Question

How do you write the partial fraction decomposition of the rational expression `(-13x+11) / (2x^3 - 2x^2 + x - 1)`?

Answer 1

`(-13x+11)/(2x^3-2x^2+x-1)=(-2/3)/(x-1)+(4/3x-35/3)/(2x^2+1)`

Explanation:

From the given `(-13x+11)/(2x^3-2x^2+x-1)`, we start by getting all the factors of the denominator

I will assume that we already know factoring, ok?

`2x^3-2x^2+x-1=(x-1)(2x^2+1)`

This is our denominator for the right sides of the equation

Let us set up the equation and the variables A, B

`(-13x+11)/(2x^3-2x^2+x-1)=A/(x-1)+(Bx+C)/(2x^2+1)`

Simplify using the LCD`=(x-1)(2x^2+1)`

`(-13x+11)/(2x^3-2x^2+x-1)=(A(2x^2+1)+(Bx+C)(x-1))/((x-1)(2x^2+1))`

Expand then simplify

`(-13x+11)/(2x^3-2x^2+x-1)=(2Ax^2+A+Bx^2-Bx+Cx-C)/((x-1)(2x^2+1))`

Rearrange from highest to lowest degree the terms in the numerator at the right side of the equation

`(-13x+11)/(2x^3-2x^2+x-1)=(2Ax^2+Bx^2-Bx+Cx+A-C)/((x-1)(2x^2+1))`

Let us match the numerical coefficients of the terms of the numerators of the left and right side of the equation

`(0*x^2+(-13)x^1+11*x^0)/(2x^3-2x^2+x-1)=((2A+B)x^2+(-B+C)x^1+(A-C)*x^0)/((x-1)(2x^2+1))`

The equations are

`2A+B=0`
`-B+C=-13`
`A-C=11`

Simultaneous solution results to

`A=-2/3`
`B=4/3`
`C=-35/3`

Our final answer

`(-13x+11)/(2x^3-2x^2+x-1)=(-2/3)/(x-1)+(4/3x-35/3)/(2x^2+1)`

God bless....I hope the explanation is useful.