## Intermediate Algebra (12th Edition)

Published by Pearson

# Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises - Page 337: 43

#### Answer

$(2xz-1)(3xz+4)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $ax^2+bx+c,$ find two numbers whose product is $ac$ and whose sum is $b$. Use these $2$ numbers to decompose the middle term of the quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ In the given expression, $6x^2z^2+5xz-4 ,$ the value of $ac$ is $6(-4)=-24$ and the value of $b$ is $5 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-24\}, \{2,-12\}, \{3,-8\}, \{4,-6\}, \{-1,24\}, \{-2,12\}, \{-3,8\}, \{-4,6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,8 \}.$ Using these $2$ numbers to decompose the middle term of the given expression results to \begin{array}{l}\require{cancel} 6x^2z^2-3xz+8xz-4 .\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} (6x^2z^2-3xz)+(8xz-4) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3xz(2xz-1)+4(2xz-1) .\end{array} Factoring the $GCF= (2xz-1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (2xz-1)(3xz+4) .\end{array}

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