Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set: 57

Answer

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

Work Step by Step

Factoring the $GCF= 3x ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 18x^3+21x^2-9x \\\\= 3x(6x^2+7x-3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3x(6x^2+7x-3) \end{array} has $ac= 6(-3)=-18 $ and $b= 7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 9,-2 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x(6x^2+9x-2x-3) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} 3x[(6x^2+9x)-(2x+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x[3x(2x+3)-(2x+3)] .\end{array} Factoring the $GCF= (2x+3) $ of the entire expression above results to \begin{array}{l}\require{cancel} 3x[(2x+3)(3x-1)] \\\\= 3x(2x+3)(3x-1) .\end{array}
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