Formula Factoring the Difference of Two Squares a2 b2 (a b)(a b) Don’t forget that we can check all factorizations by multiplying. Memorize the formula. EXAMPLE 3 In-Class Example 3 Factor completely. a) h2 64 b) 25t2 144u2 c) m2 9 49 d) y2 25 Answer: a) (h 8)(h 8) b) (5t 12u)(5t 12u) c) am 3 7 bam 3 7 b d) prime Are you writing out the steps as you are reading the example? Factor completely. a) c2 36 b) 49x2 25y2 c) t 2 4 9 d) k 2 81 Solution a) First, notice that c2 36 is the difference of two terms and those terms are perfect squares. We can use the formula a2 b2 (a b)(a b). Identify a and b. c2 36 What do you square to get c2? c (c)2 (6)2 What do you square to get 36? 6 Then, a c and b 6. Therefore, c2 36 (c 6)(c 6). b) Look carefully at 49x2 25y2. Each term is a perfect square, and they are being subtracted. Identify a and b. 49x2 25y2 What do you square to get 49x2? 7x (7x)2 (5y)2 What do you square to get 25y2? 5y Then, a 7x and b 5y. So, 49x2 25y2 (7x 5y)(7x 5y). c) Each term in t 2 4 9 is a perfect square, and they are being subtracted. t 2 4 9 (t)2 a2 3 b 2 What do you square to get t2? t So, a t and b 2 3 . Therefore, t2 4 9 What do you square to get at 2 3 b at 2 3 b. 4 9 ? 2 3 d) Each term in k2 81 is a perfect square, but the expression is the sum of two squares. This polynomial does not factor. k 2 81 (k 9)(k 9) since (k 9)(k 9) k 2 81. k 2 81 (k 9)(k 9) since (k 9)(k 9) k 2 18k 81. So, k2 81 is prime. 416 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_introductory_algebra_1e_ch4_7_10
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