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!
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
Best Answer
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.