Factor completely.
-$4 x^{3}+32 x^{2}+5 x+40$
A) $(4 x+5)(4 x-5)(x+8)$
B) $\left(4 x^{2}-5\right)(x-8)$
C) $\left(x^{2}+5\right)(4 x+8)$
D) $\left(4 x^{2}+5\right)(x+8)$

Name: Factor completely. -$4 x^{3}+32 x^{2}+5 x+40$ A) $(4 x+5)(4 x-5)(x+8)$ B) $\left(4 x^{2}-5\right)(x-8)$ C) $\left(x^{2}+5\right)(4 x+8)$ D) $\left(4 x^{2}+5\right)(x+8)$
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Question

Quizplus · Accepted Answer

The correct factorization is obtained by grouping and factoring out common terms: $4x^3 + 32x^2 + 5x + 40 = 4x^2(x + 8) + 5(x + 8) = (4x^2 + 5)(x + 8)$.

Factor Completely 4x3+32x2+5x+404 x^{3}+32 x^{2}+5 x+404x3+32x2+5x+40 A) (4x+5)(4x−5)(x+8)(4 x+5)(4 x-5)(x+8)(4x+5)(4x−5)(x+8) B) (4x2−5)(x−8)\left(4 x^{2}-5\right)(x-8)(4x2−5)(x−8)

Factor Completely $4 x^{3}+32 x^{2}+5 x+40$ A) $(4 x+5)(4 x-5)(x+8)$ B) $\left(4 x^{2}-5\right)(x-8)$