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 - Page 71: 36


$\displaystyle \frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$

Work Step by Step

$\displaystyle \frac{\frac{9x^{2}+3x-2}{12x^{2}+5x-2}}{\frac{9x^{2}-6x+1}{8x^{2}-10x-3}}=\frac{9x^{2}+3x-2}{12x^{2}+5x-2}\div\frac{9x^{2}-6x+1}{8x^{2}-10x-3}$ $=\displaystyle \frac{9x^{2}+3x-2}{12x^{2}+5x-2}\cdot\frac{8x^{2}-10x-3}{9x^{2}-6x+1}=...$ Factor each trinomial (for $ax^{2}+bx+c$ we search for factors of ac that add up to b.) $9x^{2}+3x-2=9x^{2}+6x-3x-2$ $=3x(3x+2)+-(3x+2)$ $=(3x-1)(3x+2)$ $12x^{2}+5x-2=12x^{2}+8x-3x-2$ $=4x(3x+2)-(3x+2)$ $=(4x-1)(3x+2)$ $8x^{2}-10x-3=8x^{2}-12x+2x-3$ $=4x(2x-3)+(2x-3)$ $=(4x+1)(2x-3)$ $9x^{2}-6x+1=9x^{2}-3x-3x+1$ $=3x(3x-1)-(3x-1)$ $=(3x-1)(3x-1)$ $...=\displaystyle \frac{(3x-1)(3x+2)}{(4x-1)(3x+2)}\cdot\frac{(4x+1)(2x-3)}{(3x-1)(3x-1)}=$ Common factors $(3x-1)$ and $(3x+2)$ cancel, $=\displaystyle \frac{(4x+1)(2x-3)}{(4x-1)(3x-1)}$
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