# how was this factored?

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• Dec 13th 2012, 12:10 PM
kingsolomonsgrave
how was this factored?
Attachment 26210

I can't figure out how this is true. how was this factored?
• Dec 13th 2012, 12:17 PM
Plato
Re: how was this factored?
Quote:

Originally Posted by kingsolomonsgrave
Attachment 26210
I can't figure out how this is true. how was this factored?

$x^2-x-6=(x+2)(x-3)$.
• Dec 13th 2012, 12:21 PM
kingsolomonsgrave
Re: how was this factored?
what happened to the x in the numerator?
• Dec 13th 2012, 12:24 PM
Periwinkle38
Re: how was this factored?
In order to subtract the two functions they must first have the same denominator.
If you multiply 4/(x-3) by x+2 you should get (4x+8) / (x^2-x-6)'
Then multiply 1/(x+2) by x-3 and your answer should be (x-3) / (x^2-6).

You then combine the numerator by like terms all over the denominator. It should look like (4x-x+8-(-3) / (x^2-x-6)
4x - x = 3x and 8 - (-3) = 11 (Using skip change change)
• Dec 13th 2012, 12:28 PM
skeeter
Re: how was this factored?
Quote:

Originally Posted by kingsolomonsgrave
Attachment 26210

I can't figure out how this is true. how was this factored?

Partial-Fraction Decomposition: General Techniques