How do you express 4x3−22x2+33x−6(x−3)3(x+2) in partial fractions?

1 Answer
Narad T. · Shwetank Mauria
Nov 8, 2016

The decomposition is =35(x−3)3+4225(x−3)2+308125x−3+192125x+2

Explanation:

The decomposition is 4x3−22x2+33x−6(x−3)3(x+2)=A(x−3)3+B(x−3)2+Cx−3+Dx+2
=A(x+2)+B(x−3)(x+2)+C(x−3)2(x+2)+D(x−3)3

4x3−22x2+33x−6=A(x+2)+B(x−3)(x+2)+C(x−3)2(x+2)+D(x−3)3
let x=3⇒5A=3⇒A=35
let x=−2⇒−125D=−192⇒D=192125
coefficients of x3⇒C+D=4⇒C=4−192125=308125
and putting x=0, we get −6=2A−6B+18C−27D

or ⇒6B=6+65+18⋅30825−27⋅192125

= 365+9(616125−57625)

= 365+360125

= 18025+7225=25225

⇒B=4225

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