Factor by any method.
-$$9 a ^ { 3 } - 25 a$$
A) $a ( 9 a + 1 ) ( a - 25 )$
B) $a ( 3 a - 5 ) ^ { 2 }$
C) $a ( 3 a + 5 ) ( 3 a - 5 )$
D) $\left( 3 a ^ { 2 } + 5 \right) ( 3 a - 5 )$

Question

Quizplus · Accepted Answer

The given expression can be factored by first factoring out the greatest common factor (GCF), which is $a$, resulting in $a(9a^2 - 25)$. The expression inside the parentheses is a difference of squares, which can be factored as $(3a)^2 - 5^2$, leading to the factorization $a(3a + 5)(3a - 5)$.

Factor by Any Method 9a3−25a9 a ^ { 3 } - 25 a9a3−25a

Factor by Any Method $9 a ^ { 3 } - 25 a$