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: 53

Answer

$3x^2(2x+1)(3x+2)$

Work Step by Step

Factoring the $GCF=3x^2$, then the given expression, $ 18x^4+21x^3+6x^2 $, is equivalent to $ 3x^2(6x^2+7x+2) $.\\ The two numbers whose product is $ac= 6(2)=12 $ and whose sum is $b= 7 $ are $\{ 3,4 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $ 3x^2(6x^2+7x+2) $, is \begin{array}{l}\require{cancel} 3x^2(6x^2+3x+4x+2) \\\\= 3x^2[(6x^2+3x)+(4x+2)] \\\\= 3x^2[3x(2x+1)+2(2x+1)] \\\\= 3x^2[(2x+1)(3x+2)] \\\\= 3x^2(2x+1)(3x+2) .\end{array}
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