## 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: 66

#### Answer

$5(6y-5)(2y+9)$

#### Work Step by Step

Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 220y+60y^2-225 \\\\= 60y^2+220y-225 \\\\= 5(12y^2+44y-45) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(12y^2+44y-45) \end{array} has $ac= 12(-45)=-540$ and $b= 44 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -10,54 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(12y^2-10y+54y-45) .\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} 5[(12y^2-10y)+(54y-45)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[2y(6y-5)+9(6y-5)] .\end{array} Factoring the $GCF= (6y-5)$ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(6y-5)(2y+9)] \\\\= 5(6y-5)(2y+9) .\end{array}

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