# How do you write the partial fraction decomposition of the rational expression (s^3+s-4)/( (s^2+1)(s^2+4))?

Dec 14, 2015

$\frac{{s}^{3} + s - 4}{\left({s}^{2} + 1\right) \left({s}^{2} + 4\right)} = - \frac{4}{3 \left({s}^{2} + 1\right)} + \frac{3 s + 4}{3 \left({s}^{2} + 4\right)}$

#### Explanation:

In your case, the denominator is already factorized completely, so you can start immediately.

As your factors are non-linear, the partial fraction decomposition looks like this:

Find $A$, $B$, $C$, $D$ so that

$\frac{{s}^{3} + s - 4}{\left({s}^{2} + 1\right) \left({s}^{2} + 4\right)} = \frac{A s + B}{{s}^{2} + 1} + \frac{C s + D}{{s}^{2} + 4}$

Now, solve the equation!

... multiply both sides with the denominator $\left({s}^{2} + 1\right) \left({s}^{2} + 4\right)$ in order to "get rid" of fractions...

$\iff {s}^{3} + s - 4 = \left(A s + B\right) \left({s}^{2} + 4\right) + \left(C s + D\right) \left({s}^{2} + 1\right)$

... expand the expression ...

$\iff {s}^{3} + s - 4 = A {s}^{3} + B {s}^{2} + 4 A s + 4 B + C {s}^{3} + D {s}^{2} + C s + D$

... compare the expressions with the same power of $s$....

$\iff \textcolor{b l u e}{{s}^{3}} + \textcolor{red}{0 \cdot {s}^{2}} + \textcolor{g r e e n}{s} \textcolor{p u r p \le}{- 4} = \textcolor{b l u e}{A {s}^{3}} + \textcolor{red}{B {s}^{2}} + \textcolor{g r e e n}{4 A s} + 4 B + \textcolor{b l u e}{C {s}^{3}} + \textcolor{red}{D {s}^{2}} + \textcolor{g r e e n}{C s} + \textcolor{p u r p \le}{D}$

At this point, you can split the equation in four different ones: for $\textcolor{b l u e}{{s}^{3}}$ terms, $\textcolor{red}{{s}^{2}}$ terms, $\textcolor{g r e e n}{s}$ terms and $\textcolor{p u r p \le}{\text{linear}}$ terms:

{ (color(white)(x) 1 = color(white)(x) A + C color(white)(xxx) "(I)" ), (color(white)(x) 0 = color(white)(x) B + D color(white)(xxi) "(II)" ), (color(white)(x) 1 = 4A + C color(white)(xxi) "(III)" ), (-4 = 4B + D color(white)(xx) "(IV)" ) :}

From $\text{(I)}$ and $\text{(III)}$, you can compute $A$ and $C$, e.g. by solving $\text{(I)}$ for $A$ and plugging the value into $\text{(III)}$:

$\implies A = 0$, $\text{ } C = 1$

Similarly, if you e.g. solve $\text{(II)}$ for $B$ and plug the value into $\text{(IV)}$, you will compute $B$ and $D$:

$\implies B = - \frac{4}{3}$, $\text{ } D = \frac{4}{3}$

Thus, your partial fraction decomposition looks like this:

$\frac{{s}^{3} + s - 4}{\left({s}^{2} + 1\right) \left({s}^{2} + 4\right)} = \frac{- \frac{4}{3}}{{s}^{2} + 1} + \frac{s + \frac{4}{3}}{{s}^{2} + 4}$

... expand the second fraction by $3$ to gain a nicer fraction...

$\frac{{s}^{3} + s - 4}{\left({s}^{2} + 1\right) \left({s}^{2} + 4\right)} = - \frac{4}{3 \left({s}^{2} + 1\right)} + \frac{3 s + 4}{3 \left({s}^{2} + 4\right)}$

Hope that this helped! :-)