# How do I decompose the rational expression (-x^2+9x+9)/((x-5)(x^2+4)) into partial fractions?

Nov 29, 2014

$\frac{- {x}^{2} + 9 x + 9}{\left(x - 5\right) \left({x}^{2} + 4\right)} = \frac{A}{x - 5} + \frac{B x + C}{{x}^{2} + 4}$

by finding the common denominator,

$= \frac{A \left({x}^{2} + 4\right) + \left(B x + C\right) \left(x - 5\right)}{\left(x - 5\right) \left({x}^{2} + 4\right)}$

by multiplying out the numerator,

$= \frac{A {x}^{2} + 4 A + B {x}^{2} - 5 B x + C x - 5 C}{\left(x - 5\right) \left({x}^{2} + 4\right)}$

by collecting like terms,

$= \frac{\left(A + B\right) {x}^{2} + \left(C - 5 B\right) x + \left(4 A - 5 C\right)}{\left(x - 5\right) \left({x}^{2} + 4\right)}$

by comparing the coefficients of the numerators,

$\implies \left\{\begin{matrix}A + B = - 1 \implies 5 A + 5 B = - 5 \text{ ... (Eq. 1)" \\ C-5B=9" ... (Eq. 2)" \\ 4A-5C=9" ... (Eq. 3)}\end{matrix}\right.$

by adding (Eq. 1) and (Eq. 2), and by multiplying by $5$,

$\implies 5 A + C = 4 \implies 25 A + 5 C = 20 \text{ ... (Eq. 4)}$

by adding (Eq. 3) and (Eq. 4),

$29 A = 29 \implies A = 1$

by plugging $A = 1$ into (Eq. 1),

$\implies \left(1\right) + B = - 1 \implies B = - 2$

by plugging $B = - 2$ into (Eq. 2),

$\implies C - 5 \left(- 2\right) = 9 \implies C = - 1$

Hence, we have the partial fractions

$\frac{- {x}^{2} + 9 x + 9}{\left(x - 5\right) \left({x}^{2} + 4\right)} = \frac{1}{x - 5} + \frac{- 2 x - 1}{{x}^{2} + 4}$

I hope that this was helpful.