## 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: 18

#### Answer

$5(t+3)(3t-5)$

#### Work Step by Step

Factoring the $GCF= 5 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 15t^2+20t-75 \\\\= 5(3t^2+4t-15) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 5(3t^2+4t-15) \end{array} has $ac= 3(-15)=-45$ and $b= 4 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 9,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5(3t^2+9t-5t-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} 5[(3t^2+9t)-(5t+15)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 5[3t(t+3)-5(t+3)] .\end{array} Factoring the $GCF= (t+2)$ of the entire expression above results to \begin{array}{l}\require{cancel} 5[(t+3)(3t-5)] \\\\= 5(t+3)(3t-5) .\end{array}

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