How do you write the partial fraction decomposition of the rational expression 2x−6(x−1)(x−2)2?

1 Answer
Somebody N.
Apr 9, 2018

4x−2−2(x−2)2−4x−1

Explanation:

2x−6(x−1)(x−2)2

We expect the form of the partial fractions to be:

2x−6(x−1)(x−2)2≡Ax−1+B(x−2)2+Cx−2

Adding RHS

2x−6≡A(x−2)2+B(x−1)+C(x−1)(x−2)

Because this is an identity numerators will be equal:

We are looking to find the values of A B and C.

We now set x to values that will eliminate most of the unknowns we are looking for.

2x−6≡A(x−2)2+B(x−1)+C(x−1)(x−2)

Setting x=2

2(2)−6≡A((2)−2)2+B((2)−1)+C((2)−1)((2)−2)

−2≡B

B=−2

Setting x=1

2(1)−6≡A((1)−2)2+B((1)−1)+C((1)−1)((1)−2)

−4≡A

A=−4

Setting x=0

2(0)−6≡A((0)−2)2+B((0)−1)+C((0)−1)((0)−2)

−6≡4A−B+2C

4A−B+2C=−6

We already know A and B:

4(−4)−(−2)+2C=−6

C=4

So our partial fractions are:

4x−2−2(x−2)2−4x−1

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