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: 70

Answer

$x(4x-1)(3x+1)$

Work Step by Step

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