# How do you express  (5x^2+7x-4)/(x^3+4x^2) in partial fractions?

$\textcolor{red}{\frac{5 {x}^{2} + 7 x - 4}{{x}^{3} + 4 {x}^{2}} = \frac{- 1}{x} ^ 2 + \frac{2}{x} + \frac{3}{x + 4}}$

#### Explanation:

We start from the denominator of the rational expression

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

We will use 3 variables here namely A, B, and C.
Determine the LCD$= {x}^{2} \left(x + 4\right)$

$\frac{5 {x}^{2} + 7 x - 4}{{x}^{2} \left(x + 4\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 4}$

$\frac{5 {x}^{2} + 7 x - 4}{{x}^{2} \left(x + 4\right)} = \frac{A \left(x + 4\right) + B \left({x}^{2} + 4 x\right) + C {x}^{2}}{{x}^{2} \left(x + 4\right)}$

Arrange the coefficients of ${x}^{2} , {x}^{1} , {x}^{0}$

$\frac{5 {x}^{2} + 7 {x}^{1} - 4 {x}^{0}}{{x}^{2} \left(x + 4\right)} = \frac{\left(B + C\right) {x}^{2} + \left(A + 4 B\right) {x}^{1} + 4 A \cdot {x}^{0}}{{x}^{2} \left(x + 4\right)}$

Now we can set up the equations

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

Simultaneous solution of these equations result to

$A = - 1$ and $B = 2$ and $C = 3$

and it follows

$\frac{5 {x}^{2} + 7 x - 4}{{x}^{2} \left(x + 4\right)} = \frac{A}{x} ^ 2 + \frac{B}{x} + \frac{C}{x + 4}$

$\textcolor{b l u e}{\frac{5 {x}^{2} + 7 x - 4}{{x}^{2} \left(x + 4\right)} = \frac{- 1}{x} ^ 2 + \frac{2}{x} + \frac{3}{x + 4}}$

God bless....I hope the explanation is useful.