Factor completely. If the polynomial cannot be factored, write ʺprime.ʺ
-$$2 y ^ { 3 } - 128$$
A) $2 ( y - 4 ) \left( y ^ { 2 } + 4 y + 16 \right)$
B) $2 ( y - 4 ) \left( y ^ { 2 } + 16 \right)$
C) $2 ( y + 4 ) \left( y ^ { 2 } - 4 y + 16 \right)$
D) $( 2 y - 4 ) \left( y ^ { 2 } + 4 y + 32 \right)$

Question

Quizplus · Accepted Answer

This polynomial can be factored by recognizing it as a difference of cubes: $2y^3 - 128 = 2(y^3 - 64) = 2(y^3 - 4^3)$. Factoring a difference of cubes $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ gives $2(y - 4)(y^2 + 4y + 16)$.

Factor Completely 2y3−1282 y ^ { 3 } - 1282y3−128 A) 2(y−4)(y2+4y+16)2 ( y - 4 ) \left( y ^ { 2 } + 4 y + 16 \right)2(y−4)(y2+4y+16)

Factor Completely $2 y ^ { 3 } - 128$ A) $2 ( y - 4 ) \left( y ^ { 2 } + 4 y + 16 \right)$