# How do you express  (x^2-16x+9)/(x^4+10x^2+9) in partial fractions?

Sep 23, 2017

$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{- 2 x + 1}{{x}^{2} + 1} + \frac{2 x}{{x}^{2} + 9}$

#### Explanation:

$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{{x}^{2} - 16 x + 9}{\left({x}^{2} + 1\right) \left({x}^{2} + 9\right)}$

$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{A x + B}{{x}^{2} + 1} + \frac{C x + D}{{x}^{2} + 9}$

$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{\left(A x + B\right) \left({x}^{2} + 9\right) + \left(C x + D\right) \left({x}^{2} + 1\right)}{{x}^{4} + 10 {x}^{2} + 9}$

$\textcolor{w h i t e}{\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9}} = \frac{\left(A + C\right) {x}^{3} + \left(B + D\right) {x}^{2} + \left(9 A + C\right) x + \left(9 B + D\right)}{{x}^{4} + 10 {x}^{2} + 9}$

So equating coefficients, we get:

$\left\{\begin{matrix}A + C = 0 \\ B + D = 1 \\ 9 A + C = - 16 \\ 9 B + D = 9\end{matrix}\right.$

Subtracting the first equation from the third, we get:

$8 A = - 16$

and hence $A = - 2$, $C = 2$

Subtracting the second equation from the fourth, we get:

$8 B = 8$

and hence $B = 1$, $D = 0$

So:

$\frac{{x}^{2} - 16 x + 9}{{x}^{4} + 10 {x}^{2} + 9} = \frac{- 2 x + 1}{{x}^{2} + 1} + \frac{2 x}{{x}^{2} + 9}$