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: 79



Work Step by Step

The two numbers whose product is $ac= 1(-18)=-18 $ and whose sum is $b= 3 $ are $\{ 6,-3 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ (x-4)^2+3(x-4)-18 $, is \begin{array}{l}\require{cancel} (x-4)^2+6(x-4)-3(x-4)-18 \\\\= [(x-4)^2+6(x-4)]-[3(x-4)+18] \\\\= (x-4)[(x-4)+6]-3[(x-4)+6] \\\\= [(x-4)+6][(x-4)-3] \\\\= [x-4+6][x-4-3] \\\\= (x+2)(x-7) .\end{array}
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