Factor the polynomial completely
-$9 x^{2}-24 x+16-y^{2}$
A) $(3 x-4+y)(3 x-4-y)$
B) $(3 x-4+y)(3 x+4-y)$
C) $(3 x+4+y)(3 x+4-y)$
D) Prime

Name: Factor the polynomial completely -$9 x^{2}-24 x+16-y^{2}$ A) $(3 x-4+y)(3 x-4-y)$ B) $(3 x-4+y)(3 x+4-y)$ C) $(3 x+4+y)(3 x+4-y)$ D) Prime
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Question

Quizplus · Accepted Answer

The given polynomial can be rewritten as $(3x)^2 - 2\cdot(3x)\cdot4 + 4^2 - y^2$, which is a difference of squares: $[(3x-4)^2 - y^2]$. Factoring this as a difference of squares gives $(3x-4+y)(3x-4-y)$.

Factor the Polynomial Completely - 9x2−24x+16−y29 x^{2}-24 x+16-y^{2}9x2−24x+16−y2 A) (3x−4+y)(3x−4−y)(3 x-4+y)(3 x-4-y)(3x−4+y)(3x−4−y)

Factor the Polynomial Completely
- $9 x^{2}-24 x+16-y^{2}$ A) $(3 x-4+y)(3 x-4-y)$