# How do you express (x^3+x^2+2x+1)/((x^2+1)(x^2+2)) in partial fractions?

##### 1 Answer
Jun 12, 2016

$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} = \frac{x}{{x}^{2} + 1} + \frac{1}{{x}^{2} + 2}$

#### Explanation:

Neither of the quadratic factors in the denominator have linear factors with Real coefficients.

So we are looking for a partial fraction decomposition of the form:

$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)}$

$= \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 2}$

$= \frac{\left(A x + B\right) \left({x}^{2} + 2\right) + \left(C x + D\right) \left({x}^{2} + 1\right)}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)}$

$= \frac{\left(A + C\right) {x}^{3} + \left(B + D\right) {x}^{2} + \left(2 A + C\right) x + \left(2 B + D\right)}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)}$

Equating coefficients we get the following system of linear equations:

$\left\{\begin{matrix}A + C = 1 \\ B + D = 1 \\ 2 A + C = 2 \\ 2 B + D = 1\end{matrix}\right.$

From the first and third equations we find $A = 1$, $C = 0$

From the second and fourth equations we find $B = 0$, $D = 1$

So:

$\frac{{x}^{3} + {x}^{2} + 2 x + 1}{\left({x}^{2} + 1\right) \left({x}^{2} + 2\right)} = \frac{x}{{x}^{2} + 1} + \frac{1}{{x}^{2} + 2}$