Factor by grouping.
-$t^{2}+8 t+4 t+32$
A) $(t+8)(t+4)$
B) $t(t+44)$
C) $(t-8)(t-4)$
D) $(t+8)(t-4)$

Name: Factor by grouping. -$t^{2}+8 t+4 t+32$ A) $(t+8)(t+4)$ B) $t(t+44)$ C) $(t-8)(t-4)$ D) $(t+8)(t-4)$
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Question

Quizplus · Accepted Answer

First, group the terms: $t^{2}+8t$ and $4t+32$. Factoring out the common factors gives $t(t+8)+4(t+8)$. Then, factor out the common term $(t+8)$, resulting in $(t+8)(t+4)$.

Factor by Grouping t2+8t+4t+32t^{2}+8 t+4 t+32t2+8t+4t+32 A) (t+8)(t+4)(t+8)(t+4)(t+8)(t+4) B) t(t+44)t(t+44)t(t+44) C) (t−8)(t−4)(t-8)(t-4)(t−8)(t−4) D) (t+8)(t−4)(t+8)(t-4)(t+8)(t−4)

Factor by Grouping $t^{2}+8 t+4 t+32$ A) $(t+8)(t+4)$ B) $t(t+44)$ C) $(t-8)(t-4)$ D) $(t+8)(t-4)$