Factor completely.
-$36 x-49 x^{3}$
A) $x(6-7 x)(6-7 x)$
B) $x(6+7 x)(6-7 x)$
C) $x(6+7 x)^{2}$
D) prime

Question

Quizplus · Accepted Answer

The expression can be factored by first taking out the common factor $x$, resulting in $x(36-49x^2)$. Recognizing $36$ and $49x^2$ as perfect squares, we can further factor it using the difference of squares formula $a^2 - b^2 = (a - b)(a + b)$, where $a = 6$ and $b = 7x$. This results in $x(6-7x)(6+7x)$, which matches option B.

Factor Completely 36x−49x336 x-49 x^{3}36x−49x3 A) x(6−7x)(6−7x)x(6-7 x)(6-7 x)x(6−7x)(6−7x) B) x(6+7x)(6−7x)x(6+7 x)(6-7 x)x(6+7x)(6−7x)

Factor Completely $36 x-49 x^{3}$ A) $x(6-7 x)(6-7 x)$ B) $x(6+7 x)(6-7 x)$