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: 31

Answer

$2(7y+2)(2y+1)$

Work Step by Step

Factoring the $GCF=2$, then the given expression, $ 28y^2+22y+4 $, is equivalent to \begin{array}{l}\require{cancel} 2(14y^2+11y+2) .\end{array} The two numbers whose product is $ac= 14(2)=28 $ and whose sum is $b= 11 $ are $\{ 4,7 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 2(14y^2+11y+2) $, is \begin{array}{l}\require{cancel} 2(14y^2+4y+7y+2) \\\\= 2[(14y^2+4y)+(7y+2)] \\\\= 2[2y(7y+2)+(7y+2)] \\\\= 2[(7y+2)(2y+1)] \\\\= 2(7y+2)(2y+1) .\end{array}
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