d) We cannot take out a GCF from p2 16p 64. It is a trinomial, and notice that the fi rst and last terms are perfect squares. Is this a perfect square trinomial? p2 16p 64 T T (p)2 (8)2 Does the middle term equal 2 p (8)? Yes: 2 p (8) 16p. Use a2 2ab b2 (a b)2 with a p and b 8. Then, p2 16p 64 (p 8)2. “Can I factor again?” No. It is completely factored. e) It is tempting to jump right in and try to factor 8x2 26x 20 as the product of two binomials, but ask yourself, “Can I take out a GCF?” Yes! Factor out 2. 8x2 26x 20 2(4x2 13x 10) “Can I factor again?” Yes. 2(4x2 13x 10) 2(4x 5)(x 2) “Can I factor again?” No. So, 8x2 26x 20 2(4x 5)(x 2). f ) We cannot take out a GCF from 27k3 8. Notice that 27k3 8 has two terms, so think about squares and cubes. Neither term is a perfect square and the positive terms are being added, so this cannot be the difference of squares. Is each term a perfect cube? Yes! 27k3 8 is the sum of two cubes. We will factor 27k3 8 using a3 b3 (a b)(a2 ab b2) with a 3k and b 2. 27k3 8 (3k 2)(3k)2 (3k)(2) (2)2 T T (3k)3 (2)3 (3k 2)(9k2 6k 4) “Can I factor again?” No. It is completely factored. g) Look at t2 36 and ask yourself, “Can I factor out a GCF?” No. The binomial t2 36 is the sum of two squares, so it does not factor. This polynomial is prime. YOU TRY 1 Factor completely. a) 3p2 p 10 b) 2n3 n2 12n 6 c) 4k4 36k3 32k2 d) 48 3y4 e) 8r3 125 ANSWERS TO YOU TRY EXERCISES 1) a) (3p 5)(p 2) b) (n2 6)(2n 1) c) 4k2(k 8)(k 1) d) 3(4 y2)(2 y)(2 y) e) (2r 5)(4r2 10r 25) 390 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
To see the actual publication please follow the link above