How do you express (x+13)/(x^3+2x^2-5x-6) in partial fractions?

1 Answer
May 8, 2016

$\frac{x + 3}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \frac{1}{3 \left(x - 2\right)} - \frac{1}{3 \left(x + 1\right)}$

Explanation:

Let us factorize ${x}^{3} + 2 {x}^{2} - 5 x - 6$. It is apparent that $2$ is a zero of the polynomial and hence $\left(x - 2\right)$ is a factor of ${x}^{3} + 2 {x}^{2} - 5 x - 6$. Dividing by $\left(x - 2\right)$,

${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left({x}^{2} + 4 x + 3\right)$ and hence ${x}^{3} + 2 {x}^{2} - 5 x - 6 = \left(x - 2\right) \left(x + 3\right) \left(x + 1\right)$.

Now let the partial fractions be given by

$\frac{x + 3}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \frac{x + 3}{\left(x - 2\right) \left(x + 3\right) \left(x + 1\right)} = \frac{1}{\left(x - 2\right) \left(x + 1\right)} \Leftrightarrow \frac{A}{x - 2} + \frac{B}{x + 1}$

or $\frac{1}{\left(x - 2\right) \left(x + 1\right)} \Leftrightarrow \frac{A \left(x + 1\right) + B \left(x - 2\right)}{\left(x - 2\right) \left(x + 1\right)}$

or $\frac{1}{\left(x - 2\right) \left(x + 1\right)} \Leftrightarrow \frac{\left(A + B\right) x + \left(A - 2 B\right)}{\left(x - 2\right) \left(x + 1\right)}$

Hence $A + B = 0$ and $A - 2 B = 1$ or $A = \frac{1}{3}$ and $B = - \frac{1}{3}$

Hence $\frac{x + 3}{{x}^{3} + 2 {x}^{2} - 5 x - 6} = \frac{1}{3 \left(x - 2\right)} - \frac{1}{3 \left(x + 1\right)}$