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

Dec 20, 2016

The answer is $= 2 x - 7 + \frac{1}{x + 1} + \frac{17}{x + 2}$

#### Explanation:

As the degree of the numerator is $>$ than the degree of the denominator, we do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - {x}^{2} + x + 5$$\textcolor{w h i t e}{a a a a}$∣${x}^{2} + 3 x + 2$

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} + 6 {x}^{2} + 4 x$$\textcolor{w h i t e}{a a a a a a}$∣$2 x - 7$

$\textcolor{w h i t e}{a a a a a}$$0 - 7 {x}^{2} - 3 x + 5$

$\textcolor{w h i t e}{a a a a a a a}$$- 7 {x}^{2} - 21 x - 14$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$18 x + 19$

Therefore,

$\frac{2 {x}^{3} - {x}^{2} + x + 5}{{x}^{2} + 3 x + 2} = 2 x - 7 + \frac{18 x + 19}{{x}^{2} + 3 x + 2}$

We can now do the decomposition into partial fractions

$\frac{18 x + 19}{{x}^{2} + 3 x + 2} = \frac{18 x + 19}{\left(x + 1\right) \left(x + 2\right)} = \frac{A}{x + 1} + \frac{B}{x + 2}$

$= \frac{A \left(x + 2\right) + B \left(x + 1\right)}{\left(x + 1\right) \left(x + 2\right)}$

So,

$18 x + 19 = A \left(x + 2\right) + B \left(x + 1\right)$

Let $x = - 1$, $\implies$, $1 = A$

Let $x = - 2$, $\implies ,$-17=-B#

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