# How do you express  (4x-1)/( x^2(x-4)) in partial fractions?

##### 1 Answer
Feb 5, 2017

The answer is $= \frac{\frac{1}{4}}{{x}^{2}} + \frac{- \frac{15}{16}}{x} + \frac{\frac{15}{16}}{x - 4}$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{4 x - 1}{{x}^{2} \left(x - 4\right)} = \frac{A}{{x}^{2}} + \frac{B}{x} + \frac{C}{x - 4}$

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

As the denominators are the same, we compare the numerators

$4 x - 1 = A \left(x - 4\right) + B \left(x \left(x - 4\right)\right) + C \left({x}^{2}\right)$

Let $x = 0$, $\implies$, $- 1 = - 4 A$, $\implies$, $A = \frac{1}{4}$

Let $x = 4$, $\implies$, $15 = 16 C$, $\implies$, $C = \frac{15}{16}$

Coefficients of ${x}^{2}$

$0 = B + C$, $\implies$, $B = - \frac{15}{16}$

Therefore

$\frac{4 x - 1}{{x}^{2} \left(x - 4\right)} = \frac{\frac{1}{4}}{{x}^{2}} + \frac{- \frac{15}{16}}{x} + \frac{\frac{15}{16}}{x - 4}$