# How do you write the partial fraction decomposition of the rational expression  10/[(x-1)(x^2+9)] ?

Jan 27, 2018

$\frac{1}{x - 1} + \frac{- x - 1}{{x}^{2} + 9}$

#### Explanation:

We can rewrite the function as:

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 9}$

and now we need to work out $A$ and $B$.

Multiply both sides by $\left(x - 1\right) \left({x}^{2} + 9\right)$ to get:

$10 = A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 1\right)$

Let $x = 1$ to cancel the second term and get:

$10 = A \left({1}^{2} + 9\right) = 10 A \implies A = 1$

Now let $x = 0$ to cancel the $B$, we get:

$10 = 1 \cdot \left({0}^{2} + 9\right) + C \left(0 - 1\right)$

$\implies 10 = 9 - C \implies C = - 1$

Finally, choose any other value of $x$ to get $B$, say $x = 2$

$\implies 10 = 1 \cdot \left({2}^{2} + 9\right) + \left(B \left(2\right) - 1\right) \left(2 - 1\right)$

$\implies 10 = 13 + 2 B - 1$

$\implies B = - 1$

Now putting our values in:

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