Factor the polynomial completely. $x ^ { 3 } - 8$
A) $( x - 2 ) \left( x ^ { 2 } + 2 x + 4 \right)$
B) $( x - 2 ) \left( x ^ { 2 } + 4 x + 4 \right)$
C) $( x + 2 ) \left( x ^ { 2 } - 2 x + 4 \right)$
D) $( x - 2 ) \left( x ^ { 2 } - 2 x + 4 \right)$
E) not factorable

Question

Quizplus · Accepted Answer

The polynomial $x^3 - 8$ is a difference of cubes, which can be factored as $(a^3 - b^3) = (a - b)(a^2 + ab + b^2)$. Here, $a = x$ and $b = 2$, so the factored form is $(x - 2)(x^2 + 2x + 4)$.

Factor the Polynomial Completely x3−8x ^ { 3 } - 8x3−8

Factor the Polynomial Completely $x ^ { 3 } - 8$