Question

How do you express `(x^2+8)/(x^2-5x+6)` in partial fractions?

Answer 1

`(x^2+8)/(x^2-5x+6)=1-12/(x-2)+17/(x-3)`

Explanation:

`(x^2+8)/(x^2-5x+6)`

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

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

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

`=1+A/(x-2)+B/(x-3)`

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

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

Equating coefficients we get:

`{ (A+B=5), (3A+2B=-2) :}`

Subtract twice the first equation from the second to find:

`A = -12`

Then substitute this value of `A` into the first equation to find:

`B = 17`

So:

`(x^2+8)/(x^2-5x+6)=1-12/(x-2)+17/(x-3)`