# Partial Fractions - $\frac{x^3}{x^2 + 12x +36}$

Ok, so I know that since the numerator has a higher power that long division is needed. So after doing that, the main fraction is $\frac{-6x-36}{x^2 + 12x + 36}$. I think that's right. But my problem is that after you factor the denominator they're both equal to $(x+6)$. So how would you use partial fraction decomposition?

• I don't understand, is your denominator $x^2 + 12x + 36$ or $x^2 + 6x + 9$? By the way, your long division doesn't look quite correct. – user49685 Mar 23 '14 at 4:44
• Sorry that was a typo, it's x^2+12x+36. – Mahina Mar 23 '14 at 13:53

After the observation that the denominator is equal to $(x+6)^2$ (which was mentioned in other answer), this problem can be solved relatively easy using the substitution $y=x+6$.

\begin{align*} \frac{x^3}{(x+6)^2} &=\frac{(y-6)^3}{y^2}=\\ &=\frac{y^3-3\cdot 6 y^2+3\cdot 6^2 y - 6^3}{y^2}=\\ &=\frac{y^3-18y^2+108y-216}{y^2}=\\ &=y-18+\frac{108}y-\frac{216}{y^2}=\\ &=x-12+\frac{108}{x+6}-\frac{216}{(x+6)^2} \end{align*}

You can check the result using WolframAlpha.

$$\frac{x^3}{x^2+12x+36}=\frac{x^3}{(x+6)^2}=x+A+\frac B{x+6}+\frac C{(x+6)^2}$$

• So I don't have to use long division? – Mahina Mar 23 '14 at 4:37
• Of course you have to use long division to lower the degree of the numerator first. That is why there's a term $x + A$ in lab bhattacharjee's answer, that's the result when you divide the numerator by the denominator. – user49685 Mar 23 '14 at 4:39
• @Katie, just multiply out and compare the coefficients of $x^2,x,x^0$ to find $A,B,C$ – lab bhattacharjee Mar 23 '14 at 4:39
• @Katie, You don't have to use long division. We can find the highest power with coefficient as $$\frac{1\cdot x^3}{1\cdot x^2}=1\cdot x^1$$, then $A_0x^{1-1}$ and so on. Then mathworld.wolfram.com/PartialFractionDecomposition.html – lab bhattacharjee Mar 23 '14 at 4:43

$$\begin{array}{rcrrrrrrr} & & x & -12 \\ & & -- & --- & --- & --- & \\ x^2+12x+36 & ) & x^3 \\ & & x^3 & 12x^2 & 36x \\ & & -- & --- & --- \\ & & & -12x^2 & -36x \\ & & & -12x^2 & -144x & -432 \\ & & & --- & --- & --- \\ & & & & 108x & 432 \\ \end{array}$$

And so

$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108x+432}{(x+6)^2}$$

where

\begin{align} \dfrac{108x+432}{(x+6)^2} &= \dfrac{A}{x+6} + \dfrac{B}{(x+6)^2} \\ 108x + 432 &= A(x+6) + B \end{align}

$$\text{Let x=-6 and you get B = -216.}$$

$$\text{Let B = -216 and you get}$$

\begin{align} 108x + 432 &= A(x+6) - 216 \\ A(x+6) &= 108x + 648 \\ A &= 108 \end{align}

And so

$$\dfrac{x^3}{(x+6)^2} = x - 12 + \dfrac{108}{x+6} - \dfrac{216}{(x+6)^2}$$