Question

How do you write the partial fraction decomposition of the rational expression `(x-2) /( x^2+4x+3)`?

Answer 1

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

Explanation:

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

`color(red) "Factorize the Denominator"`
`x^2+4x+3 = (x+1)(x+3)`

Our expression becomes

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

This expression can be written as

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

Equating the Numerators
`x-2 = A(x+2)+B(x+1)`

Now we have to solve for `A` and `B`

Let us take `x=-1` for making `x+1 =0` and thus removing `B`.

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

Now let us use `x=-2`

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

`B=4`

Substituting the value of `A` and `B` in

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

Our final answer is

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

Rewriting

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