## 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: 27

#### Answer

$(y+3)(20y-1)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 20y^2+59y-3 \end{array} has $ac= 20(-3)=-60$ and $b= 59 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 60,-1 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 20y^2+60y-1y-3 .\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} (20y^2+60y)-(y+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 20y(y+3)-(y+3) .\end{array} Factoring the $GCF= (y+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (y+3)(20y-1) .\end{array}

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