## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson

# Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set - Page 970: 37

#### Answer

$x=\left\{ -\dfrac{5}{2},7 \right\}$

#### Work Step by Step

Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} 4x^2-18x=70 \\\\ 4x^2-18x-70=0 \\\\ \dfrac{4x^2-18x-70}{2}=\dfrac{0}{2} \\\\ 2x^2-9x-35=0 .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the given $\text{ equation }$ \begin{array}{l}\require{cancel} 2x^2-9x-35=0 \end{array} has $ac= 2(-35)=-70$ and $b= -9 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -14,5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 2x^2-14x+5x-35=0 .\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^2-14x)+(5x-35)=0 .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 2x(x-7)+5(x-7)=0 .\end{array} Factoring the $GCF= (x-7)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x-7)(2x+5)=0 .\end{array} Equating each factor to zero (Zero Product Property), and then isolating the variable, the solutions are $x=\left\{ -\dfrac{5}{2},7 \right\} .$

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