Question

How do you write the partial fraction decomposition of the rational expression `(7x^2 - 12x + 11) / (2x^3 - 5x^2 + x +2)`?

Answer 1

The answer is `=-2/(x-1)+5/(2x+1)+3/(x-2)`

Explanation:

We must factorise the denominator

Let `f(x)=2x^3-5x^2+x+2`

`f(1)=2-5+1+2=0`

Therefore, `(x-1)` is a factor of `f(x)`

To find the other factors, we have to do a long division

`color(white)(aaaa)` `2x^3-5x^2+x+2` `color(white)(aaaa)` `∥` `x-1`

`color(white)(aaaa)` `2x^3-2x^2` `color(white)(aaaaaaaaaaa)` `∥` `2x^2-3x-2`

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

`color(white)(aaaaaaaa)` `-3x^2+3x`

`color(white)(aaaaaaaaaaa)` `0-2x+2`

`color(white)(aaaaaaaaaaaaa)` `-2x+2`

`color(white)(aaaaaaaaaaaaaa)` `-0+0`

Therefore,

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

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

Now, we can perform our decomposition into partial fractions

`(7x^2-12x+11)/(2x^3-5x^2+x+2)=(7x^2-12x+11)/((x-1)(2x+1)(x-2))`

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

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

Therefore,

`7x^2-12x+11=A(2x+1)(x-2)+B(x-1)(x-2)+C(2x+1)(x-1)`

Let `x=1`, `=>`, `6=-3A`, `=>`, `A=-2`

Let `x=2`, `=>`, `15=5C`, `=>`, `C=3`

Coefficients of `x^2`, `=>`, `7=2A+B+2C`

`B=7-2A-2C=7+4-6=5`

So,

`(7x^2-12x+11)/(2x^3-5x^2+x+2)=-2/(x-1)+5/(2x+1)+3/(x-2)`