Factor completely.
-$98 x^{2}-50 y^{2}$
A) $2(7 x-5 y)^{2}$
B) $(14 x+10 y)(7 x-5 y)$
C) $(7 x+5 y)(14 x-10 y)$
D) $2(7 x+5 y)(7 x-5 y)$

Name: Factor completely. -$98 x^{2}-50 y^{2}$ A) $2(7 x-5 y)^{2}$ B) $(14 x+10 y)(7 x-5 y)$ C) $(7 x+5 y)(14 x-10 y)$ D) $2(7 x+5 y)(7 x-5 y)$
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Question

Quizplus · Accepted Answer

This expression is a difference of squares, which can be factored as $a^2 - b^2 = (a+b)(a-b)$. Here, $98x^2 = (7x)^2$ and $50y^2 = (5y)^2$, so the expression becomes $(7x)^2 - (5y)^2$, which factors into $2(7x+5y)(7x-5y)$.

Factor Completely 98x2−50y298 x^{2}-50 y^{2}98x2−50y2 A) 2(7x−5y)22(7 x-5 y)^{2}2(7x−5y)2 B) (14x+10y)(7x−5y)(14 x+10 y)(7 x-5 y)(14x+10y)(7x−5y)

Factor Completely $98 x^{2}-50 y^{2}$ A) $2(7 x-5 y)^{2}$ B) $(14 x+10 y)(7 x-5 y)$