Factor the polynomial completely. If the polynomial cannot be factored, write "prime."
-$16 x - x ^ { 5 }$
A) $x \left( x ^ { 2 } + 4 \right) ^ { 2 }$
B) $x \left( 4 + x ^ { 2 } \right) ( 2 - x ) ( 2 + x )$
C) $x \left( x ^ { 2 } - 4 \right) ^ { 2 }$
D) prime

Question

Quizplus · Accepted Answer

The polynomial $16x - x^5$ can be factored by first factoring out the greatest common factor, which is $x$, giving $x(16 - x^4)$. Recognizing $16 - x^4$ as a difference of squares, it can be further factored into $(4^2 - (x^2)^2)$, which is $(4 - x^2)(4 + x^2)$. The term $(4 + x^2)$ cannot be factored further over the real numbers, but $(4 - x^2)$ can be factored once more into $(2 - x)(2 + x)$, leading to the final factorization of $x(4 + x^2)(2 - x)(2 + x)$.

Factor the Polynomial Completely 16x−x516 x - x ^ { 5 }16x−x5

Factor the Polynomial Completely $16 x - x ^ { 5 }$