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: 36


$(2y-5)(8y-3) $

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 --- $16y^{2}-46y+15 =...$ Always start by searching for a GCF ... (there are none other than 1). 1. $\quad ac=+240\qquad $ 2. $\quad$sum = $-46 \quad$... factors: $-40$ and $-6$ 3. $\quad 16y^{2}-46y+15 = (16y^{2}-40y)+(-6y+15)$ 4. $\quad$... $= 8y(2y-5)+(-3)(2y-5)=(2y-5)(8y-3)$ Check by FOIL $F:\quad 16y^{2}$ $O:\quad -6y$ $I:\quad -40y$ $L:\quad +15$ $(2y-5)(8y-3) $ = $16y^{2}-46y+15$
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