Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set: 71

Answer

$(3a-5b)(4a-3b)$

Work Step by Step

The two numbers whose product is $ac= 12(15)=180 $ and whose sum is $b= -29 $ are $\{ -20,-9 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 12a^2-29ab+15b^2 $, is \begin{array}{l}\require{cancel} 12a^2-20ab-9ab+15b^2 \\\\= (12a^2-20ab)-(9ab-15b^2) \\\\= 4a(3a-5b)-3b(3a-5b) \\\\= (3a-5b)(4a-3b) .\end{array}
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