Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Common Core End-of-Course Assessment - Page 793: 9


Factor: $\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x+3)(2x-3)}{3x(2x+{3})}$ $\frac{2x-3}{3x}$

Work Step by Step

1. Factor out the numerator and denominator ${4x^2-9}$= $(2x)^{2}-3^{2}$ = $(2x+3)(2x-3)$ ${6x^{2}+{9x}}$=${3x}{(2x+3)}$ $\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x^2+3)(2x-3)}{3x(2x+{3})}$ 2. Cancel out the common term: ${2x+3}$ $\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{(2x+3)(2x-3)}{3x(2x+{3})}$ $\frac{4x^2-9}{6x^{2}+{9x}}$=$\frac{2x-3}{3x}$ 3. Final answer is $\frac{2x-3}{3x}$
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