Simplifying third degree polynomials I am trying to figure out this simple problem but I have no idea how to do it.
$\dfrac {2x^3 - 5x^2 -4x -3}{2x^3 + x^2 - 18x -9}$
I know that I can keep guessing at random to find a factor in the terms to factor them out but that seems like it would be far too time intensive, especially considering that if this was on a test it would likely be the first step of the problem and I can't spend 30 minutes just setting the problem up.
 A: Simply in what sense? Removing common factors?
There are only 8 things to try to find a rational root of the numerator. It shouldn't take 30 minutes to work out what those 8 are and to test them.  (note that it can be proven that if they have a common factor, then not only does the factor have integer coefficients, but both cubics also have a rational root)
Or, you can use the Euclidean algorithm to find the (polynomial) greatest common divisor of your two polynomials.
e.g. to find the gcd of $x^2 - 2x + 1$ and $x^2 - 1$, you can compute:
$$ \begin{align} (x^2-2x+1, x^2-1) &= (x^2 - 2x + 1 - (x^2 - 1), x^2 - 1)
\\ &= (-2x + 2, x^2 - 1)
\\ &= (x-1, x^2 - 1)
\\ &= (x-1, x^2 - 1 - x(x-1))
\\ &= (x-1, x-1)
\\ &= x-1
\end{align}$$
(note the "size" of a polynomial here is determined by its degree, and I was "lazy" and avoided doing a full division-with-remainder calculation)
A: Any common divisor of $2x^3 - 5x^2 -4x -3$ and $2x^3 + x^2 - 18x -9$ should also be a common divisor of their difference.
$$(2x^3 + x^2 - 18x -9) - (2x^3 - 5x^2 -4x -3) 
  = 6x^2 - 14x - 6 = 2(3x^2 - 7x -3)$$
$3x^2 - 7x -3$ does not have any rational-coefficient factors and it does not divide evenly into $2x^3 - 5x^2 -4x -3$. That is enough information to conclude that 
$$\gcd(2x^3 - 5x^2 -4x -3, 2x^3 + x^2 - 18x -9) = 1$$
