Note If the sum of two squares does not contain a common factor, then it cannot be factored. YOU TRY 2 Factor completely. a) r2 25 b) 49p2 121q2 c) x2 25 144 d) h2 1 Remember that sometimes we can factor out a GCF fi rst. And, after factoring once, ask yourself, “Can I factor again?” EXAMPLE 4 Factor completely. a) 128t 2t3 b) 5x2 45 c) h4 16 Solution a) Ask yourself, “Can I take out a common factor?” Yes. Factor out 2t. 128t 2t3 2t(64 t2) Now ask yourself, “Can I factor again?” Yes. 64 t2 is the difference of two squares. Identify a and b. 64 t2 T T (8)2 (t)2 So, a 8 and b t. 64 t2 (8 t)(8 t). Therefore, 128t 2t3 2t(8 t)(8 t). (8 t)(8 t) is not the same as (t 8)(t 8) because subtraction is not commutative. While 8 t t 8, 8 t does not equal t 8. You must write the terms in the correct order. Another way to see that they are not equivalent is to multiply (t 8)(t 8). (t 8)(t 8) t 2 64. This is not the same as 64 t 2. b) Ask yourself, “Can I take out a common factor?” Yes. Factor out 5. 5(x2 9) “Can I factor again?” No; x2 9 is the sum of two squares. Therefore, 5x2 45 5(x2 9). c) The terms in h4 16 have no common factors, but they are perfect squares. Identify a and b. h4 16. T T What do you square to get h4? h2 (h)2 (4)2 What do you square to get 16? 4 So, a h2 and b 4. Therefore, h4 16 (h2 4)(h2 4). Can we factor again? 382 CHAPTER 7 Factoring Polynomials www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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