SOLUTION: F(x)= 3x-2 over x^2+2x-24 - x-3 over x^2-9 So far I have (x+6)(x+4) along with (x+3)(x-3) so if the x-3's cancel each other you're left with 3x-2 over (x+6)(x+4)(x+3)...is

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Question 1026867: F(x)= 3x-2 over x^2+2x-24 - x-3 over x^2-9
So far I have (x+6)(x+4) along with (x+3)(x-3) so if the x-3's cancel each other you're left with 3x-2 over (x+6)(x+4)(x+3)...is that the final?
Found 2 solutions by rothauserc, stanbon:
Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website!
x^2 +2x -24 = (x+6) * (x-4)
:
x^2 -9 = (x+3) * (x-3)
:
(3x-2) / ((x+6) * (x-4)) - (x-3) / ((x+3)*(x-3))
:
(3x-2) / ((x+6) * (x-4)) - (1 / (x+3))
:
((3x-2)*(x+3)) - ((x+6)*(x-4)) / ((x+6)*(x-4)*(x+3))
:
(3x^2+7x-6) - (x^2+2x-24)) / ((x+6)*(x-4)*(x+3))
:
*****************************************
F(x) = (2x^2+5x+18) / ((x+6)*(x-4)*(x+3))
*****************************************
:

Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
F(x)= 3x-2 over x^2+2x-24 - x-3 over x^2-9
------
F(x) = (3x-2)/[(x+6)(x-4)] - (x-3)/[(x+3)(x-3)]
----
F(x) = (3x-2)/[(x+6)(x-4)] - 1/[(x+3)
------
F(x) = [(x+3)(3x-2)-(x+6)(x-4)]/[(x+3)(x+6)(x-4)]
------
F(x) = [3x^2 +7x -6 - x^2 - 2x +24]/[(x+3)(x^2-2x-24)]
------
F(x) = [2x^2 +5x + 18]/[(x+3)(x^2-2x-24)]
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Cheers,
Stan H.
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