Factor completely.
-$6 x ^ { 2 } y ^ { 2 } + 7 x y ^ { 2 } - 20 y ^ { 2 }$
A) $y ^ { 2 } ( x - 4 ) ( 6 x + 5 )$
B) $( 3 x - 4 y ) ( 2 x + 5 y )$
C) $( 2 x - 4 y ) ( 3 x + 5 y )$
D) $y ^ { 2 } ( 3 x - 4 ) ( 2 x + 5 )$

Question

Quizplus · Accepted Answer

The expression can be factored by first taking out the common factor $y^2$, resulting in $y^2(6x^2 + 7x - 20)$. Factoring the quadratic expression inside the parentheses gives $y^2(3x - 4)(2x + 5)$.

Factor Completely 6x2y2+7xy2−20y26 x ^ { 2 } y ^ { 2 } + 7 x y ^ { 2 } - 20 y ^ { 2 }6x2y2+7xy2−20y2

Factor Completely $6 x ^ { 2 } y ^ { 2 } + 7 x y ^ { 2 } - 20 y ^ { 2 }$