For instance, consider this polynomial $x^3-3x^2+2x-39270$. I happen to know that $35$ is a root, and thus I can use synthetic division to factor out $(x-35)$. But if I would attempt to find a root using the rational root theorem, I would have too many options, since the constant is such a big number.
Is there a faster and more efficient way of doing this?
On
If you depress the polynomial (cancel the quadratic term by shifting the variable, $x:=y+1$), you get
$$y^3-y-39270=0.$$
If you assume that $y^3\gg y$ and neglect $y$, $y^3=39270$ yields $y=33.99\cdots$, and searching nearby integer solutions you immediately get
$$34^3-34=29270.$$
This will work whenever the solution is known to be integer and the linear coefficient is small.
Have you omitted some context from your question, and $x^3 - 3x^2 + 2x - 39270$ was not made up by you?
The reason I ask is that the zeros of this cubic polynomial can be found by solving $x(x^2 - 3x + 2) = 39270,$ which is equivalent to $x(x-1)(x-2) = 39270,$ and thus one is led to consider the possibility that 39270 is the product of three consecutive integers, each of which is roughly the cube root of 39270.
Since $32^3 = 32 \times 32^2 = 32 \times 2^{10},$ which is approximately $32 \times 1000 = 32000,$ one is led to look for three consecutive integers in the vicinity of $32$ such that their product ends in $0.$ Three possibilities arise, namely $(33)(34)(35)$ and $(34)(35)(36)$ and $(35)(36)(37).$ A quick check shows that $(33)(34)(35) = 39270,$ and thus $x = 35$ is a zero of your cubic polynomial.