## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - Chapter Review Exercises - Page 288: 50

#### Answer

$(5b+1)(b-3)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the given expression, $5b^2-14b-3 ,$ find two numbers whose product is $ac$ and whose sum is $b$ in the quadratic expression $ax^2+bx+c.$ Use these $2$ numbers to decompose the middle term of the given quadratic expression and then use factoring by grouping. $\bf{\text{Solution Details:}}$ Using factoring of trinomials, the value of $ac$ in the trinomial expression above is $5(-3)=-15$ and the value of $b$ is $-14 .$ The $2$ numbers that have a product of $ac$ and a sum of $b$ are $\left\{ 1,-15 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 5b^2+b-15b-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} (5b^2+b)-(15b+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} b(5b+1)-3(5b+1) .\end{array} Factoring the $GCF= (5b+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} (5b+1)(b-3) .\end{array}

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