Factor completely.
-$$x ^ { 2 } - 6 x + 9 - y ^ { 2 }$$
A) $( x + 3 - y ) ^ { 2 }$
B) $( x - 3 + y ) ( x - 3 - y )$
C) $( x - 3 - y ) ^ { 2 }$
D) $( x + 3 + y ) ( x + 3 - y )$

Name: Factor completely. -\[x ^ { 2 } - 6 x + 9 - y ^ { 2 }\] A) $( x + 3 - y ) ^ { 2 }$ B) $( x - 3 + y ) ( x - 3 - y )$ C) $( x - 3 - y ) ^ { 2 }$ D) $( x + 3 + y ) ( x + 3 - y )$
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Question

Quizplus · Accepted Answer

The given expression is a difference of squares: $x^2 - 6x + 9 - y^2 = (x-3)^2 - y^2$, which can be factored as $(x-3+y)(x-3-y)$.

Factor Completely x2−6x+9−y2x ^ { 2 } - 6 x + 9 - y ^ { 2 }x2−6x+9−y2

Factor Completely $x ^ { 2 } - 6 x + 9 - y ^ { 2 }$