## 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: 71

#### Answer

$3(3x-4y)^2$

#### Work Step by Step

Factoring the $GCF= 3 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 27x^2-72xy+48y^2 \\\\= 3(9x^2-24xy+16y^2) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 3(9x^2-24xy+16y^2) \end{array} has $ac= 9(16)=144$ and $b= -24 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -12,-12 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 3(9x^2-12xy-12xy+16y^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} 3[(9x^2-12xy)-(12xy-16y^2)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3[3x(3x-4y)-4y(3x-4y)] .\end{array} Factoring the $GCF= (3x-4y)$ of the entire expression above results to \begin{array}{l}\require{cancel} 3[(3x-4y)(3x-4y)] \\\\= 3(3x-4y)^2 .\end{array}

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