# Finding the partial fraction decomposition of $\frac{4s^2 - 5s + 2}{s^2(s^2 +9)}$

I am trying to find the partial fraction decomposition of $\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}$ into something of the form $A\dfrac{1}{s} + B\dfrac{1}{s^2} + C\dfrac{1}{s^2+9} + D\dfrac{s}{s^2 + 9}$.

I am unable to factor the numerator into linear terms and how to factor the denominator into linear terms in a useful way is unclear to me since if I just break $s^2$ into $s \cdot s$ then it doesn't seem very helpful. Thank you in advance.

• The only reason you may want to try to factor the numerator is if it will allow you to cancel something in the denominator (to simplify the form). It is not hard to see that there will be nothing to cancel... so not useful to try factor (the numerator). – TravisJ Jun 7 '15 at 22:20
• The denominator is already factored as much as necessary. $s^{2}=(s-0)^{2}$... so a repeated linear term... and $s^{2}+9$ is an irreducible quadratic. – TravisJ Jun 7 '15 at 22:22
• Sorry for the editing. I've been confusing myself as to what the question is asking. This should be the proper form now. – letsmakemuffinstogether Jun 7 '15 at 22:28
• Yes, this is good form now. Do you know how to finish? – TravisJ Jun 7 '15 at 22:28
• So I tried letting: $A(s(s^2+9)) + B(s^2 + 9) + Cs^2 + Ds^2$. Then it would seem that, matching coefficients to powers of $s$, we have $(A)s^3 + (B+C+D)s^2 + (A9)s + B9 = 4s^2 -5s + 2$. Is this on the right track? – letsmakemuffinstogether Jun 7 '15 at 22:30

$$\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+9}$$ Multiply both sides by $s^2(s^2-9):$ $${4s^2 - 5s + 2}=({As+B})({s^2}+9)+{(Cs+D)}s^2$$ Combine terms: $$4s^2-5s+2=(A+C)s^3+(B+D)s^2+(9A)s+9B$$ Solve for coefficients: $$9B=2\implies B=\frac 29$$ $$\left(\frac 29+D\right)=4\implies D=\frac{34}9$$ $$9A=-5\implies A=-\frac{5}9$$ $$-\frac 59+C=0\implies C=\frac 59$$