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: 82


$\displaystyle \frac{1}{9}(3x+1)^{2}$

Work Step by Step

Write all coefficients with a common denominator: $\displaystyle \frac{9}{9}x^{2}+\frac{6}{9}x+\frac{1}{9}$ ...and factor out the gcf, $\displaystyle \frac{1}{9}...$ $...=\displaystyle \frac{1}{9}(9x^{2}+6x+1)$ ... breaking $6x$ into $3x+3x$, we can factor in groups: $... =\displaystyle \frac{1}{9}[(9x^{2}+3x)+(3x+1)]$ $... =\displaystyle \frac{1}{9}[3x(3x+1)+1(3x+1)]$ = $\displaystyle \frac{1}{9}(3x+1)(3x+1)$ = $\displaystyle \frac{1}{9}(3x+1)^{2}$
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