# How do you express  (5x^2 - 25x + 28) / (x^2(x-7)) in partial fractions?

Dec 21, 2017

$\frac{5 {x}^{2} - 25 x + 28}{{x}^{2} \left(x - 7\right)} = \frac{3}{x} - \frac{4}{x} ^ 2 + \frac{2}{x - 7}$

#### Explanation:

Given:

$\frac{5 {x}^{2} - 25 x + 28}{{x}^{2} \left(x - 7\right)}$

$= \frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 7}$

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

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

So:

$\left\{\begin{matrix}A + C = 5 \\ - 7 A + B = - 25 \\ - 7 B = 28\end{matrix}\right.$

From the third equation, we find that $B = - 4$

Substituting this value of $B$ in the second equation, we find $A = 3$

Then substituting this value of $A$ into the first equation, we find $C = 2$

So:

$\frac{5 {x}^{2} - 25 x + 28}{{x}^{2} \left(x - 7\right)} = \frac{3}{x} - \frac{4}{x} ^ 2 + \frac{2}{x - 7}$