Ask yourself, “Can I factor again?” Examine 4x2 25y2. It has two terms that are being subtracted, and each term is a perfect square. 4x2 25y2 is the difference of squares. 4x2 25y2 (2x 5y)(2x 5y) (2x)2 (5y)2 8x2 50y2 2(4x2 25y2) 2(2x 5y)(2x 5y) “Can I factor again?” No. It is completely factored. b) Look at t2 t 56. “Can I factor out a GCF?” No. Think of two numbers whose product is 56 and sum is 1. The numbers are 8 and 7. t 2 t 56 (t 8)(t 7) “Can I factor again?” No. It is completely factored. c) We have to factor a2b 9b 4a2 36. “Can I factor out a GCF?” No. Notice that this polynomial has four terms. When a polynomial has four terms, think about factoring by grouping. a2b 9b 4a2 36 b(a2 9) 4(a2 9) Take out the common factor from each pair of terms. (a2 9)(b 4) Factor out (a2 9) using the distributive property. Examine (a2 9)(b 4) and ask yourself, “Can I factor again?” Yes! (a2 9) is the difference of two squares. Factor again. (a2 9)(b 4) (a 3)(a 3)(b 4) “Can I factor again?” No. So, a2b 9b 4a2 36 (a 3)(a 3)(b 4). Note Seeing four terms is a clue to try factoring by grouping. d) We cannot take out a GCF from k2 12k 36. It is a trinomial, and notice that the fi rst and last terms are perfect squares. Is this a perfect square trinomial? k2 12k 36 (k)2 (6)2 Does the middle term equal 2 k 6? Yes. 2 k 6 12k Use a2 2ab b2 (a b)2 with a k and b 6. Then, k2 12k 36 (k 6)2. “Can I factor again?” No. It is completely factored. e) It is tempting to jump right in and try to factor 15p2 51p 18 as the product of two binomials, but ask yourself, “Can I take out a GCF?” Yes! Factor out 3. 15p2 51p 18 3(5p2 17p 6) “Can I factor again?” Yes. 3(5p2 17p 6) 3(5p 2)( p 3) “Can I factor again?” No. So, 15p2 51p 18 3(5p 2)(p 3). Answer: a) 5(3m 2n)(3m 2n) b) (w 12)(w 2) c) (y 1)(y 1)(x 7) d) (b 4)2 e) 2(4r 3)(r 1) f) prime After each step, what should you ask yourself? www.mhhe.com/messersmith Putting It All Together 421
messersmith_power_introductory_algebra_1e_ch4_7_10
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