Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.4 - Factoring Trinomials - Exercise Set - Page 361: 52



Work Step by Step

$8y^{2}+10y+3=$ ... Searching for two factors of $ac=+24$ whose sum is $b=+10,$ we find$\qquad+6$ and $+4.$ Rewrite the middle term and factor in pairs: $=8y^{2}+4y+6y+3=$ $=4y(2y+1)+3(2y+1)$ = $(2y+1)(4y+3)$
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