YOU TRY 3 Try using 2 and 9. 3t2 29t 18 (3t 2)(t 9) Correct! 2t (27t) 29t These must both be 29t. So, 3t2 29t 18 (3t 2)(t 9). Check by multiplying. Factor completely. a) 2k2 17k 8 b) 6z2 23z 20 EXAMPLE 5 In-Class Example 5 Factor completely. a) 18k2 21k 30 b) 4h2 h 18 Answer: a) 3(6k 5)(k 2) b) (4h 9)(h 2) Factor completely. a) 16a2 62a 8 b) 2c2 3c 20 Solution a) Ask yourself, “Can I take out a common factor?” Yes! 16a2 62a 8 2(8a2 31a 4) Now, try to factor 8a2 31a 4. To get a product of 8a2, we can try either 8a and a or 4a and 2a. Let’s start by trying 8a and a. 8a2 31a 4 (8a )(a ) List pairs of integers that multiply to 4: 4 and 1, 4 and 1, 2 and 2. Try 4 and 1. Do not put 4 in the same binomial as 8a since then it would be possible to factor out 2. But, 2 does not factor out of 8a2 31a 4. Put the 4 in the same binomial as a. 8a2 31a 4 (8a 1)(a 4) a 32a 31a Correct Don’t forget that the very fi rst step was to factor out a 2. Therefore, 16a2 62a 8 2(8a2 31a 4) 2(8a 1)(a 4) Check by multiplying. b) Since the coeffi cient of the squared term is negative, begin by factoring out 1. (There is no other common factor except 1.) 2c2 3c 20 1(2c2 3c 20) Try to factor 2c2 3c 20. To get a product of 2c2, we will use 2c and c in the binomials. 2c2 3c 20 (2c )(c ) We need pairs of integers whose product is 20. They are 1 and 20, 1 and 20, 2 and 10, 2 and 10, 4 and 5, 4 and 5. Do not start with 1 and 20 or 1 and 20 because the middle term, 3c, is not very large. Using 1 and 20 or 1 and 20 would likely result in a larger middle term. Think about 2 and 10 and 2 and 10. These will not work because if we put any of these numbers in the factor containing 2c, then it will be possible to factor out 2. Be sure to read the reasoning in this example, and apply it to the You Try! SECTION 7.3 Factoring Trinomials of t www.mhhe.com/messersmith he Form ax2 bx c (a 1) 409
messersmith_power_introductory_algebra_1e_ch4_7_10
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