Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.2 - Factoring Trinomials Whose Leading Coefficient is 1 - Exercise Set - Page 436: 81


$\displaystyle \frac{1}{25}(5x-1)^{2}$

Work Step by Step

$\displaystyle \frac{25}{25}x^{2}-\frac{10}{25}x+\frac{1}{25}$ ...and factor out the gcf, $\displaystyle \frac{1}{25}...$ $...=\displaystyle \frac{1}{25}(25x^{2}-10x+1)$ ... breaking $-10x$ into $-5x-5x$, we can factor in groups: $... =\displaystyle \frac{1}{25}[(25x^{2}-5x)+(-5x+1)]$ $... =\displaystyle \frac{1}{25}[5x(5x-1)-1(5x-1)]$ = $\displaystyle \frac{1}{25}(5x-1)(5x-1)$ = $\displaystyle \frac{1}{25}(5x-1)^{2}$
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