2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user49685 Mar 23 '14 at 4:44
  • $\begingroup$ Sorry that was a typo, it's x^2+12x+36. $\endgroup$ – Mahina Mar 23 '14 at 13:53
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ So I don't have to use long division? $\endgroup$ – Mahina Mar 23 '14 at 4:37
  • $\begingroup$ 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. $\endgroup$ – user49685 Mar 23 '14 at 4:39
  • $\begingroup$ @Katie, just multiply out and compare the coefficients of $x^2,x,x^0$ to find $A,B,C$ $\endgroup$ – lab bhattacharjee Mar 23 '14 at 4:39
  • $\begingroup$ @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 $\endgroup$ – lab bhattacharjee Mar 23 '14 at 4:43
0
$\begingroup$

\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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.