Discriminant of the polynomial $f(x)=4x^3-ax-b$ 
Definition. The discriminant of the polynomial $f(x)=4(x-x_1)(x-x_2)(x-x_3)$ is the product $16\{(x_2-x_1)(x_3-x_2)(x_3-x_1)\}^2$. 

How to prove that the discriminant of $f(x)=4x^3-ax-b$ is $a^3-27b^2$.
Any help would be appreciated.
 A: We know the discriminant of a cubic polynomial of the form $ax^3 + bx^2 + cx + d$ is $\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$. For $f(x)=4x^3-ax-b$, we just plug in the coefficients and get $\Delta = -4(4)(-a)^3 -27(4^2)(-b)^2$ which simplifies to  $\Delta = 16(\color{blue}{a^3 - 27b^2})$.
Edit: I realize now we are going by the definition. In that case, how I'd do it is start with the definition of the discriminant for a polynomial and find that it is $\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$ as I mentioned earlier. Then plug in the numbers. You'll then get the right answer, up to a constant factor.
This document is a nice derivation of the discriminant of a monic cubic polynomial (highest degree term is $1$). You can  follow the same approach in your case (where the highest degree term is $4$). It's a roundabout way of doing it, but it'll work.
A: You should read about the sylvester Sylvester matrix.
To be precise, $f(x) = 4x^3-ax−b$ gives $f'(x) = 12 x^2 - a$. So the discriminant is the determinant
$$ \Delta(a,b) = \left|\begin{array}{ccccc} 
4 & 0 & -a & -b & 0 \\
0 & 4 & 0 & -a & -b\\
12 & 0 & -a & 0 & 0\\
0 & 12 & 0 & -a & 0\\
0 & 0 &12 & 0 & -a
\end{array}\right|
 = -64(a^3 - 27b^2).$$
A: The discriminant shows the behavior of the roots (or zeros in the graph). It shows whether the graph has positive, negative, or imaginary roots. You can prove it by solving the quadratic  and plugin $a,b,c$ into the discriminant to see your results. 
In other words, since $a=4$, $b=a$, $c=-b$, we can say say that $a=4$, $b=a=4$, $c=-b=-(a)=-4$. Thus, we can plug it into the discriminant and figure out if the polynomial has either positive, negative, or imaginary roots. And then you can plug it into the quadratic formula.
If you need more, just let me know.
