College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.7 - Rational Expressions - R.7 Assess Your Understanding: 91

Answer

$\dfrac{3x^2+2x}{9x^2+6x+1}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \dfrac{(3x+1)\cdot2x-x^2\cdot3}{(3x+1)^2} ,$ use the Distributive Property first. Then remove the grouping symbols and combine like terms. Finally, use special products to simplify the denominator. $\bf{\text{Solution Details:}}$ Using the Distributive Property which is given by $a(b+c)=ab+ac,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(3x\cdot2x+1\cdot2x)-x^2\cdot3}{(3x+1)^2} \\\\= \dfrac{(6x^2+2x)-3x^2}{(3x+1)^2} .\end{array} Removing the grouping symbols and then combining like terms, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{6x^2+2x-3x^2}{(3x+1)^2} \\\\= \dfrac{3x^2+2x}{(3x+1)^2} .\end{array} Using the square of a binomial which is given by $(a+b)^2=a^2+2ab+b^2$ or by $(a-b)^2=a^2-2ab+b^2,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{3x^2+2x}{(3x)^2+2(3x)(1)+(1)^2} \\\\= \dfrac{3x^2+2x}{9x^2+6x+1} .\end{array}
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