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: 80



Work Step by Step

The two numbers whose product is $ac= 1(-8)=-8 $ and whose sum is $b= -2 $ are $\{ -4,2 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ (x-3)^2-2(x-3)-8 $, is \begin{array}{l}\require{cancel} (x-3)^2-4(x-3)+2(x-3)-8 \\\\= [(x-3)^2-4(x-3)]+[2(x-3)-8] \\\\= (x-3)[(x-3)-4]+2[(x-3)-4] \\\\= [(x-3)-4][(x-3)+2] \\\\= [x-3-4][x-3+2] \\\\= (x-7)(x-1) .\end{array}
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