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

Published by Pearson

# Chapter 5 - Polynomials and Factoring - Mid-Chapter Review - Mixed Review - Page 335: 11

#### Answer

$(3x-1)(4x+1)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 12x^2-x-1 \end{array} has $ac= 12(-1)=-12$ and $b= -1 .$ 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} 12x^2-4x+3x-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} (12x^2-4x)+(3x-1) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 4x(3x-1)+(3x-1) .\end{array} Factoring the $GCF= (3x-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (3x-1)(4x+1) .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.