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 - Page 304: 69



Work Step by Step

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