Question

How do you convert this to partial fractions?

`3/(x^2-x-12)`

Answer 1

`color(blue)(-3/(7(x+3)) + 3/(7(x-4)))`

Explanation:

`3/(x^2-x-12)`

Factor the denominator first:

`3/((x+3)(x-4))`

With linear factors we expect the form to be:

`3/((x+3)(x-4)) -= A/(x+3) + B/(x-4)`

Adding the fractions:

`3/((x+3)(x-4))-= (A(x-4)+B(x+3))/((x+3)(x-4))`

Because this is an identity, it follows that the coefficients on the right hand side will be identical to the coefficients on the left hand side. This will help us find A and B.

For the numerator:

`3 -= A(x-4)+B(x+3)`

Expanding:

`3 -= Ax-4A+Bx+3B`

Equating coefficients of `x`

`[x]=0`

`0 = A+B`

equating constants:

`[3]`

`3 = -4A+3B`

We now need to solve the two equations simultaneously.

`A+B=0color(white)(*)[1]`

`-4A+3B=3color(white)(*)[2]`

Multiply [1] by 4 and add to [2]:

`7B=3=>B=3/7`

Substituting in [1]:

`A + 3/7=0=>A=-3/7`

So our partial fractions are:

`color(blue)(-3/(7(x+3)) + 3/(7(x-4)))`