## 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: 15

#### Answer

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

#### Work Step by Step

Factoring the $GCF= 2 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 6x^2-10x-4 \\\\= 2(3x^2-5x-2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 2(3x^2-5x-2) \end{array} has $ac= 3(-2)=-6$ and $b= -5 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -6,1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2(3x^2-6x+1x-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} 2[(3x^2-6x)+(x-2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2[3x(x-2)+(x-2)] .\end{array} Factoring the $GCF= (x-2)$ of the entire expression above results to \begin{array}{l}\require{cancel} 2[(x-2)(3x+1)] \\\\= 2(x-2)(3x+1) .\end{array}

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