Can someone factor u^3-v^3-v^2v+uv^2 and x^2-y^2-4y-4
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Can someone factor u^3-v^3-v^2v+uv^2 and x^2-y^2-4y-4

[From: ] [author: ] [Date: 11-10-24] [Hit: ]
If so,Hoping this helps!PS. The question answered above is wrong because they answered it assuming the eqution goes like,........
Step by step please. Thanks

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the third term is u^2v, right?
If so, put in standard order first:

u^3-v^3-u^2v+uv^2=u^3-u^2v+uv^2-v^3

Now factor by grouping:
= u^2(u-v) +v^2(u-v)

=(u-v)(u^2+v^2)
------------------------Edit
If you actually did mean u^3-v^3-v^3+uv^2
Then (u^3-v^3)+v^2(u-v)
= (u-v)(u^2+uv+v^2) +v^2(u-v)
= (u-v)(u^2+uv+v^2+v^2)
=(u-v)(u^2+uv+2v^2)

2) x^2-y^2-4y-4= x^2-(y^2+4y+4)

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

Factor as difference of squares:

[x+(y+2)][x-(y+2)]

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

Hoping this helps!

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Equation: u^3-v^3-v^2v+uv^2

answer: u^3-v^3-(v^2)(v)+uv^2
u^3-v^3-v^3+uv^2
(u-v)(v^2+uv+2v^2)

Equation: x^2-y^2-4y-4

answer: (x-y-2)(x+y+2)

PS. The question answered above is wrong because they answered it assuming the eqution goes like,... u^2v..., but in real, you have written the equation as ...v^2V...
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