (5x 4 3) 2 a) (2y 3) 2 b) (4a 3b) 2 c) (x 3 +5) 2 Solution: Using the pattern (a ± b) 2 = a 2 ± 2ab + b 2, we expand each binomially mentally, writing down the answer without any intermediate steps. a) (2y 3) 2 = 4y 2 12y+9 b) (4a 3b) 2 = 16a 2 24ab+9b 2 c) (x 3 +5) 2 = x 6 +10x Answer: 25x 8 30x 4 +9 Now, because factoring is unmultiplying, it should be a simple matter to reverse the process of Example 3. EXAMPLE 4. Factor each of the following trinomials: Factor: 25x 8 30x 4 +9 a) 4y 2 12y+9 b) 16a 2 24ab+9b 2 c) x 6 +10x Solution: Because of the work already done in Example 3, it is a simple task to factor each of these trinomials. a) 4y 2 12y+9 = (2y 3) 2 b) 16a 2 24ab+9b 2 = (4a 3b) 2 c) x 6 +10x = (x 3 +5) 2 Answer: (5x 4 3) 2

3 442 CHAPTER 6. FACTORING Eachofthe trinomials in Example 4is an exampleofaperfect square trinomial. Perfect square trinomial. If a trinomial a 2 + 2ab + b 2 is the square of a binomial, as in (a+b) 2, then the trinomial is called a perfect square trinomial. So, how does one recognize a perfect square trinomial? If the first and last terms of a trinomial are perfect squares, then you should suspect that you may be dealing with a perfect square trinomial. However, you also have to have the correct middle term in order to have a perfect square trinomial. Factor: 16x 2 +72x+81 List of Squares n n EXAMPLE 5. Factor each of the following trinomials: a) 9x 2 42x+49 b) 49a 2 +70ab+25b 2 c) 4x 2 37x+9 Solution: Note that the first and last terms of each trinomial are perfect squares. a) In the trinomial 9x 2 42x+49, note that (3x) 2 = 9x 2 and 7 2 = 49. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 9x 2 42x+49 factors as follows: 9x 2 42x+49? = (3x 7) 2 However, we must check to see if the middle term is correct. Multiply 3x and 7, then double: 2(3x)(7) = 42x. Thus, the middle term is correct and therefore 9x 2 42x+49 = (3x 7) 2. b) Inthetrinomial49a 2 +70ab+25b 2, notethat(7a) 2 = 49a 2 and(5b) 2 = 25b 2. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 49a 2 +70ab+25b 2 factors as follows: 49a 2 +70ab+25b 2? = (7a+5b) 2 However, we must check to see if the middle term is correct. Multiply 7a and 5b, then double: 2(7a)(5b) = 70ab. Thus, the middle term is correct and therefore 49a 2 +70ab+25b 2 = (7a+5b) 2. c) In the trinomial 4x 2 37x+9, note that (2x) 2 = 4x 2 and (3) 2 = 9. Hence, the first and last terms are perfect squares. Taking the square roots, we suspect that 4x 2 37x+9 factors as follows: 4x 2 37x+9? = (2x 3) 2

4 6.5. FACTORING SPECIAL FORMS 443 However, we must check to see if the middle term is correct. Multiply 2x and 3, then double: 2(2x)(3) = 12x. However, this is not the middle term of 4x 2 37x+9, so this factorization is incorrect! We must find another way to factor this trinomial. Comparing 4x 2 37x+9 with ax 2 +bx+c, we need a pair of integers whose product is ac = 36 and whose sum is b = 37. The integer pair 1 and 36 comes to mind. Replace the middle term as a sum of like terms using this ordered pair. 4x 2 37x+9 = 4x 2 x 36x+9 37x = x 36x. = x(4x 1) 9(4x 1) Factor by grouping. = (x 9)(4x 1) Factor out 4x 1. This example clearly demonstrates how important it is to check the middle term. Answer: (4x+9) 2 Remember the first rule of factoring! The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF. EXAMPLE 6. Factor each of the following trinomials: Factor: 4x 3 24x 2 36x a) 2x 3 y +12x 2 y 2 +18xy 3 b) 4x 5 +32x 4 64x 3 Solution: Remember, first factor out the GCF. a) In the trinomial 2x 3 y + 12x 2 y xy 3, we note that the GCF of 2x 3 y, 12x 2 y 2, and 18xy 3 is 2xy. We first factor out 2xy. 2x 3 y +12x 2 y 2 +18xy 3 = 2xy(x 2 +6xy +9y 2 ) We now note that the first and last terms of the resulting trinomial factor are perfect squares, so we take their square roots and factors as follows. = 2xy(x+3y) 2 Of course, the last factorization is correct only if the middle term is correct. Because 2(x)(3y) = 6xy matches the middle term of x 2 +6xy+9y 2, we do have a perfect square trinomial and our result is correct.

5 444 CHAPTER 6. FACTORING b) In the trinomial 4x 5 +32x 4 64x 3, we note that the GCF of 4x 5, 32x 4, and 64x 3 is 4x 3. We first factor out 4x 3. 4x 5 +32x 4 64x 3 = 4x 3 ( x 2 +8x 16) However, the first and third terms of x 2 +8x 16 are negative, and thus are not perfect squares. Let s begin again, this time factoring out 4x 3. 4x 5 +32x 4 64x 3 = 4x 3 (x 2 8x+16) This time the first and third terms of x 2 8x+16 are perfect squares. We take their square roots and write: = 4x 3 (x 4) 2 Again, this last factorization is correct only if the middle term is correct. Because 2(x)(4) = 8x, we do have a perfect square trinomial and our result is correct. Answer: 4x(x+3) 2 The Difference of Squares The second special product form we learned in Chapter 5 was the difference of squares. The difference of squares. Here is the difference of squares rule. (a+b)(a b) = a 2 b 2 If you are multiplying two binomials which have the exact same terms in the First positions and the exact same terms in the Last positions, but one set is separated by a plus sign while the other set is separated by a minus sign, then multiply as follows: 1. Square the first term: a 2 2. Square the second term: b 2 3. Place a minus sign between the two squares.

6 6.5. FACTORING SPECIAL FORMS 445 EXAMPLE 7. Expand each of the following: a) (3x+5)(3x 5) b) (a 3 2b 3 )(a 3 +2b 3 ) Solution: We apply the difference of squares pattern to expand each of the given problems. Expand: (4x 3y)(4x+3y) a) In (3x + 5)(3x 5), we have the exact same terms in the First and Last positions, with the first set separated by a plus sign and the second set separated by a minus sign. a) Square the first term: (3x) 2 = 9x 2 b) Square the second term: 5 2 = 25 c) Place a minus sign between the two squares. Hence: (3x+5)(3x 5) = 9x 2 25 b) In (a 3 2b 3 )(a 3 +2b 3 ), we have the exact same terms in the First and Last positions, withthefirstsetseparatedbyaminussignandthesecond set separated by a plus sign. a) Square the first term: (a 3 ) 2 = a 6 b) Square the second term: (2b 3 ) 2 = 4b 6 c) Place a minus sign between the two squares. Hence: (a 3 2b 3 )(a 3 +2b 3 ) = a 6 4b 6 Answer: 16x 2 9y 2 Because factoring is unmultiplying, is should be a simple matter to reverse the process of Example 7. EXAMPLE 8. Factor each of the following: a) 9x 2 25 b) a 6 4b 6 Solution: Because of the work already done in Example 7, it is a simple matter to factor (or unmultiply ) each of these problems. Factor: 81x 2 49 a) 9x 2 25 = (3x+5)(3x 5) b) a 6 4b 6 = (a 3 2b 3 )(a 3 +2b 3 )

7 446 CHAPTER 6. FACTORING Answer: (9x+7)(9x 7) In each case, note how we took the square roots of each term, then separated one set with a plus sign and the other with a minus sign. Because of the commutative property of multiplication, it does not matter which one you make plus and which one you make minus. Always remember the first rule of factoring. The first rule of factoring. The first step to perform in any factoring problem is factor out the GCF. Factor: 4x 4 16x 2 EXAMPLE 9. Factor: x 3 9x Solution: In x 3 9x, the GCF of x 3 and 9x is x. Factor out x. Answer: 4x 2 (x+2)(x 2) x 3 9x = x(x 2 9) Note that x 2 9 is now the difference of two perfect squares. Take the square roots of x 2 and 9, which are x and 3, then separate one set with a plus sign and the other set with a minus sign. = x(x+3)(x 3) Factoring Completely Sometimes after one pass at factoring, factors remain that can be factored further. You must continue to factor in this case. Factor: x 4 81 EXAMPLE 10. Factor: x 4 16 Solution: In x 4 16, we have the difference of two squares: (x 2 ) 2 = x 4 and 4 2 = 16. First, we take the square roots, then separate one set with a plus sign and the other set with a minus sign. x 4 16 = (x 2 +4)(x 2 4)

8 6.5. FACTORING SPECIAL FORMS 447 Notethatx 2 +4isthesum oftwosquaresanddoesnotfactorfurther. However, x 2 4 is the difference of two squares. Take the square roots, x and 2, then separate one set with a plus sign and the other set with a minus sign. Done. We cannot factor further. = (x 2 +4)(x+2)(x 2) Answer: (x 2 +9)(x+3)(x 3) Nonlinear Equations Revisited Remember, if an equation is nonlinear, the first step is to make one side equal to zero by moving all terms to one side of the equation. Once you ve completed this important first step, factor and apply the zero product property to find the solutions. EXAMPLE 11. Solve for x: 25x 2 = 169 Solve for x: 16x 2 = 121 Solution: Make one side equal to zero, factor, then apply the zero product property. 25x 2 = x = 0 Original equation. Subtract 169 from both sides. Note that we have two perfect squares separated by a minus sign. This is the difference of squares pattern. Take the square roots, making one term plus and one term minus. (5x+13)(5x 13) = 0 Use difference of squares to factor. Use the zero product property to complete the solution, setting each factor equal to zero and solving the resulting equations. 5x+13 = 0 or 5x 13 = 0 x = 13 5 x = 13 5 Hence, the solutionsof 25x 2 = 169arex = 13/5and x = 13/5. We encourage readers to check each of these solutions. Answer: 11/4, 11/4

9 448 CHAPTER 6. FACTORING Solve for x: 25x 2 = 80x 64 One can also argue that the only number whose square is zero is the number zero. Hence, one can go directly from to (7x 9) 2 = 0 7x 9 = 0. Hence, the only solution of 49x = 126x is x = 9/7. EXAMPLE 12. Solve for x: 49x = 126x Solution: Make one side equal to zero, factor, then apply the zero product property. 49x = 126x 49x 2 126x+81 = 0 Original equation. Subtract 126x from both sides. Note that the first and last terms of the trinomial are perfect squares. Hence, it make sense to try and factor as a perfect square trinomial, taking the square roots of the first and last terms. (7x 9) 2 = 0 Factor as a perfect square trinomial. Of course, be sure to check the middle term. Because 2(7x)(9) = 126x, the middle term is correct. Because (7x 9) 2 = (7x 9)(7x 9), we can use the zero product property to set each factor equal to zero and solve the resulting equations. 7x 9 = 0 or 7x 9 = 0 x = 9 7 x = 9 7 Answer: 8/5 Hence, the only solution of 49x = 126x is x = 9/7. We encourage readers to check this solution. Solve for x: 5x = x x EXAMPLE 13. Solve for x: 2x 3 +3x 2 = 50x+75 Solution: Make one side equal to zero, factor, then apply the zero product property. 2x 3 +3x 2 = 50x+75 Original equation. 2x 3 +3x 2 50x 75 = 0 Make one side zero. This is a four-term expression, so we try factoring by grouping. Factor x 2 out of the first two terms, and 25 out of the second two terms. x 2 (2x+3) 25(2x+3) = 0 Factor by grouping. (x 2 25)(2x+3) = 0 Factor out 2x+3.

10 6.5. FACTORING SPECIAL FORMS 449 Complete the factorization by using the difference of squares to factor x (x+5)(x 5)(2x+3) = 0 Use difference of squares to factor. Finally, use the zero product property. Set each factor equal to zero and solve for x. x+5 = 0 or x 5 = 0 or 2x+3 = 0 x = 5 x = 5 x = 3 2 Hence, the solutions of 2x 3 +3x 2 = 50x+75 are x = 5, x = 5, and x = 3/2. We encourage readers to check each of these solutions. Answer: 6, 6, 1/5 Let s solve another nonlinear equation, matching the algebraic and graphical solutions. EXAMPLE 14. Solve the equation x 3 = 4x, both algebraically and graphi- Solve the equation x 3 = 16x cally, then compare your answers. both algebraically and graphically, then compare Solution: Note that we have a power of x larger than one, so the equation your answers. x 3 = 4x is nonlinear. Make one side zero and factor. x 3 = 4x x 3 4x = 0 x(x 2 4) = 0 x(x+2)(x 2) = 0 Original equation. Nonlinear. Make one side zero. Factor out GCF. Apply difference of squares. Note that we now have a product of three factors that equals zero. The zero product property says that at least one of these factors must equal zero. x = 0 or x+2 = 0 or x 2 = 0 x = 2 x = 2 Hence, the solutions of x 3 = 4x are x = 0, x = 2, and x = 2. Graphical solution. Load y = x 3 and y = 4x into Y1 and Y2 in the Y= menu of your calculator. Select 6:ZStandard from the ZOOM menu to produce the graph in Figure Although the image in Figure 6.26 shows all three points of intersection, adjusting the WINDOW parameters as shown in Figure 6.27, then pressing the GRAPH button will produce a nicer view of the points of intersection, as shown in the figure on the right in Figures 6.27.

11 450 CHAPTER 6. FACTORING Figure 6.26: Sketching y = x 3 and y = 4x. Figure 6.27: Adjusting the viewing window. Use the 5:intersect tool from the CALC menu to find the three points of intersection. Press the ENTER key in response to First curve, then press ENTER again in response to Second curve, then use the left-arrow key to move your cursor close to the leftmost point of intersection and press ENTER in response to Guess. The result is shown in the first image on the left in Figure Repeat the process to find the remaining points of intersection. The results are shown in the last two images in Figure Figure 6.28: Finding the points of intersection. Thus, the graphical solutions are x = 2, x = 0, and x = 2. Reporting the solution on your homework: Duplicate the image in your calculator s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves. Label the horizontal and vertical axes with x and y, respectively (see Figure 6.29). PlaceyourWINDOWparametersattheendofeachaxis(seeFigure 6.29). Label the graph with its equation (see Figure 6.29).

12 6.5. FACTORING SPECIAL FORMS 451 Drop dashed vertical lines through each x-intercept. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation x 3 = 4x (see Figure 6.29). y 15 y = x 3 y = 4x Answer: 4, 0, x 15 Figure 6.29: Reporting your graphical solution on your homework. Finally, note that the graphical solutions x = 2, x = 0, and x = 2 match our algebraic solutions exactly.

13 452 CHAPTER 6. FACTORING Exercises In Exercises 1-8, expand each of the given expressions. 1. (8r 3t) 2 2. (6a+c) 2 3. (4a+7b) 2 4. (4s+t) 2 5. (s 3 9) 2 6. (w 3 +7) 2 7. (s 2 +6t 2 ) 2 8. (7u 2 2w 2 ) 2 In Exercises 9-28, factor each of the given expressions s 2 +60st+36t u 2 +24uv+16v v 2 60vw+25w b 2 42bc+9c a 4 +18a 2 b 2 +81b u 4 144u 2 w 2 +81w s 4 28s 2 t 2 +4t a 4 12a 2 c 2 +9c b 6 112b x 6 10x r r a 6 16a s 3 t 20s 2 t 2 +20st r 3 t 12r 2 t 2 +3rt a 3 c+8a 2 c 2 +2ac x 3 z 60x 2 z 2 +50xz b b 2 75b c c 2 80c 27. 5u 5 30u 4 45u z 5 36z 4 27z 3 In Exercises 29-36, expand each of the given expressions. 29. (21c+16)(21c 16) 30. (19t+7)(19t 7) 31. (5x+19z)(5x 19z) 32. (11u+5w)(11u 5w) 33. (3y 4 +23z 4 )(3y 4 23z 4 ) 34. (5x 3 +z 3 )(5x 3 z 3 ) 35. (8r 5 +19s 5 )(8r 5 19s 5 ) 36. (3u 3 +16v 3 )(3u 3 16v 3 )

14 6.5. FACTORING SPECIAL FORMS 453 In Exercises 37-60, factor each of the given expressions x b v r x 2 576y y 2 81z r 2 289s a 2 144b r 6 256t x z u 10 25w a 6 81c y 5 242y y 5 147y a 3 b 324ab b 3 c 1875bc x 3 z 1156xz u 3 v 507uv t 4 4t z 5 256z x x x x 4 1 In Exercises 61-68, factor each of the given expressions completely. 61. z 3 +z 2 9z u 3 +u 2 48u x 3 2x 2 y xy 2 +2y x 3 +2x 2 z 4xz 2 8z r 3 3r 2 t 25rt 2 +75t b 3 3b 2 c 50bc 2 +75c x 3 +x 2 32x r 3 2r 2 r+2 In Exercises 69-80, solve each of the given equations for x x 3 +7x 2 = 72x x 3 +7x 2 = 32x x 3 +5x 2 = 64x x 3 +4x 2 = 49x x = 264x x = 874x x 2 = x 2 = x 2 = x 2 = x = 608x x = 136x

15 454 CHAPTER 6. FACTORING In Exercises 81-84, perform each of the following tasks: i) Use a strictly algebraic technique to solve the given equation. ii) Use the 5:intersect utility on your graphing calculator to solve the given equation. Report the results found using graphing calculator as shown in Example x 3 = x 82. x 3 = 9x 83. 4x 3 = x 84. 9x 3 = x Answers 1. 64r 2 48rt+9t a 2 +56ab+49b 2 5. s 6 18s s 4 +12s 2 t 2 +36t 4 9. (5s+6t) (6v 5w) (a 2 +9b 2 ) (7s 2 2t 2 ) (7b 3 8) (7r 3 +8) st(s 2t) ac(2a+c) b(4b 5) u 3 (u+3) c x 2 361z y 8 529z r s (19x+23)(19x 23) 39. (4v +13)(4v 13) 41. (13x+24y)(13x 24y) 43. (23r+17s)(23r 17s) 45. (7r 3 +16t 3 )(7r 3 16t 3 ) 47. (6u 5 +5w 5 )(6u 5 5w 5 ) 49. 2y 3 (6y +11)(6y 11) 51. 4ab(19a+9b)(19a 9b) 53. 4xz(12x+17z)(12x 17z) 55. 4t 2 (12t+1)(12t 1) 57. (9x 2 +16)(3x+4)(3x 4) 59. (9x 2 +4)(3x+2)(3x 2) 61. (z +3)(z 3)(z +1) 63. (x+y)(x y)(x 2y) 65. (r+5t)(r 5t)(r 3t) 67. (x+4)(x 4)(2x+1)

16 6.5. FACTORING SPECIAL FORMS x = 6, 6, x = 8, 8, x = x = 13 4, x = 5 3, x = x = 0, 1,1 83. x = 0, 1/2,1/2

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