Since there is a minus sign in front of 14n, n2 14n 49 fi ts the pattern of a2 2ab b2 (a b)2 with a n and b 7. Therefore, n2 14n 49 (n 7)2. Check by multiplying. b) From 4p3 24p2 36p we can begin by taking out the GCF of 4p. 4p3 24p2 36p 4p(p2 6p 9) T T What do you square to get p2? p (p)2 (3)2 What do you square to get 9? 3 Does the middle term equal 2 p 3? Yes: 2 p 3 6p. 4p3 24p2 36p 4p(p2 6p 9) 4p(p 3)2 Check by multiplying. c) We cannot take out a common factor. Since the fi rst and last terms of 9k2 30k 25 are perfect squares, let’s see if this is a perfect square trinomial. 9k2 30k 25 T T What do you square to get 9k2? 3k (3k)2 (5)2 What do you square to get 25? 5 Does the middle term equal 2 3k 5? Yes: 2 3k 5 30k. Therefore, 9k2 30k 25 (3k 5)2. Check by multiplying. d) We cannot take out a common factor. The fi rst and last terms of 4c2 20c 9 are perfect squares. Is this a perfect square trinomial? 4c2 20c 9 T T What do you square to get 4c2? 2c (2c)2 (3)2 What do you square to get 9? 3 Does the middle term equal 2 2c 3? No: 2 2c 3 12c This is not a perfect square trinomial. Applying a method from Section 7.2 we fi nd that the trinomial does factor, however. 4c2 20c 9 (2c 9)(2c 1). Check by multiplying. Don’t forget to ask yourself, “Can I factor out a GCF?” YOU TRY 1 Factor completely. a) w2 8w 16 b) a2 20a 100 c) 4d 2 36d 81 2 Factor the Difference of Two Squares Another common type of factoring problem is a difference of two squares. Some examples of these types of binomials are y2 9, 25m2 16n2, 64 t2, and h4 16. Notice that in each binomial, the terms are being subtracted, and each term is a perfect square. In Section 6.4, we saw that (a b)(a b) a2 b2. If we reverse the procedure, we get the factorization of the difference of two squares. 380 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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