Factoring a polynomial of degree 5 I was looking at the solutions to a question about BIBO stability. The transfer function was 
$$\dfrac{s^2+1}{s^5+2s^4+4s^3+7s^2+3s+5}=\dfrac{1}{s^3+2s^2+3s+5}$$
So the lecturer factorised the bottom polynomial and cancelled accordingly, but it isn't explained in the solutions. Is there a general method for factorising polynomials of degree 5?

There exists as simple method to find quadratic factors of monic polynomials of any degree that doesn't involve anything more than straightforward high school level algebra. For 5th degrees polynomials this works as follows. We want to factorize the function:
$$f(x) = x^5 + a x^4 + b x^3 + c x^2 + d x + e$$
where the coefficients are integers. Suppose this has a quadratic factor $g(x) = x^2 -p x - q$, then reducing $f(x)$ modulo $g(x)$ yields:
$$\begin{split}
&\left(a p^3+2 a p q+b p^2+b q+c p+d+p^4+3 p^2 q+q^2\right) x\\
& + a p^2 q+a q^2+b p q+c q+e+p^3 q+2 p q^2
\end{split}$$
We want to choose $p$ and $q$ such that this becomes identical to zero. Let's denote the coefficient of $x$ as $A_1$ and the constant term as $A_2$. Since we know that $q$ will have to divide $e$, which will limit the possible values that $q$ can take, we choose to eliminate $p$ in favor of $q$. We can do this easily by setting $A_1$ and $A_2$ equal to zero and then eliminating the highest powers of $p$ moving on to lower and lower powers until $p$ has been completely eliminated.
Using $A_2=0$ you can express $p^3$ into lower powers of $p$ and $q$. Multiplying that expression by $p$, yields an expression for $p^4$, which you can substitute in the equation $A_1 = 0$, this then not only eliminates the $p^4$ term, it also eliminates the $p^3$ term there. We find the equation $A_3 = 0$ with:
$$A_3 = -e p + d q + b q^2 + a p q^2 + p^2 q^2 + q^3$$ 
We can then use this equation to eliminate $p^3$ and $p^2$ from $A_2$. This yields the equation $A_4 = 0$ where:
$$A_4 = e^2 p - d e q - b e q^2 - a e p q^2 - d p q^3 + c q^4 + a q^5 + p q^5$$
Since this is linear in $p$ we can easily express $p$ in terms of $q$:
$$p = -q\frac{a q^4+c q^3 -b e q-d e }{q^5-d q^3 -a e q^2+e^2}\tag{1}$$
Substituting this in one of the other equations above will yield a polynomial equation for $q$. The intermediary equation where $p^3$ was eliminated from $A_3$ is the most suitable, it will yield the lowest degree non-trivial equation. This yields:
$$
\begin{split}
&q^{10} +b q^9 + (a c - d)q^8 + \left(c^2 + a^2 d - 2 b d - a e\right) q^7 + \left(a c d - d^2 + a^3 e - 3 a b e + c e\right)q^6\\
& + \left(b d^2 + a^2 c e - 2 b c e - 2 a d e + 2 e^2\right)q^{5}+ \left(d^3 + a b d e - 3 c d e - a^2 e^2 + b e^2\right)q^{4}\\
&+ \left(a d^2 e + b^2 e^2 - 2 a c e^2 - d e^2\right)q^{3}+ \left(b d e^2 - a e^3\right)q^{2}+c e^3 q + e^4 = 0
\end{split}\tag{2}
$$
Since $q$ is an integer the rational root theorem applies, but we already knew that $q$ has to divide $e$, so one can just check the possibilities, any solution can be plugged into Eq. (1) to see of this yields an integer $p$.
Plugging in the coefficients into Eq. (2) yields:
$$q^{10}+ 4 q^9+ 11 q^8 +  27 q^7- 12 q^6 - 114 q^5- 168 q^4 - 285 q^3  + 50 q^2 + 875 q +625=0$$
and we see that $q = -1$ is the only solution. Eq. (1) for the given polynomial becomes:
$$p = -q\frac{2 q^4+7 q^3-20 q-15}{q^5-3 q^3-10 q^2+25}$$
and we see that $p = 0$ for $q = -1$, so the quadratic factor is $x^2 - p x - q = x^2 + 1$.
