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 6.3 - Factoring Trinomials Whose Leading Coefficient is Not 1 - Exercise Set - Page 444: 55

Answer

$(5a+3b)(3a-2b)$

Work Step by Step

Factoring by grouping: 1. Multiply the leading coefficient, a, and the constant, c. 2. Find the factors of ac whose sum is b. 3. Rewrite the middle term, bx, as a sum or difference using the factors from step 2. 4. Factor by grouping --- $15a^{2}-ab-6b^{2} =...$ Always start by searching for a GCF ... (there are none other than 1). 1. $\quad$"$ac$"$=-90b^{2} $ 2. $\quad$sum = $-b\quad$... factors: $-10b$ and $+9b$ 3. $\quad$ $15a^{2}-ab-6b^{2} = (15a^{2}-10ab)+( 9ab-6b^{2})$ 4. $\quad$... $= 5a(3a-2b)+3b(3a-2b) = (5a+3b)(3a-2b) $ Check by FOIL $F:\quad 15a^{2}$ $O:\quad -10ab$ $I:\quad +9ab$ $L:\quad -6b^{2}$ $(5a+3b)(3a-2b)$ = $15a^{2}-ab-6b^{2}$
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