# How do you express (x^2+8)/(x^2-5x+6) in partial fractions?

Apr 21, 2016

$\frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} = 1 - \frac{12}{x - 2} + \frac{17}{x - 3}$

#### Explanation:

$\frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6}$

$= \frac{\left({x}^{2} - 5 x + 6\right) + \left(5 x + 2\right)}{{x}^{2} - 5 x + 6}$

$= 1 + \frac{5 x + 2}{{x}^{2} - 5 x + 6}$

$= 1 + \frac{5 x + 2}{\left(x - 2\right) \left(x - 3\right)}$

$= 1 + \frac{A}{x - 2} + \frac{B}{x - 3}$

$= 1 + \frac{A \left(x - 3\right) + B \left(x - 2\right)}{{x}^{2} - 5 x + 6}$

$= 1 + \frac{\left(A + B\right) x - \left(3 A + 2 B\right)}{{x}^{2} - 5 x + 6}$

Equating coefficients we get:

$\left\{\begin{matrix}A + B = 5 \\ 3 A + 2 B = - 2\end{matrix}\right.$

Subtract twice the first equation from the second to find:

$A = - 12$

Then substitute this value of $A$ into the first equation to find:

$B = 17$

So:

$\frac{{x}^{2} + 8}{{x}^{2} - 5 x + 6} = 1 - \frac{12}{x - 2} + \frac{17}{x - 3}$