Question

How do you express `(x^3-x^2-5x) /( x^2-3x+2)` in partial fractions?

Answer 1

The answer is `=x+2+5/(x-1)-6/(x-2)`

Explanation:

As the degree of the numerator is greater than the degree of the denominator, perform a long division first

`color(white)(aaaa)` `x^3-x^2-5x+0` `color(white)(aaaa)` `|` `x^2-3x+2`

`color(white)(aaaa)` `x^3-3x^2+2x` `color(white)(aaaaaaa)` `|` `x+2`

`color(white)(aaaa)` `0x^3+2x^2-7x+0`

`color(white)(aaaaaaaa)` `+2x^2-6x+4`

`color(white)(aaaaaaaa)` `+0x^2-x-4`

Therefore,

`(x^3-x^2-5x)/(x^2-3x+2)=x+2-(x+4)/(x^2-3x+2)`

Perform the decomposition into partial fractions

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

`=A/(x-1)+B/(x-2)`

`=(A(x-2)+B(x-1))/((x-1)(x-2))`

Compare the numerators

`x+4=A(x-2)+B(x-1)`

Let `x=1`, `=>`, `5=-A`

Let `x=2`, `=>`, `6=B`

Finally,

`(x^3-x^2-5x)/(x^2-3x+2)=x+2+5/(x-1)-6/(x-2)`

Answer 2

`\frac{x^3-x^2-5x}{x^2-3x+2}=x+2+5/{x-1}-6/{x-2}`

Explanation:

Given rational function:

`\frac{x^3-x^2-5x}{x^2-3x+2}`

`=x+2-\frac{x+4}{x^2-3x+2}`

`=x+2-\frac{x+4}{(x-1)(x-2)}`

`=x+2-(\frac{A}{x-1}+B/{x-2})`

Comparing corresponding coefficients to find `A=-5` & `B=6`

`=x+2-(\frac{-5}{x-1}+6/{x-2})`

`=x+2+5/{x-1}-6/{x-2}`