# How do you write the partial fraction decomposition of the rational expression  (x^2+8)/(x^2-5x+6)?

Dec 15, 2015

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

#### Explanation:

Since the degree of the denominator is not more than the degree of the numerator, we first long divide the expression to obtain that

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

The denominator ${x}^{2} + 8$ is an irreducible quadratic factor and hence has a partial fraction form given by

$\frac{- 5 x - 2}{{x}^{2} + 8} = \frac{A x + B}{{x}^{2} + 8}$

By comparing terms in the numerator of these equivalent fractions, it is then clear that $A = - 5 \mathmr{and} B = - 2$.

Hence the partial fraction decomposition for the original expression is

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