How do you express 2x−1(x−1)3(x−2) in partial fractions?

1 Answer
Narad T.
Jul 14, 2017

The answer is =−1(x−1)3−3(x−1)2−3x−1+3x−2

Explanation:

Let's perform the decomposition into partial fractions

2x−1(x−1)3(x−2)=A(x−1)3+B(x−1)2+Cx−1+Dx−2

=A(x−2)+B(x−1)(x−2)+C(x−1)2(x−2)+D(x−1)3(x−1)3(x−2)

The denominators are the same, we compare the numerators

2x−1=A(x−2)+B(x−1)(x−2)+C(x−1)2(x−2))+D(x−1)3

Let x=1, ⇒, 1=−A, ⇒, A=−1

Let x=2, ⇒, 3=D

Coefficients of x3

0=C+D, ⇒, C=−D=−3

Let x=0, −1=−2A+2B−2C−D

2B=2A+2C+D−1=−2−6+3−1=−6

B=−3

Therefore,

2x−1(x−1)3(x−2)=−1(x−1)3−3(x−1)2−3x−1+3x−2

Related questions
See all questions in Partial Fraction Decomposition (Linear Denominators)
Impact of this question
9074 views around the world
You can reuse this answer
Creative Commons License