Discriminant of a polynomial of arbitrary degree and positivity

Question

Maybe a trivial question: Is there any relation between the discriminant of a polynomial $$p(x)=a_n x^n + \ldots +a_0$$ with an arbitrary degree $n$ and the positivity of it for all $x$, just like quadratic case?

Answer 1

I assume you mean polynomials with real coefficients.

The discriminant tells us about the nature (real or complex) of the roots. Since complex roots occurs in pairs, when $n=2$, having no real roots means the polynomial is always positive or always negative.

Already when $n=3$, something like this cannot be true because every cubic has at least one real zero. Nevertheless, the discriminant tells us about the nature of the roots, see Wikipedia:

• If $\Delta > 0$, then the equation has three distinct real roots.
• If $\Delta = 0$, then the equation has a multiple root and all of its roots are real.
• If $\Delta < 0$, then the equation has one real root and two non-real complex conjugate roots.

When $n=4$, the sign of the discriminant is not enough to give complete information about the nature of the roots. See Wikipedia.

When $n\ge 5$, things get even more complicated.

Answer 2

This answer turned out somewhat long due to the derivation of the fact that the discriminant has sign $(-1)^s$. The point is that the sign of the discriminant and the congruence of $n$ modulo $4$ may sometimes be sufficient to decide whether the polynomial has any real roots.

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For a monic polynomial of degree $n$ with roots $\alpha_1, \ldots, \alpha_n$ (counting multiplicities), the discriminant is given by $$\Delta = \prod\limits_{i \lt j}^n(\alpha_i - \alpha_j)^2$$

If the polynomial has coefficients in $\mathbb{R}$ then all non-real roots come in complex conjugate pairs.

Write $n = r + 2s$ where $r$ is the number of real roots and $s$ is the number of complex conjugate pairs. Denote the real roots by $x_1, \ldots, x_r$ and the complex roots by $y_1 \pm iz_1, \ldots, y_s \pm iz_s$.

Any term of the form $(x_i - x_j)^2$ where $i \lt j$ is a square of a real number and hence is positive.

Whenever a term of the form $(x_i - y_j - iz_j)^2$ occurs, there will also be a term of the form $(x_i - y_j + iz_j)^2$ (its complex conjugate) and the combined contribution to the product is $\left((x_i - y_j)^2+z_j^2\right)^2$ which is again a positive number.

Terms of the form $(y_i \pm iz_i - y_j \pm iz_j)^2$ where $i \neq j$ will also occur in complex conjugate pairs whose combined contribution to the product is again positive.

So we are left to consider terms of the form $(y_i + iz_i - y_i + iz_i)^2$ which are their own complex conjugates. These give a contribution of $-4z_i^2$ to the product. There are $s$ such terms (for each pair of non-real roots).

We conclude that $\text{sgn}\, \Delta = (-1)^s$.

Therefore the sign of the discriminant is $(-1)^s$ where $n = r + 2s$ and $r$ is the number of real roots, $s$ is the number of pairs of complex conjugate roots.

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To get back to the question, the positivity of $p(x)$ means that $r = 0$.

The sign of the discriminant tells us whether $s$ is even or odd.

If $\Delta \gt 0$ then $s$ is even.
If $\Delta \lt 0$ then $s$ is odd.
If $\Delta = 0$ then the polynomial has repeated roots and we cannot decide using this method.

Knowing the degree $n$ and the parity of $s$, we do not get a lot of information about $r$.

But we do know that $r = n - 2s$.

If $\Delta \gt 0$ then $s = 2k$ and $r = n - 4k$ for some $k = 0, 1, \ldots$. In order for $r = 0$ to be possible we need $n = 4k$, i.e. the degree must be a multiple of $4$. In such a case, both $r = 0$ and $r \gt 0$ are possible (i.e. not distinguishable by the sign of the discriminant).

If $\Delta \lt 0$ then $s = 2k + 1$ and $r = n - 4k - 2$ for some $k = 0, 1, \ldots$. In order for $r = 0$ to be possible we need $n = 4k + 2$, i.e. $n \equiv 2 \pmod 4$. In such a case, if $n = 2$ we are forced to conclude $r = 0$ because $r$ cannot be negative, otherwise if $n \gt 2$ we cannot decide using this method.

In conclusion we have the following decision procedure:

If $n$ is odd then the polynomial always has a real root (from Intermediate Value Theorem).
If $\Delta \gt 0$ and $n$ is not divisible by $4$ then the polynomial always has a real root.
If $\Delta \lt 0$ and $n$ is divisible by $4$ then the polynomial always has a real root.
If $\Delta \lt 0$ and $n = 2$ then the polynomial does not have any real roots.
If $\Delta = 0$ and $n = 2$ then the polynomial has a repeated root which must be real.

In any other case, the test is inconclusive. Perhaps more information can be deduced using the value of $\Delta$ rather that just its sign.