Factor.
-$108 x ^ { 2 } + 63 x y + 9 y ^ { 2 }$
A) $( 3 x + y ) ( 9 x + 3 y )$
B) $( 27 x + 9 y ) ( 4 x + y )$
C) $9 ( 3 x + y ) ( 4 x + y )$
D) $9 ( 3 x - y ) ( 4 x - y )$

Question

Quizplus · Accepted Answer

This expression can be factored by first taking out the greatest common factor (GCF), which is 9. This leaves us with $9(12x^2 + 7xy + y^2)$. The trinomial inside the parentheses can be factored further into $(3x + y)(4x + y)$, resulting in the final factorization of $9(3x + y)(4x + y)$.

Factor - 108x2+63xy+9y2108 x ^ { 2 } + 63 x y + 9 y ^ { 2 }108x2+63xy+9y2

Factor
- $108 x ^ { 2 } + 63 x y + 9 y ^ { 2 }$