Factoring Copyright Cengage Learning All rights reserved 6

Factoring Copyright © Cengage Learning. All rights reserved. 6

SECTION 6. 3 More Trinomials to Factor Copyright © Cengage Learning. All rights reserved.

Objectives A Use Diamond method to factorize a trinomial whose leading coefficient is other than 1. B Use Box method to factor a trinomial whose leading coefficient is a number other than 1. 3

Method 1: Diamond Method of Factoring Trinomials 4

Diamond Method 13 + 15. Factor 2 x 2 + 13 x Draw a diamond Compute 2 × 15 = 30 30 3 10 13 Next, think of two numbers whose product is 30, and whose sum will be 13. 10 and 3 will work! 5

Diamond Method Factor 2 x 2 + 13 x + 15. 30 3 10 13 2 x 2 + 10 x + 3 x +15 = (2 x 2 + 10 x) + (3 x + 15) = 2 x(x + 5) + 3(x + 5) = (2 x + 3)(x + 5) This method works pretty well for small numbers. 6

Example 3 Factor 42 y 3 + 35 y 2 – 42 y. Solution We begin by factoring out the greatest common factor of 7 y. 42 y 3 + 35 y 2 – 42 y = 7 y(6 y 2 + 5 5 y – 6) Compute Draw a diamond 6 × (-6) = -36 -4 9 5 Next, think of two numbers whose product is -36, and whose sum will be 5. -4 and 9 will work! 7

Example 3 - Solution cont’d Factor 6 y 2 + 5 y – 6. -36 -4 9 5 (6 y 2 + 5 y – 6) = (6 y 2 – 4 y) + (9 y – 6) = 2 y(3 y – 2) + 3(3 y – 2) = (2 y + 3)(3 y – 2) Final answer: 42 y 3 + 35 y 2 – 42 y = 7 y(2 y + 3)(3 y – 2) 8

Diamond Method This method is difficult for large numbers, for instance: Factor 20 x 2 + 37 x + 15 20 × 15 = 300 37 Now we need to think of two numbers whose product is 300 and whose sum is 37. Not easy! 9

Method 2: Box Method of Factoring Trinomials 10

Box Method This method works better for large numbers, for instance: Factor 20 x 2 + 37 x + 15 First draw a box, 4 x 3 5 x 5 then find two numbers whose product is 20. Say 4 and 5. Next, find two numbers whose product is 15. Say 3 and 5. 11

Exercise 33 - Solution Factor 20 x 2 + 37 x + 15 Now cross multiply to check if we can get 37 x 15 x 4 x 3 20 x + 15 x = 35 x 5 x 5 20 x No, it is not equal to 37 x. So try to switch the 3 and 5. 12

Box Method Factor 20 x 2 + 37 x + 15 Let us switch the 3 and 5. 25 x 4 x 5 x 3 12 x + 25 x = 37 x 5 12 x Cross multiply again to check Yes, it does check. So the answer is 20 x 2 + 37 x + 15 = (4 x + 5)(5 x + 3). 13

Exercise 33 Use Box Method to factor 15 x 2 – 67 x + 38 Solution: Draw a box. 3 x 2 5 x 19 First find two numbers whose product is 15. Say 3 and 5. Next, find two numbers whose product is 38. Say 2 and 19. 14

Exercise 33 - Solution Use Box Method to factor 15 x 2 – 67 x + 38 Now cross multiply to check if we can get – 67 x 10 x 3 x 5 x 2 10 x + 57 x = 67 x 19 57 x No, it is not equal to - 67 x. We need some negative numbers. So change 2 to -2 and 19 to -19. 15

Exercise 33 - Solution Use Box Method to factor 15 x 2 – 67 x + 38 -10 x 3 x -2 -10 x + (-57 x) = - 67 x 5 x -19 -57 x Now cross multiply again to check if we can get – 67 x Yes, it checks. So the answer is 15 x 2 – 67 x + 38 = (3 x – 2)(5 x – 19) 16