# How do you write the partial fraction decomposition of the rational expression  (2x^3-x^2+x+5)/(x^2+3x+2)?

Aug 7, 2016

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{{x}^{2} + 3 x + 2} = 2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$

#### Explanation:

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{{x}^{2} + 3 x + 2}$

$= \frac{\textcolor{b l u e}{2 {x}^{3} + 6 {x}^{2} + 4 x} \textcolor{g r e e n}{- 7 {x}^{2} - 21 x - 14} + 18 x + 19}{{x}^{2} + 3 x + 2}$

$= \frac{\textcolor{b l u e}{2 x \left({x}^{2} + 3 x + 2\right)} \textcolor{g r e e n}{- 7 \left({x}^{2} + 3 x + 2\right)} + 18 x + 19}{{x}^{2} + 3 x + 2}$

$= 2 x - 7 + \frac{18 x + 19}{{x}^{2} + 3 x + 2}$

Focusing on the remaining rational expression:

$\frac{18 x + 19}{{x}^{2} + 3 x + 2}$

$= \frac{18 x + 19}{\left(x + 1\right) \left(x + 2\right)}$

$= \frac{A}{x + 1} + \frac{B}{x + 2}$

Using Heaviside's cover-up method we find:

$A = \frac{18 \left(- 1\right) + 19}{\left(- 1\right) + 2} = \frac{1}{1} = 1$

$B = \frac{18 \left(- 2\right) + 19}{\left(- 2\right) + 1} = \frac{- 17}{- 1} = 17$

So:

$\frac{18 x + 19}{{x}^{2} + 3 x + 2} = \frac{1}{x + 1} + \frac{17}{x + 2}$

Putting it all together:

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{{x}^{2} + 3 x + 2} = 2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$