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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 62

#### Answer

$24x^3(3x-2)(2x-1)$

#### Work Step by Step

Factoring the $GCF= 12x^3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 144x^5-168x^4+48x^3 \\\\= 24x^3(6x^2-7x+2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 24x^3(6x^2-7x+2) \end{array} has $ac= 6(2)=12$ and $b= -7 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -4,-3 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 24x^3(6x^2-4x-3x+2) .\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} 24x^3[(6x^2-4x)-(3x-2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 24x^3[2x(3x-2)-(3x-2)] .\end{array} Factoring the $GCF= (7x+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} 24x^3[(3x-2)(2x-1)] \\\\= 24x^3(3x-2)(2x-1) .\end{array}

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