# How do you write the fractional decomposition of (5x^2 + 7x + 3)/(x^3 + x)?

##### 1 Answer
Feb 16, 2017

$\frac{5 {x}^{2} + 7 x + 3}{{x}^{3} + x} = \textcolor{g r e e n}{\frac{3}{x} + \frac{1 + \frac{7}{2} i}{x + i} + \frac{1 - \frac{7}{2} i}{x - i}}$

#### Explanation:

First we need to find the linear factors of the denominator (note that 2 of these factors only exist as complex values):
${x}^{3} + x = x \left({x}^{2} + 1\right) = \textcolor{b l u e}{x \left(x + i\right) \left(x - i\right)}$

We need to find values $A , B , \mathmr{and} C$ such that
$\frac{5 {x}^{2} + 7 x + 3}{{x}^{3} + x} = \frac{A}{x} + \frac{B}{x + i} + \frac{C}{x - i}$

$\textcolor{w h i t e}{\text{XXXXX}} = \frac{A \left(x + i\right) \left(x - i\right) + B \left(x\right) \left(x - i\right) + C \left(x\right) \left(x + i\right)}{x \left(x + i\right) \left(x - i\right)}$

$\textcolor{w h i t e}{\text{XXXXX}} = \frac{A {x}^{2} + A + B {x}^{2} - B i x + C {x}^{2} - C i x}{{x}^{3} + x}$

rarr5x^2+7x+3=((A+B+C)x^2+(-Bi+Ci)x+A
rarrcolor(white)("XXXXX"){(color(white)("X")A+B+C=5),(color(white)("X")-Bi+Ci=7),(color(white)("X")A=3):}

Combining  and  we have
$\textcolor{w h i t e}{\text{XXXXX")color(white)("X}} B + C = 2$

and from  we can
$\textcolor{w h i t e}{\text{XXXXX")color(white)("X}} B - C = 7 i$

Adding  and  then dividing by 2:
$\textcolor{w h i t e}{\text{XXXXX")color(white)("X}} B = 1 + \frac{7}{2} i$

Then substituting back into 
$\textcolor{w h i t e}{\text{XXXXX}} C = 1 - \frac{7}{2} i$