# How do you express (x^2 - 3x + 2) / (4x^3 + 11x^2) in partial fractions?

Dec 26, 2016

The answer is $= \frac{\frac{2}{11}}{x} ^ 2 + \frac{- \frac{41}{121}}{x} + \frac{\frac{285}{121}}{4 x + 11}$

#### Explanation:

We start the decomposition into partial fractions

$\frac{{x}^{2} - 3 x + 2}{4 {x}^{3} + 11 {x}^{2}} = \frac{{x}^{2} - 3 x + 2}{{x}^{2} \left(4 x + 11\right)}$

$= \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{4 x + 11}$

$= \frac{A \left(4 x + 11\right) + B x \left(4 x + 11\right) + C {x}^{2}}{{x}^{2} \left(4 x + 11\right)}$

Therefore,

${x}^{2} - 3 x + 2 = A \left(4 x + 11\right) + B x \left(4 x + 11\right) + C {x}^{2}$

Let, $x = 0$, $\implies$, $2 = 11 A$, $\implies$, $A = \frac{2}{11}$

Coefficients of $- 3 = 4 A + 11 B$

$11 B = - 3 - 4 A = - 3 - \frac{8}{11}$

$B = - \frac{41}{121}$

Coefficients of ${x}^{2}$, $1 = 4 B + C$

$C = 1 - 4 B$

$C = 1 + \frac{164}{121}$

$C = \frac{285}{121}$

So,

$\frac{{x}^{2} - 3 x + 2}{4 {x}^{3} + 11 {x}^{2}} = \frac{\frac{2}{11}}{x} ^ 2 + \frac{- \frac{41}{121}}{x} + \frac{\frac{285}{121}}{4 x + 11}$