Solve 6th degree polynomial: $(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30$ I came across what seems to be a very difficult "solve for $x$" type of
problem, primarily because there should be $6$ real roots of this problem:
$$(x^2 - 3x - 4)(x^2 - 5x + 6)(x^2 + 2x) + 30 = 0.$$
My first step was to (tediously) expand this into the degree $6$ polynomial
that it is:
$$x^6-6 x^5+x^4+36 x^3-20 x^2-48 x+30 = 0.$$
I would imagine that the first step would be to simplify this into a set of
smaller roots, i.e.
$$(ax + b)(cx^5 + dx^4 + ex^3 + fx^2 + gx + h) = 0,$$
and then perform a similar factorization on the polynomial of degree $5$.
However, after trying hard to find the constant terms for this factorization,
I can never quite get the correct factorization. Any recommendations on
problems like these? I would really like to solve this without using Wolfram.

Let $P(x)$ denote the LHS. Note that the first summand factors nicely:
$$
  P(x)=(x-2)(x-3)(x-4)x(x+1)(x+2)+30.
$$
From this it is clear that $P(x)$ is symmetric around $x=1$. Explicitly, setting $y=x-1$ we have
$$\begin{eqnarray*}
  P(x)&=&(y-1)(y-2)(y-3)(y+1)(y+2)(y+3)+30\\
    &=&(y^2-1)(y^2-4)(y^2-9)+30.
\end{eqnarray*}$$
Note that this is a cubic in $y^2$, so it is possible to solve for $y^2$ using the formula for roots of a cubic, and hence find the roots of $P$ in terms of radicals.
