## Precalculus: Mathematics for Calculus, 7th Edition

$(a+1)^2(a+3)(a-1)$
$Factor$ $the$ $expression$ $completely:$ $(a^2+2a)^2 - 2(a^2+2a) - 3$ 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 $(a^2+2a+1)$ = $(a+1)(a+1)$ = $(a+1)^2$ $(a^2+2a-3)$ = $(a+3)(a-1)$ $= (a+1)^2(a+3)(a-1)$