## 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: 31

#### Answer

$-(x-3)(2x+5)$

#### Work Step by Step

Rearranging the terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -2x^2+15+x \\\\= -2x^2+x+15 .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -2x^2+x+15 \end{array} has $ac= -2(15)=-30$ and $b= 1 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 6,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -2x^2+6x-5x+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} (-2x^2+6x)-(5x-15) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -2x(x-3)-5(x-3) .\end{array} Factoring the $GCF= (x-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-3)(-2x-5) \\\\= -(x-3)(2x+5) .\end{array}

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