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 362: 87

Answer

$(y^5+3)(2y^5+1)$.

Work Step by Step

The given expression is $=2y^{10}+7y^5+3$ We can write $=2(y^5)^2+7x^5+3$ Substitute $y^5=u$. $=2(u)^2+7u+3$ Simplify. $=2u^2+7u+3$ Rewrite $7u$ as $6u+1u$. $=2u^2+6u+1u+3$ Factor. $=2u(u+3)+1(u+3)$ $=(u+3)(2u+1)$. Substitute back $u=y^5$. $=(y^5+3)(2y^5+1)$.
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