## 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: 39

#### Answer

$(x+1)(3x+1)$

#### Work Step by Step

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

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