# How do you write the partial fraction decomposition of the rational expression  (7x^2 - 12x + 11) / (2x^3 - 5x^2 + x +2)?

##### 1 Answer
Dec 28, 2016

The answer is $= - \frac{2}{x - 1} + \frac{5}{2 x + 1} + \frac{3}{x - 2}$

#### Explanation:

We must factorise the denominator

Let $f \left(x\right) = 2 {x}^{3} - 5 {x}^{2} + x + 2$

$f \left(1\right) = 2 - 5 + 1 + 2 = 0$

Therefore, $\left(x - 1\right)$ is a factor of $f \left(x\right)$

To find the other factors, we have to do a long division

$\textcolor{w h i t e}{a a a a}$$2 {x}^{3} - 5 {x}^{2} + x + 2$$\textcolor{w h i t e}{a a a a}$∥$x - 1$

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

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

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

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

$\textcolor{w h i t e}{a a a a a a a a a a a a a}$$- 2 x + 2$

$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$$- 0 + 0$

Therefore,

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

$= \left(x - 1\right) \left(2 x + 1\right) \left(x - 2\right)$

Now, we can perform our decomposition into partial fractions

$\frac{7 {x}^{2} - 12 x + 11}{2 {x}^{3} - 5 {x}^{2} + x + 2} = \frac{7 {x}^{2} - 12 x + 11}{\left(x - 1\right) \left(2 x + 1\right) \left(x - 2\right)}$

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

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

Therefore,

$7 {x}^{2} - 12 x + 11 = A \left(2 x + 1\right) \left(x - 2\right) + B \left(x - 1\right) \left(x - 2\right) + C \left(2 x + 1\right) \left(x - 1\right)$

Let $x = 1$, $\implies$, $6 = - 3 A$, $\implies$, $A = - 2$

Let $x = 2$, $\implies$, $15 = 5 C$, $\implies$, $C = 3$

Coefficients of ${x}^{2}$, $\implies$, $7 = 2 A + B + 2 C$

$B = 7 - 2 A - 2 C = 7 + 4 - 6 = 5$

So,

$\frac{7 {x}^{2} - 12 x + 11}{2 {x}^{3} - 5 {x}^{2} + x + 2} = - \frac{2}{x - 1} + \frac{5}{2 x + 1} + \frac{3}{x - 2}$