# How do you express x+22 div x^2+2x-8 in partial fractions?

##### 1 Answer
Mar 9, 2016

Partial fractions of $\frac{x + 2}{\left(x - 2\right) \left(x + 4\right)}$ are $\frac{4}{x - 2} - \frac{3}{x + 4}$

#### Explanation:

To express $\frac{x + 22}{{x}^{2} + 2 x - 8}$ in partial fractions, first let us factorize the denominator $\left({x}^{2} + 2 x - 8\right)$. This can be done as follows:

x^2+2x-8=x^2+4x-2x-8)=x(x+4)-2(x+4)=(x-2)(x+4)

Hence $\frac{x + 22}{{x}^{2} + 2 x - 8}$ can be written as $\frac{x + 22}{\left(x - 2\right) \left(x + 4\right)}$.

Let the partial fractions be $\frac{x + 22}{\left(x - 2\right) \left(x + 4\right)} = \frac{A}{x - 2} + \frac{B}{x + 4}$

Simplifying RHS, we get $\frac{A \left(x + 4\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x + 4\right)}$ or

$\frac{\left(A + B\right) x + \left(4 A - 2 B\right)}{\left(x - 2\right) \left(x + 4\right)}$ and as this is equivalent to$\frac{x + 2}{\left(x - 2\right) \left(x + 4\right)}$, we have

$A + B = 1$ and $4 A - 2 B = 22$. From first we get, $B = 1 - A$ and putting this in second we get $4 A - 2 \left(1 - A\right) = 22$ or $6 A = 24$ or $A = 4$ and hence $B = 1 - 4 = - 3$

Hence partial fractions of $\frac{x + 2}{\left(x - 2\right) \left(x + 4\right)}$ are $\frac{4}{x - 2} - \frac{3}{x + 4}$