Don’t forget that the very fi rst step was to factor out a 2. Therefore, 12d Check by multiplying. b) Since the coeffi cient of the squared term is negative, begin by factoring out 1. (There is no other common factor except 1.) Try to factor 2h2 9h 56. To get a product of 2h2 we will use 2h and h in the binomials. We need pairs of integers so that their product is 56. The ones that come to mind quickly involve 7 and 8 and 1 and 56: 7 and 8, 7 and 8, 1 and 56, 1 and 56. There are other pairs; if these do not work, we will list others. Do not start with 1 and 56 or 1 and 56 because the middle term, 9h, is not very large. Using 1 and 56 or 1 and 56 would likely result in a larger middle term. Try 7 and 8. Do not put 8 in the same binomial as 2h since then it would be possible to factor out 2. 2h2 9h 56 (2h 7)(h 8) 7h 16h 9h This must equal 9h. Incorrect! Only the sign of the sum is incorrect. Change the signs in the binomials to get the correct sum. 7h (16h) 9h Remember that we factored out 1 to begin the problem. Correct! 2h2 9h 56 1(2h2 9h 56) (2h 7)(h 8) Check by multiplying. YOU TRY 7 Factor completely. 2 46d 8 2(6d 2 23d 4) 2(6d 1)(d 4). 2h2 9h 56 1(2h2 9h 56) 2h2 9h 56 (2h )(h ) 2h2 9h 56 (2h 7)(h 8) a) 15b2 55b 30 b) 4p2 3p 10 We have seen two methods for factoring ax2 bx c (a 1): factoring by grouping and factoring by trial and error. In either case, remember to begin by taking out a common factor from all terms whenever possible. 6 Factor Using Substitution Some polynomials can be factored using a method called substitution. We will illustrate this method in Example 9. 374 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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