Decompose $\frac{x^4 + 5}{x^5 + 6x^3}$ (partial fraction decomposition)

Question

Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.

$$\frac{x^4 + 5}{x^5 + 6x^3}$$

So I factored the denominator to be $x^3(x^2+6)$ and here is the answer i got:

$\frac{Ax+B}{x^3}+\frac{Cx+D}{x^2+6}$

But this isn't the correct answer. Any help is appreciated!

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Also have another one:

$\frac{5}{(x^2 − 16)^2}$

Here is what I got: $\frac{A}{x+4}+\frac{Bx+C}{\left(x+4\right)^2}+\frac{D}{x-4}+\frac{Ex+F}{\left(x-4\right)^2}$ which is also wrong... if anyone could help me that would be great

Answer 1

The correct way is $$\frac{x^4 + 5}{x^3(x^2 + 6)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+6}$$ and $$\frac{5}{(x^2 − 16)^2}=\frac{A}{x+4}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-4}+\frac{D}{\left(x-4\right)^2}$$ In general, if the denominator has degree $n$, then the numerator is given a degree $(n-1)$. This is because for any higher degree, one can divide and get a remainder where the numerator has a degree less than the denominator. Moreover, the numerator can have a degree less than $(n-1)$ if the coefficients of appropriate powers of $x$ are $0$.

When the denominator is a $k^{th}$ power of a polynomial in $x$, the numerator can be split into $k$ expressions each being some multiple of the polynomial in the denominator, which in turn can be split into $k$ different fractions. This gives rise to the form of expression used above.

Answer 2

Question 1: $$ x^{4}+5=f(x)\left(x^{2}+6\right)+x^{3} g(x) $$ Let $ \quad \displaystyle \frac{x^{4}+5}{x^{3}\left(x^{2}+6\right)} \equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x^{3}}+\frac{f(x)}{x^{2}+6}$, then

$$x^{4}+5 \equiv A x^{2}\left(x^{2}+6\right)+B x\left(x^{2}+6\right)+C\left(x^{2}+6\right)+x^{3} f(x) \cdots (1)$$

Putting $x=0$ yields $5=6 C\Rightarrow C=\dfrac{5}{6}.$

Rearranging (1) yields $$ \begin{aligned} x^{4}+5-\frac{5}{6}\left(x^{2}+6\right) & \equiv A x^{2}\left(x^{2}+6\right)+B x\left(x^{2}+6\right)+x^{3} f(x) \\ \frac{1}{6} x\left(6 x^{2}-5\right) & \equiv A x\left(x^{2}+6\right)+B\left(x^{2}+6\right)+x^{2} f(x)\cdots (2) \end{aligned} $$ Putting $x=0$ in (2) yields $B=0$.

Rearranging (2) yields $$ \frac{1}{6}\left(6 x^{2}-5\right)=A\left(x^{2}+6\right)+x f(x) \cdots (3) $$ Putting $x=0$ in (3) yields $ \displaystyle -\frac{5}{6}=6 A \Rightarrow A=-\frac{5}{36}$. $\begin{aligned}\text{ Then rearranging (3) yields } & \\ x^{2} f(x) &=\frac{1}{6}\left(6 x^{2}-5\right)+\frac{5}{36}\left(x^{2}+6\right) =\frac{41}{6} x^{2} \\ \therefore \quad f(x) &=\frac{41}{6} \\ \therefore \frac{x^{4}+5}{x^{3}\left(x^{2}+6\right)} &=-\frac{5}{36 x}+\frac{5}{6 x^{3}}+\frac{41}{6\left(x^{2}+6\right)} .\end{aligned}$

Question 2:

For this particular fraction, we can do it in a simpler way. $$ \begin{aligned} \frac{5}{\left(x^{2}-16\right)^{2}} &=5\left[\frac{1}{(x+4)(x-4)}\right]^{2} \\ &=5\left[\frac{1}{8}\left(\frac{1}{x-4}-\frac{1}{x+4}\right)\right]^{2} \\ &=\frac{5}{64}\left[\frac{1}{(x-4)^{2}}-\frac{2}{(x-4)(x+4)}+\frac{1}{(x+4)^{2}}\right] \\ &=\frac{5}{64}\left[\frac{1}{(x-4)^{2}}-\frac{1}{4}\left(\frac{1}{x-4}-\frac{1}{x+4}\right)+\frac{1}{(x+4)^{2}}\right] \\ &=\frac{5}{256}\left[\frac{4}{(x-4)^{2}}-\frac{1}{x-4}+\frac{1}{x+4}+\frac{4}{(x+4)^{2}}\right] \end{aligned} $$