How do you factor completely $x^2 - y^2 +xz - yz$ ?

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Answer 1 · 2016-05-05T22:24:46

$x^2-y^2+xz-yz=(x-y)(x+y+z)$

$x^2-y^2+xz-yz$

$=(x^2-y^2)+(xz-yz)$

$=(x-y)(x+y)+(x-y)z$

$=(x-y)((x+y)+z)$

$=(x-y)(x+y+z)$

Note the step where we replace $(x^2-y^2)$ with $(x-y)(x+y)$

The identity:

$x^2-y^2=(x-y)(x+y)$

is known as the difference of squares identity.

How do you factor completely x^2 - y^2 +xz - yzx^2 - y^2 +xz - yz?