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 - Page 326: 61



Work Step by Step

Factoring the $GCF= 3x ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 168x^3+45x^2+3x \\\\= 3x(56x^2+15x+1) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3x(56x^2+15x+1) \end{array} has $ac= 56(1)=56 $ and $b= 15 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 8,7 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3x(56x^2+8x+7x+1) .\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[(56x^2+8x)+(7x+1)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3x[8x(7x+1)+(7x+1)] .\end{array} Factoring the $GCF= (7x+1) $ of the entire expression above results to \begin{array}{l}\require{cancel} 3x[(7x+1)(8x+1)] \\\\= 3x(7x+1)(8x+1) .\end{array}
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