# 6.6 Factoring Strategy

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1 456 CHAPTER 6. FACTORING 6.6 Factoring Strategy When you are concentrating on factoring problems of a single type, after doing a few you tend to get into a rhythm, and the remainder of the exercises, because they are similar, seem to flow. However, when you encounter a mixture of factoring problems of different types, progress is harder. The goal of this section is to set up a strategy to follow when attacking a general factoring problem. If it hasn t already been done, it is helpful to arrange the terms of the given polynomial in some sort of order (descending or ascending). Then you want to apply the following guidelines. Factoring Strategy. These steps should be followed in the order that they appear. 1. Factor out the greatest common factor (GCF). 2. Look for a special form. a) If you have two perfect squares separated by a minus sign, use the difference of squares pattern to factor: a 2 b 2 = (a+b)(a b) b) If you have a trinomial whose first and last terms are perfect squares, you should suspect that you have a perfect square trinomial. Take the square roots of the first and last terms and factor as follows. a 2 +2ab+b 2 = (a+b) 2 Be sure to check that the middle term is correct. 3. If you have a trinomial of the form ax 2 + bx+c, use the ac-method to factor. 4. If you have a four-term expression, try to factor by grouping. Once you ve applied the above strategy to the given polynomial, it is quite possible that one of your resulting factors will factor further. Thus, we have the following rule. Factor completely. The factoring process is not complete until none of your remaining factors can be factored further. This is the meaning of the phrase, factor completely.

2 6.6. FACTORING STRATEGY 457 Finally, a very good word of advice. Check your factoring by multiplying. Once you ve factored the given polynomial completely, it is a very good practice to check your result. If you multiply to find the product of your factors, and get the original given polynomial as a result, then you know your factorization is correct. It s a bit more work to check your factorization, but it s worth the effort. It helps to eliminate errors and also helps to build a better understanding of the factoring process. Remember, factoring is unmultiplying, so the more you multiply, the better you get at factoring. Let s see what can happen when you don t check your factorization! Warning! The following solution is incorrect! Factor: 2x 4 +8x 2. Solution: Factor out the GCF. 2x 4 +8x 2 = 2x 2 (x 2 +4) = 2x 2 (x+2) 2 Note that this student did not bother to check his factorization. Let s do that for him now. Check: Multiply to check. Remember, when squaring a binomial, there is a middle term. 2x 2 (x+2) 2 = 2x 2 (x 2 +4x+4) = 2x 4 +8x 3 +8x 2 This is not the same as the original polynomial 2x 4 + 8x 2, so the student s factorization is incorrect. Had the student performed this check, he might have caught his error, provided of course, that he multiplies correctly during the check. The correct factorization follows. 2x 4 +8x 2 = 2x 2 (x 2 +4) The sum of squares does not factor, so we are finished. Check: Multiply to check. 2x 2 (x 2 +4) = 2x 4 +8x 2 This is the same as the original polynomial 2x 4 +8x 2, so this factorization is correct.

3 458 CHAPTER 6. FACTORING 4x 7 +64x 3 EXAMPLE 1. 3x 6 +3x 2 Solution: The first rule of factoring is Factor out the GCF. The GCF of 3x 6 and 3x 2 is 3x 2, so we could factor out 3x 2. 3x 6 +3x 2 = 3x 2 ( x 4 +1) This is perfectly valid, but we don t like the fact that the second factor starts with x 4. Let s factor out 3x 2 instead. 3x 6 +3x 2 = 3x 2 (x 4 1) The second factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign. = 3x 2 (x 2 +1)(x 2 1) The sum of squares does not factor. But the last factor is the difference of two squares. Take the square roots, separating one pair with a plus sign, one pair with a minus sign. Check: Multiply to check the result. = 3x 2 (x 2 +1)(x+1)(x 1) 3x 2 (x 2 +1)(x+1)(x 1) = 3x 2 (x 2 +1)(x 2 1) = 3x 2 (x 4 1) = 3x 6 +3x 2 Answer: 4x 3 (x 2 +4)(x+2)(x 2) The factorization checks. 3a 2 b 4 +12a 4 b 2 12a 3 b 3 EXAMPLE 2. x 3 y +9xy 3 +6x 2 y 2 Solution: The first rule of factoring is Factor out the GCF. The GCF of x 3 y, 9xy 3, and 6x 2 y 2 is xy, so we factor out xy. x 3 y +9xy 3 +6x 2 y 2 = xy(x 2 +9y 2 +6xy) Let s order that second factor in descending powers of x. = xy(x 2 +6xy +9y 2 )

4 6.6. FACTORING STRATEGY 459 The first and last terms of the trinomial factor are perfect squares. We suspect we have a perfect square trinomial, so we take the square roots of the first and last terms, check the middle term, and write: Thus, x 3 y +9xy 3 +6x 2 y 2 = xy(x+3y) 2. Check: Multiply to check the result. = xy(x+3y) 2 xy(x+3y) 2 = xy(x 2 +6xy +9y 2 ) = x 3 y +6x 2 y 2 +9xy 3 Except for the order, this result is the same as the given polynomial. The factorization checks. Answer: 3a 2 b 2 (2a b) 2 EXAMPLE 3. 2x 3 48x+20x 2 Solution: In the last example, we recognized a need to rearrange our terms after we pulled out the GCF. This time, let s arrange our terms in descending powers of x right away. 27x 3 3x 4 60x 2 Now, let s factor out the GCF. 2x 3 48x+20x 2 = 2x 3 +20x 2 48x = 2x(x 2 +10x 24) The last term of the trinomial factor is not a perfect square. Let s move to the ac-method to factor. The integer pair 2,12 has a product equal to ac = 24 and a sum equal to b = 10. Because the coefficient of x 2 is one, this is a drop in place situation. We drop our pair in place and write: Thus, 2x 3 48x+20x 2 = 2x(x 2)(x+12). = 2x(x 2)(x+12) Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up. 2x(x 2)(x+12) = 2x(x 2 +10x 24) = 2x 3 +20x 2 48x Except for the order, this result is the same as the given polynomial. The factorization checks. Answer: 3x 2 (x 4)(x 5)

5 460 CHAPTER 6. FACTORING 8x 2 +14xy 15y 2 EXAMPLE 4. 2a 2 13ab 24b 2 Solution: There is no common factor we can factor out. We have a trinomial, butthefirstandlasttermsarenotperfectsquares,solet sapplytheac-method. Ignoring the variables for a moment, we need an integer pair whose product is ac = 48 and whose sum is 13. The integer pair 3, 16 comes to mind (if nothing comes to mind, start listing integer pairs). Break up the middle term into a sum of like terms using the integer pair 3, 16, then factor by grouping. 2a 2 13ab 24b 2 = 2a 2 +3ab 16ab 24b 2 = a(2a+3b) 8b(2a+3b) = (a 8b)(2a+3b) Thus, 2a 2 13ab 24b 2 = (a 8b)(2a+3b). Check: Multiply to check the result. We use the FOIL method shortcut and mental calculations to speed things up. (a 8b)(2a+3b) = 2a 2 13ab 24b 2 Answer: (2x+5y)(4x 3y) This result is the same as the given polynomial. The factorization checks. 36x 3 +60x 2 +9x EXAMPLE 5. 30x 4 +38x 3 20x 2 Solution: The first step is to factor out the GCF, which in this case is 2x 2. 30x 4 +38x 3 20x 2 = 2x 2 (15x 2 +19x 10) The first and last terms of the trinomial factor are not perfect squares, so let s moveagaintotheac-method. Comparing15x 2 +19x 10with ax 2 +bx+c,note that ac = (15)( 10) = 150. We need an integer pair whose product is 150 and whose sum is 19. The integer pair 6 and 25 satisfies these requirements. Because a 1, this is not a drop in place situation, so we need to break up the middle term as a sum of like terms using the pair 6 and 25. = 2x 2 (15x 2 6x+25x 10) Factor by grouping. Factor 3x out of the first two terms and 5 out of the third and fourth terms. = 2x 2 (3x(5x 2)+5(5x 2))

6 6.6. FACTORING STRATEGY 461 Finally, factor out the common factor 5x 2. = 2x 2 (3x+5)(5x 2) Thus, 30x 4 +38x 3 20x 2 = 2x 2 (3x+5)(5x 2). Check: Multiply to check the result. Use the FOIL method to first multiply the binomials. Distribute the 2x 2. 2x 2 (3x+5)(5x 2) = 2x 2 (15x 2 +19x 10) = 30x 4 +38x 3 20x 2 This result is the same as the given polynomial. The factorization checks. Answer: 3x(6x + 1)(2x + 3) EXAMPLE 6. 8x 5 +10x 4 72x 3 90x 2 Solution: Each of the terms is divisible by 3x 3. Factor out 3x 3. 15x 6 33x 5 240x x 3 = 3x 3[ 5x 3 11x 2 80x+176 ] 15x 6 33x 5 240x x 3 The second factor is a four-term expression. Factor by grouping. = 3x 3[ x 2 (5x 11) 16(5x 11) ] = 3x 3 (x 2 16)(5x 11) The factorx 2 16isadifferenceoftwosquares. Takethe squareroots, separate one pair with a plus, one pair with a minus. = 3x 3 (x+4)(x 4)(5x 11) Thus, 15x 6 33x 5 240x x 3 = 3x 3 (x+4)(x 4)(5x 11). Check: Multiply to check the result. 3x 3 (x+4)(x 4)(5x 11) = 3x 3 (x 2 16)(5x 11) = 3x 3 (5x 3 11x 2 80x+176 = 15x 6 33x 5 240x x 3 This result is the same as the given polynomial. The factorization checks. Answer: 2x 2 (x 3)(x+3)(4x+5)

7 462 CHAPTER 6. FACTORING Using the Calculator to Assist the ac-method When using the ac-method to factor ax 2 +bx+c and ac is a very large number, then it can be difficult to find a pair whose product is ac and whose sum in b. For example, consider the trinomial: 12y 2 11y 36 Weneedanintegerpairwhoseproductisac = 432andwhosesumis b = 11. We begin listing integer pair possibilities, but the process quickly becomes daunting. 1, 432 2, 216 Note that the numbers in the second column are found by dividing ac = 432 by the number in the first column. We ll now use this fact and the TABLE feature on our calculator to pursue the desired integer pair. 1. Enter the expression -432/X into Y1 in the Y= menu (see the first image in Figure 6.30). 2. Above the WINDOW button you ll see TBLSET. Use the 2nd key, then press the WINDOW button to access the menu shown in the second image of Figure Set TblStart=1, Tbl=1, then highlight AUTO for both the independent and dependent variables. 3. Above the GRAPH button you ll see TABLE. Use the 2nd key, then press the GRAPH button to access the table shown in the third image in Figure Use the up- and down-arrow keys to scroll through the contents of the table. Note that you can ignore most of the pairs, because they are not both integers. Pay attention only when they are both integers. In this case, remember that you are searching for a pair whose sum is b = 11. Note that the pair 16, 27 shown in the third image of Figure 6.30 is the pair we seek. Figure 6.30: Using the TABLE feature to assist the ac-method.

8 6.6. FACTORING STRATEGY 463 Now we can break the middle term of 12y 2 11y 36 into a sum of like terms using the ordered pair 16, 27, then factor by grouping. 12y 2 11y 36 = 12y 2 +16y 27y 36 = 4y(3y+4) 9(3y+4) = (4y 9)(3y +4) Check: Use the FOIL method shortcut and mental calculations to multiply. The factorization checks. (4y 9)(3y +4) = 12y 2 11y 36

9 464 CHAPTER 6. FACTORING Exercises In Exercises 1-12, factor each of the given polynomials completely y 4 z 2 144y 2 z s 4 t 4 242s 2 t x 7 z 5 363x 5 z r 5 s 2 80r 3 s u 7 162u x 4 320x v v a 9 48a x 6 300x y 5 18y u 7 w 3 2u 3 w y 8 z 4 3y 4 z 8 In Exercises 13-24, factor each of the given polynomials completely a 6 210a a v 7 560v v a 5 b a 4 b a 3 b u 6 v u 5 v u 4 v b 5 +4b 4 +2b v 6 +30v 5 +75v z 4 4z 3 +2z u 6 40u u x x x b 4 +84b 3 +18b b 4 c 5 240b 3 c b 2 c a 5 c 4 180a 4 c 5 +50a 3 c 6 In Exercises 25-36, factor each of the given polynomials completely a 5 +5a 4 210a y 5 9y 4 12y y 6 39y y y 7 27y 6 +42y z 4 +12z 3 135z a 4 40a 3 45a a 6 +64a a x 4 +64x x z 4 +33z 3 +84z a 6 +65a a z 7 75z z y 4 27y 3 +24y 2

10 6.6. FACTORING STRATEGY 465 In Exercises 37-48, factor each of the given polynomials completely b 3 22b 2 +30b 38. 4b 6 22b 5 +30b u 4 w 5 3u 3 w 6 20u 2 w x 5 z 2 +9x 4 z 3 30x 3 z x 4 y 5 +50x 3 y 6 +50x 2 y s 4 t 3 +62s 3 t 4 +40s 2 t x 3 +9x 2 30x 44. 6v 4 +2v 3 20v u 6 +34u 5 +30u a 4 +29a 3 +30a a 4 c 4 35a 3 c 5 +25a 2 c x 6 z 5 39x 5 z 6 +18x 4 z 7 In Exercises 49-56, factor each of the given polynomials completely y 5 +15y 4 108y 3 135y b 8 +12b 7 324b 6 432b x 6 z 5 +6x 5 z 6 144x 4 z 7 96x 3 z u 7 w 3 +9u 6 w 4 432u 5 w 5 324u 4 w z z 5 2z 4 3z x x 6 6x 5 9x a 6 c a 5 c 4 4a 4 c 5 10a 3 c a 8 c 4 +32a 7 c 5 3a 6 c 6 2a 5 c 7 In Exercises 57-60, use your calculator to help factor each of the given trinomials. Follow the procedure outline in Using the Calculator to Assist the ac-method on page x 2 +61x x 2 62x x 2 167x x 2 +x 144 Answers 1. 4y 2 z 2 (11y+6z)(11y 6z) 3. 3x 5 z 5 (x+11)(x 11) 5. 2u 5 (u+9)(u 9) 7. 3v 4 (v 2 +25)(v +5)(v 5) 9. 3x 4 (x+10)(x 10) 11. 2u 3 w 3 (25u 2 +w 2 )(5u+w)(5u w) 13. 3a 4 (5a 7) a 3 b 3 (2a+3b) b 3 (b+1) z 2 (z 1) x 2 (9x+5) b 2 c 5 (5b 8c) a 3 (a 6)(a+7)

11 466 CHAPTER 6. FACTORING 27. 3y 4 (y 8)(y 5) 29. 3z 2 (z 5)(z +9) 31. 4a 4 (a+9)(a+7) 33. 3z 2 (z +7)(z +4) 35. 5z 5 (z 6)(z 9) 37. 2b(2b 5)(b 3) 39. u 2 w 5 (u 4w)(2u+5w) 41. 2x 2 y 5 (3x+5y)(2x+5y) 43. 3x(4x 5)(x+2) 45. 2u 4 (4u+5)(u+3) 47. a 2 c 4 (4a 5c)(3a 5c) 49. 3y 2 (y +3)(y 3)(4y +5) 51. 3x 3 z 5 (x+4z)(x 4z)(3x+2z) 53. z 3 (6z +1)(6z 1)(2z +3) 55. 2a 3 c 3 (6a+c)(6a c)(2a+5c) 57. (2x+15)(3x+8) 59. (15x 8)(4x 9)

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### FACTOR POLYNOMIALS by SPLITTING FACTOR POLYNOMIALS by SPLITTING THE IDEA FACTOR POLYNOMIALS by SPLITTING The idea is to split the middle term into two pieces. Say the polynomial looks like c + bx + ax 2. Further more suppose the DL METHOD Activity 23 Math 40 Factoring using the BOX Team Name (optional): Your Name: Partner(s): 1. (2.) Task 1: Factoring out the greatest common factor Mini Lecture: Factoring polynomials is our focus now. Factoring GCF/ Factor by Grouping (Student notes) Factoring is to write an expression as a product of factors. For example, we can write 10 as (5)(2), where 5 and 2 are called factors of 10. We can also do this