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: 126

Answer

$(a+1)^2(a+3)(a-1)$

Work Step by Step

$Factor$ $the$ $expression$ $completely:$ $(a^2+2a)^2 - 2(a^2+2a) - 3$ We can identify the expression is in trinomial form, so we can factor it to two binomials $(a^2+2a)^2 - 2(a^2+2a) - 3$ = $((a^2+2a)+1)((a^2+2a)-3)$ Simplify the binomials $= (a^2+2a+1)(a^2+2a-3)$ Since we simplified, the two binomials turned into two trinomials, so we factor those two trinomials to two binomials each $(a^2+2a+1)$ = $(a+1)(a+1)$ = $(a+1)^2$ $(a^2+2a-3)$ = $(a+3)(a-1)$ $= (a+1)^2(a+3)(a-1)$
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