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: 47



Work Step by Step

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