## 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: 19

#### Answer

$(2-3x)(5-4x)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 10-23x+12x^2 \end{array} has $ac= 10(12)=120$ and $b= -23 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -15,-8 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 10-15x-8x+12x^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} (10-15x)-(8x-12x^2) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5(2-3x)-4x(2-3x) .\end{array} Factoring the $GCF= (2-3x)$ of the entire expression above results to \begin{array}{l}\require{cancel} (2-3x)(5-4x) .\end{array}

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