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

Jan 21, 2017

Real solution:

${x}^{2} / \left({x}^{2} + x + 4\right) = 1 - \frac{x + 4}{{x}^{2} + x + 4}$

Complex solution:

${x}^{2} / \left({x}^{2} + x + 4\right) = 1 - \frac{\frac{1}{2} - \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} - \frac{\sqrt{15}}{2} i} - \frac{\frac{1}{2} + \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} + \frac{\sqrt{15}}{2} i}$

#### Explanation:

Note that the discriminant of ${x}^{2} + x + 4$ is negative:

$\Delta = {1}^{2} - 4 \left(1\right) \left(4\right) = 1 - 16 = - 15 < 0$

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 $\frac{A x + B}{{x}^{2} + x + 4}$, which we can do as follows:

${x}^{2} / \left({x}^{2} + x + 4\right) = \frac{\left({x}^{2} + x + 4\right) - \left(x + 4\right)}{{x}^{2} + x + 4} = 1 - \frac{x + 4}{{x}^{2} + x + 4}$

$\textcolor{w h i t e}{}$
Complex solution

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

$\frac{x + 4}{{x}^{2} + x + 4} = \frac{x + 4}{\left(x + \frac{1}{2} - \frac{\sqrt{15}}{2} i\right) \left(x + \frac{1}{2} + \frac{\sqrt{15}}{2} i\right)}$

$\textcolor{w h i t e}{\frac{x + 4}{{x}^{2} + x + 4}} = \frac{A}{x + \frac{1}{2} - \frac{\sqrt{15}}{2} i} + \frac{B}{x + \frac{1}{2} + \frac{\sqrt{15}}{2} i}$

$\textcolor{w h i t e}{\frac{x + 4}{{x}^{2} + x + 4}} = \frac{A \left(x + \frac{1}{2} + \frac{\sqrt{15}}{2} i\right) + B \left(x + \frac{1}{2} - \frac{\sqrt{15}}{2} i\right)}{{x}^{2} + x + 4}$

Equating coefficients, we find:

$\left\{\begin{matrix}A + B = 1 \\ \frac{1}{2} \left(A + B\right) + \frac{\sqrt{15}}{2} i \left(A - B\right) = 4\end{matrix}\right.$

Subtracting $\frac{1}{2}$ of the first equation from the second we get:

$\frac{\sqrt{15}}{2} i \left(A - B\right) = \frac{7}{2}$

Multiplying both sides by $- \frac{2}{\sqrt{15}} i$ we get:

$A - B = - \frac{7}{15} \sqrt{15} i$

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

$A = \frac{1}{2} \left(1 - \frac{7}{15} \sqrt{15} i\right) = \frac{1}{2} - \frac{7}{30} \sqrt{15} i$

$B = \frac{1}{2} \left(1 + \frac{7}{15} \sqrt{15} i\right) = \frac{1}{2} + \frac{7}{30} \sqrt{15} i$

So:

$\frac{x + 4}{{x}^{2} + x + 4} = \frac{\frac{1}{2} - \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} - \frac{\sqrt{15}}{2} i} + \frac{\frac{1}{2} + \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} + \frac{\sqrt{15}}{2} i}$

And:

${x}^{2} / \left({x}^{2} + x + 4\right) = 1 - \frac{\frac{1}{2} - \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} - \frac{\sqrt{15}}{2} i} - \frac{\frac{1}{2} + \frac{7}{30} \sqrt{15} i}{x + \frac{1}{2} + \frac{\sqrt{15}}{2} i}$