Question

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

Answer 1

Real solution:

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

Complex solution:

`x^2/(x^2+x+4) = 1-(1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)-(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)`

Explanation:

Note that the discriminant of `x^2+x+4` is negative:

`Delta = 1^2-4(1)(4) = 1-16 = -15 < 0`

So this quadratic is irreducible over the real numbers.

So (assuming we want real coefficients), the best we can do is split the given rational expression into a polynomial part and an expression of the form `(Ax+B)/(x^2+x+4)`, which we can do as follows:

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

`color(white)()`
Complex solution

If we want a partial fraction decomposition with Complex coefficients, then we can proceed as follows:

`(x+4)/(x^2+x+4) = (x+4)/((x+1/2-sqrt(15)/2i)(x+1/2+sqrt(15)/2i))`

`color(white)((x+4)/(x^2+x+4)) = A/(x+1/2-sqrt(15)/2i)+B/(x+1/2+sqrt(15)/2i)`

`color(white)((x+4)/(x^2+x+4)) = (A(x+1/2+sqrt(15)/2i)+B(x+1/2-sqrt(15)/2i))/(x^2+x+4)`

Equating coefficients, we find:

`{ (A+B = 1), (1/2(A+B)+sqrt(15)/2i(A-B) = 4) :}`

Subtracting `1/2` of the first equation from the second we get:

`sqrt(15)/2i(A-B) = 7/2`

Multiplying both sides by `-2/sqrt(15)i` we get:

`A-B = -7/15sqrt(15)i`

Adding or subtracting the first equation and dividing by `2` we find:

`A = 1/2(1-7/15sqrt(15)i) = 1/2-7/30sqrt(15)i`

`B = 1/2(1+7/15sqrt(15)i) = 1/2+7/30sqrt(15)i`

So:

`(x+4)/(x^2+x+4) = (1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)+(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)`

And:

`x^2/(x^2+x+4) = 1-(1/2-7/30sqrt(15)i)/(x+1/2-sqrt(15)/2i)-(1/2+7/30sqrt(15)i)/(x+1/2+sqrt(15)/2i)`