## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 5 - Section 5.6 - Factoring Trinomials - Exercise Set - Page 303: 37

#### Answer

$(5x+2)(5x+8)$

#### Work Step by Step

Let $z=(5x+1)$. Then the given expression, $(5x+1)^2+8(5x+1)+7$, is equivalent to $z^2+8z+7$. The two numbers whose product is $ac= 1(7)=7$ and whose sum is $b= 8$ are $\{ 1,7 \}$. Using these two numbers to decompose the middle term, then the factored form of the expression, $z^2+8z+7$, is \begin{array}{l}\require{cancel} z^2+z+7z+7 \\\\= (z^2+z)+(7z+7) \\\\= z(z+1)+7(z+1) \\\\= (z+1)(z+7) .\end{array} Since $z=(5x+1)$, then, \begin{array}{l} (z+1)(z+7) \\\\= (5x+1+1)(5x+1+7) \\\\= (5x+2)(5x+8) .\end{array}

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