How can one solve polynomial equations with constant terms that have a high number of factors?

Question

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?

Answer 1

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 $$ (5 t^2 - 5 t + a)^2 - d( bt +c )^2 $$

Answer 2

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.