Question 113750
decompose into a partial fraction

3x^2+15x+8/(x+1)(x+2)(x-3)
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3x^2+15x+8/(x+1)(x+2)(x-3) = A/(x+1) + B/(x+2) + C/(x-3)

= A(x+2)(x-3) + B(x+1)(x-3) + C(x+1)(x+2 / ......

= A(x^2-x-6) + B(x^2-2x-3) + C(x^2+3x+2)/ ....

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A+B+C = 3
-A-2B+3C = 15
-6A-3B+2C = 8
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Using the Matrix function of a TI-83 I get:
A = 1
B = -2 
C = 4
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Cheers,
Stan H.