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?
We are looking for $$f(s):=s^5+2s^4+4s^3+7s^2+3s+5 = (s^2+1)(s^3+as^2+bs+c)$$
The coefficient of $s^4$ on the RHS is $a$, which must be $2$, from the LHS.
The coefficient of $s^3$ on the RHS is $1+b$, which must be $4$, from the LHS, so that $b=3$.
The coefficient of $s^2$ on the RHS is $c+a$, which must be $7$, from the LHS, so that $c=5$ since $a=2$. Therefore you've found $a,b,c$.
Notice that in general factoring polynomials of degree $\geq 5$ is difficult. There are algorithms as Berlekamp–Zassenhaus algorithm, but no "simple" way.
Here is how to find the factor $s^2+1$. First, I would try to find rational roots. By the rational root theorem, such a root $x=p/q$ (with $p,q$ coprime integers) satisfies $p \mid 5, q \mid 1$, so that $x \in \{±1,±5\}$. But none of these $4$ values is a root of $f(s)$.
Then we try to find a (monic) factor $x^2+ux+v$ of degree $2$, and if none of them works, then $f(s)$ is irreducible in $\Bbb Q[X]$ because it has degree $5$. Notice that $f$ is monic, so that irreducibility over $\Bbb Q$ is equivalent to irreducibility over $\Bbb Z$, i.e. we can assume $u,v \in \Bbb Z$.
You write as before $$s^5+2s^4+4s^3+7s^2+3s+5 = (s^2+us+v)(s^3+as^2+bs+c)$$ to get $$u+a=2,\quad v+ua+b=4, \quad c+ub+av=7, \quad bv+cu=3, \quad cv=5$$ We know that $v$ divides $5$, so it is $±1,±5$.
Because $f(i)=0$, $s^2+1$ divides $f(s)$ and then we can try $u=0$ (which is equivalent to finding imaginary roots of $f(s)$ if $v>0$, since $f$ would have the factor $s^2+v$). Then $bv=3$ gives $v=\pm 1$, so that $v=1$ since $v=-1 \implies s^2-1$ divides $f \implies f(1)=0$, not possible as mentioned before.
The simplest is to use Euclidean division of $$ s^5+2s^4+4s^3+7s^2+3s+5 $$ by $s^2+1$.
The algorithm is the following :
Input : $P$ (the polynomial to be divided) $D$ (the divisor)
Running data : $(Q_n,R_n)$
Initialization : $(Q_0,R_0)=(P,0)$
Process : If $d:=\deg(Q_n)-\deg(D)\geq 0$ then
$$
R_{n+1}=Q_n-t.s^{d}D;\ Q_{n+1}:=Q_n+t.s^{d};
$$
$t$ is such that $\deg(R_{n+1})<\deg(R_n)$
else STOP
Here, the sequence of $(Q_n,R_n)$ (quotients/remainders) begins like that
Do not hesitate to interact.
There is no algebraic expression for general quintic equations over the rationals in terms of radicals; this statement is known as the Abel–Ruffini theorem. This result also holds for equations of higher degrees. But, however your quintic $s^5 + 2s^4 + 4s^3 + 7s^2 + 3s+ 5$ can be easily factorised by $s^2 + 1$ using Euclidean division.
Some quintics may be solved in terms of radicals. However, the solution is generally too complex to be used in practice. Instead, numerical approximations are calculated using root-finding algorithm for polynomials.
You can read more about solving Quintic functions here. Hope it helps.
Both long division and Euclidean division will be effective as mentioned by other answers.
Adding to the great answers abovementioned, I want to address to the question of whether there exists a general method to solve polynomial equations with degree 5 or above.
https://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem
Abel-Ruffini Theorem clearly states that you cannot find a general equation to solve polynomial equations higher than 5, which means there is no general rule to factorize such a polynomial excepts those special ones like $x^5-1$.
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 $x^2 - p x - q = x^2 + 1$.
