# How do you write the partial fraction decomposition of the rational expression (x+6)/(x^3-3x^2-4x+12)?

##### 1 Answer
Mar 24, 2016

Partial fractions are $\frac{1}{5 \left(x + 2\right)} - \frac{2}{\left(x - 2\right)} + \frac{9}{5 \left(x - 3\right)}$

#### Explanation:

As function in the denominator f)x)=x^3-3x^2-4x+12 has $2 , - 2$ and $3$ as zeros, we have ${x}^{3} - 3 {x}^{2} - 4 x + 12 = \left(x - 2\right) \left(x + 2\right) \left(x - 3\right)$ and we can write partial fractions as

$\frac{x + 6}{{x}^{3} - 3 {x}^{2} - 4 x + 12} \Leftrightarrow \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C}{x - 3}$ or

$\frac{x + 6}{{x}^{3} - 3 {x}^{2} - 4 x + 12} \Leftrightarrow \frac{A \left(x - 2\right) \left(x - 3\right) + B \left(x + 2\right) \left(x - 3\right) + C \left(x + 2\right) \left(x - 3\right)}{\left(x - 2\right) \left(x + 2\right) \left(x - 3\right)}$ or

$\frac{A \left({x}^{2} - 5 x + 6\right) + B \left({x}^{2} - x - 6\right) + C \left({x}^{2} - 4\right)}{\left(x - 2\right) \left(x + 2\right) \left(x - 3\right)}$ or

$\frac{{x}^{2} \left(A + B + C\right) + x \left(- 5 A - B\right) + \left(6 A - 6 B - 4 C\right)}{\left(x - 2\right) \left(x + 2\right) \left(x - 3\right)}$

Comparing like terms we have three terms $A + B + C = 0$, $- 5 A - B = 1$ and $6 A - 6 B - 4 C = 6$

From second equation we get $B = - 1 - 5 A$ and putting this in third equation we get $6 A - 6 \left(- 1 - 5 A\right) - 4 C = 6$ or $6 A + 6 + 30 A - 4 C = 6$ or $4 C = 36 A$ or $C = 9 A$.

Now putting these in first we get $A - 1 - 5 A + 9 A = 0$ or $5 A = 1$ or $A = \frac{1}{5}$. Hence $B = - 2$ and $C = \frac{9}{5}$

Hence partial fractions are $\frac{1}{5 \left(x + 2\right)} - \frac{2}{\left(x - 2\right)} + \frac{9}{5 \left(x - 3\right)}$