SOLUTION: determine the partial fraction expansion of the rational functionh(x)= 1/(x-1)^2 (x^2+2)

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Question 983476: determine the partial fraction expansion of the rational functionh(x)= 1/(x-1)^2 (x^2+2)
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Including all needed parentheses, h(x)= 1/((x-1)^2 (x^2+2))


Two constant variables for the linear denominator, and one linear variable expression for the quadratic denominator.
The reason for two variables for the linear denominator is that this denominator occurs as factor twice (see the exponent, "2").




Ignoring the h(x) member, multiply the other two members by the original denominator.

1=A%28x-1%29%28x%5E2%2B2%29%2BB%28x%5E2%2B2%29%2B%28Cx%2BD%29%28x-1%29

Continue with the simplifications.
Most of the rest of the steps through....

A%28x%5E3%2B2x-x%5E2-2%29%2BBx%5E2%2B2B%2BCx%5E2%2BDx-Cx-D
Ax%5E3%2B2Ax-Ax%5E2-2%2BBx%5E2%2B2B%2BCx%5E2%2BDx-Cx-D
Ax%5E3%2B%28B%2BC-A%29x%5E2%2B%282A-C%2BD%29x%2B%282B-D%29

Left member has only one constant term for an x^0 coefficient.
highlight%28A=0%29

system%28A=0%2CB%2BC-A=0%2C2A-C%2BD=0%2C2B-D=1%29
system%28B%2BC=0%2CC-D=0%2C2B-D=1%29
----C=D----
system%28B%2BC=0%2C2B-C=1%29
----C=2B----useless.
---B=-C---
2%28-C%29-C=1
-2C-C=1
-3C=1
highlight%28C=-%281%2F3%29%29

B%2BC=0
B=-C
highlight%28B=1%2F3%29

Already determined C and D are equal.
highlight%28D=-%281%2F3%29%29