Everytime when I encounter such problem, I would decompose it straight away. In the solution sheet, I have no idea why does does those polynomials would be relative prime, and is it important that it has to be relative prime to employ partial fraction?
Anyone comments would be appreaciated!
This solution is based upon the following statement:
Theorem: If $g(x)$ and $h(x)$ are two relatively prime polynomials over a field $K$, if $a$ is the degree of $g(x)$ and $b$ that of $h(x)$, and if $f(x)$ is an arbitrary polynomial of degree less than $a+b$, then an identity $f(x)=r(x)g(x)+s(x)h(x)$ exists, where $r(x)$ is of degree $<b$, and $s(x)$ of degree $<a$.
Note that it follows from this theorem that$$\frac{f(x)}{g(x)h(x)}=\frac{r(x)}{h(x)}+\frac{s(x)}{g(x)};$$so, this gives as a partial fraction decomposition of $\frac{f(x)}{g(x)h(x)}$.
On the other hand, if you have a finite set of polynomials, then they are relatively prime if and only if no two of them have a common root.