1. ## Discriminant question

Using the Theorem on Symmetric Functions or otherwise,
show that the discriminant of the quartic polynomial x^4 + qx + r
is 27q^4 + 256r^3.

Now, using the Symmetric Functions theorem, it's obvious that q= -s3 and r=s4 (where s3 and s4 are the third and fourth elementary symmetric polynomials). How do I proceed from here?

Many thanks.

2. Originally Posted by KSM08
Using the Theorem on Symmetric Functions or otherwise,
show that the discriminant of the quartic polynomial x^4 + qx + r
is 27q^4 + 256r^3.

Now, using the Symmetric Functions theorem, it's obvious that q= -s3 and r=s4 (where s3 and s4 are the third and fourth elementary symmetric polynomials). How do I proceed from here?
The discriminant is given by $\Delta = \prod_{i, where $x_1,\,x_2,\,x_3,\,x_4$ are the roots of the polynomial (over the complex field, say).

For a quartic polynomial, the discriminant is a symmetric function of the roots, of degree 12. But q has degree 3 and r has degree 4. The symmetric function theorem says that $\Delta$ must be a function of the elementary symmetric functions, and in this example the only available such functions are q and r (because $s_1$ and $s_2$ are 0). So $\Delta$ must be of the form $mq^4+nr^3$, for some constants m and n.

To find m and n, evaluate $\Delta$ in some special cases. For example, if q=0 and r=–1 then the equation becomes x^4=1, with roots $\pm1,\,\pm i$. From the definition of $\Delta$, you can check that $\Delta = -256$ in this case. So n=256. Similarly, if q = –1 and r=0 then the equation is $x^4-x=0$, with roots $0,\,1,\,(-1\pm\sqrt3i)/2$. I haven't tried to check this, but presumably $\Delta$ works out as 27 in this case.