How can one solve polynomial equations with constant terms that have a high number of factors? Today I was given this question on a test:

$x^4 - 20x^3 - 20x^2 + 1500x - 9000 = 0$. Find the value(s) for $x$.

I know how to solve these types of equations. I must record the positive and negative integer factors of the y-intercept or constant term, use guess-and-check to find one factor, and then use synthetic division to factor the equation.
However, there are a very high number of factors for 9000. I know that Descarte's Rule of Signs can help determine whether to consider the positive/negative factors, but even then the number is very high. I guessed and checked as much as I could before realizing I would not be able to solve the question in time.
Is there a quicker way of finding the first x-intercept or zero?
 A: try $$  x^4 - 20 x^3  - 20 x^2 + 1200 x - 16000  $$
with the suggested $x = 10 t,$   then divide the result by $2000,$ we get
$$  5 t^4 - 10 t^3 - t^2 + 6 t -8$$
To write as a difference of squares, we multiply back by 5,
$$  25 t^4 - 50 t^3 - 5t^2 + 30 t -40$$
Next I wrote this as
$$  (5 t^2 - 5 t + a)^2 - ( bt +c )^2  $$  and tried, to begin, with integer $(a,b,c)$   and that works.  It turned out that one could arrange $b=0$  for this problem.
If that had not worked, I would have hoped for $x$  values of the form $\sqrt{v \pm \sqrt w}, $    by allowing an extra coefficient $d > 0$  in
$$  (5 t^2 - 5 t + a)^2 - d( bt +c )^2  $$
A: While not a solution to OP's question as stated, if we suppose that the coefficient of $x^2$ is in error, and that a coefficient of $-50$ rather than $-20$ was intended, then the resulting polynomial graph is symmetric about the vertical line $x=5$.
In general it can be shown with a bit of algebraic manipulation that a quartic $P(x)=Ax^4+Bx^3+Cx^2+Dx+E$ is symmetric with respect to the vertical line
$x=h$ if and only if $h=-\frac{B}{4A}$ while $B(B^2-4AC)+8A^2D=0$.
These conditions are met for
$$x^4 - 20x^3 - 50x^2 + 1500x - 9000 = 0$$
Letting $x=t+5$ we find that
$$ P(t+5)=t^4-200t^2-4625 $$
whose roots are easily found.
