Question

How do you express `(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)` as partial fractions?

Answer 1

`(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) = x^2 + 3/(x-2) - (x+1)/(x^2+1)`

Explanation:

First note that:

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

Is the numerator divisible by either of these factors?

It is not divisible by `(x-2)` since `5` is odd.

How about `(x^2+1)` ?

`x^5-2x^4+x^3+x+5 = (x^2+1)(x^3-2x^2+2)+(x+3)`

So there are no common factors of the numerator and denominator.

Next note that:

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

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

`color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + A/(x-2) + (Bx+C)/(x^2+1)`

`color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + (A(x^2+1) + (Bx+C)(x-2))/(x^2-2x^2+x-2)`

`color(white)((x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2)) = x^2 + ((A+B)x^2+(-2B+C)x+(A-2C))/(x^2-2x^2+x-2)`

Equating coefficients we get the following system of linear equations:

`{(A+B=2), (-2B+C=1), (A-2C=5) :}`

Subtracting the third equation from the first, we get:

`B+2C = -3`

Adding twice this to the second equation, we find:

`5C = -5`

Hence:

`C = -1`

Substituting this value of `C` into the second equation we find:

`-2B-1=1`

Hence:

`B=-1`

Then from the first equation we find:

`A = 2-B = 3`

So:

`(x^5-2x^4+x^3+x+5)/(x^3-2x^2+x-2) = x^2 + 3/(x-2) - (x+1)/(x^2+1)`