Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.3 - Algebraic Expressions - 1.3 Exercises: 124

Answer

$ y^4(y+2)^3(y+1)^2$

Work Step by Step

$Factor$ $the$ $expression$ $completely:$ $y^4(y+2)^3 + y^5(y+2)^4$ Factor out $y^4(y+2)^3$ from both terms $y^4(y+2)^3[1 + y(y+2)]$ Distribute the y to $(y+2)$ in $[1 + y(y+2)]$ $= y^4(y+2)^3[1 + y^2 + 2y]$ $ = y^4(y+2)^3[y^2 + 2y + 1]$ Factor the trinomial in the bracket into binomials $y^2 + 2y + 1$ = $(y+1)(y+1)$ = $(y+1)^2$ $ = y^4(y+2)^3(y+1)^2$
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