Computing discriminant of $x^4 + px + 1$, need help to find a mistake

Question

$f(x) = x^4 + px + q \in \mathbb{K}[x]$, I want to find the discriminant $D(f)$.

First I note that $D(f) = aq^3 + bp^4$ because all the symmetric functions of the roots except $s_3 = -p, \; s_4 = q$ are zero.

By taking $x_1 =0 $ and $x_2 = x_3 = x_4 = 1$ I obtain $0 = a \cdot 0 + bp^4 \Rightarrow b = 0 $.

Now take $x_1 = x_2 = x_3 = x_4 = 1$ then $0 = a \cdot 1 + 0 = 0$.

So the discriminant is identically zero which is not true.

Where is the mistake?

Answer 1

Taking $x_1=0$ and $x_2=x_3=x_4=1$ yields the polynomial $$(x-x_1)(x-x_2)(x-x_3)(x-x_4)=x^4-3x^3+3x^2-x,$$ which is not of the form $x^4+px+q$. The same is true for your other choice of roots.

Answer 2

COMMENT.- According to WolframMathWorld (see attached figure), the discriminant of your quartic polynomial in which $(a_0,a_1,a_2,a_3,a_4)=(q,p,0,0,1)$ is $$\Delta=-3^3p^4+2^8q^3$$