Note If the sum of two squares does not contain a common factor, then it cannot be factored. YOU TRY 2 Factor completely. a) m2 100 b) 4c2 81d 2 c) h2 64 25 d) p2 49 Remember that sometimes we can factor out a GCF fi rst. And, after factoring once, ask yourself, “Can I factor again?” EXAMPLE 4 In-Class Example 4 Factor completely. a) 24 6x2 b) 3r2 75 c) x4 81 Answer: a) 6(2 x)(2 x) b) 3(r2 25) c) (x2 9)(x 3)(x 3) Factor completely. a) 300p 3p3 b) 7w2 28 c) x4 81 Solution a) Ask yourself, “Can I take out a common factor?” Yes. Factor out 3p. 300p 3p3 3p(100 p2) Now ask yourself, “Can I factor again?” Yes. 100 p2 is the difference of two squares. Identify a and b. 100 p2 (10)2 (p)2 So, a 10 and b p. 100 p2 (10 p)(10 p). Therefore, 300p 3p3 3p(10 p)(10 p). (10 p)(10 p) is not the same as (p 10)(p 10) because subtraction is not commutative. While 10 p p 10, 10 p does not equal p 10. You must write the terms in the correct order. Another way to see that they are not equivalent is to multiply (p 10)(p 10). (p 10)(p 10) p2 100. This is not the same as 100 p2. b) Look at 7w2 28. Ask yourself, “Can I take out a common factor?” Yes. Factor out 7. 7(w 2 4) “Can I factor again?” No, because w2 4 is the sum of two squares. Therefore, 7w2 28 7(w2 4). c) The terms in x4 81 have no common factors, but they are perfect squares. Identify a and b. x4 81 (x2)2 (9)2 What do you square to get x4? x2 What do you square to get 81? 9 What is the first question you should always ask yourself when you are trying to factor a polynomial? www.mhhe.com/messersmith SECTION 7.4 Factoring Special Trinomials and Binomials 417
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