## College Algebra (11th Edition)

Published by Pearson

# Chapter R - Review Exercises - Page 76: 79

#### Answer

$\dfrac{x+1}{x+4}$

#### Work Step by Step

Factoring the expressions and then cancelling the common factors between the numerator and the denominator, the given expression, $\dfrac{x^2+x-2}{x^2+5x+6}\div\dfrac{x^2+3x-4}{x^2+4x+3} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{x^2+x-2}{x^2+5x+6}\cdot\dfrac{x^2+4x+3}{x^2+3x-4} \\\\= \dfrac{(x+2)(x-1)}{(x+2)(x+3)}\cdot\dfrac{(x+3)(x+1)}{(x+4)(x-1)} \\\\= \dfrac{(\cancel{x+2})(\cancel{x-1})}{(\cancel{x+2})(\cancel{x+3})}\cdot\dfrac{(\cancel{x+3})(x+1)}{(x+4)(\cancel{x-1})} \\\\= \dfrac{x+1}{x+4} .\end{array}

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