## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.4 Factoring Polynomials - 3.4 Exercises: 73

#### Answer

$(4x-3)(6x-5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $24x^2-38x+15 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $24(15)=360$ and the value of $b$ is $-38 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ -18,-20 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 24x^2-18x-20x+15 .\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^2-18x)-(20x-15) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 6x(4x-3)-5(4x-3) .\end{array} Factoring the $GCF= (4x-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (4x-3)(6x-5) .\end{array}

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