Polynomials – Can a Trinomial with Degree Exceeding Four Have Only Real Roots?

Question

If there's no general proof about trinomials and upper bound on roots with degree arbitrary,

For $a,b,c$ real numbers, and $n$ in the integers, can $\pm ax^{2n}+\pm bx^n\pm c$ or $\pm ax^{2n+1}\pm bx^n\pm c$, or $ \pm ax^{2n+1} \pm bx^{2n}\pm c$, or $\pm ax^{2n+2} \pm bx^{2n}\pm c$ have only real roots for $n>1$?

The fundamental theorem of algebra implies

• Every univariate polynomial of positive degree $n$ with complex coefficients can be factorized as
  $$
  c\left(x-r_1\right) \cdots\left(x-r_n\right)
  $$
  where $c, r_1, \ldots, r_n$ are complex numbers.
  The $n$ complex numbers $r_1, \ldots, r_n$ are the roots of the polynomial. If a root appears in several factors, it is a multiple root, and the number of its occurrences is, by definition, the multiplicity of the root.
  The proof that this statement results from the previous ones is done by recursion on $n$ : when a root $r_1$ has been found, the polynol division by $x-r_1$ provides a polynomial of degree $n-1$ whose roots are the other roots of the given polynomial.

If n=1, then the answer is at most two.

If n=2 the answer is at most four if I'm assuming true you can write y=(x-3)(x-2)(x+3)(x-4) some how such that the cubic and linear term cancels out and you're only left with even exponents.

If n=k, the answer is at most 2k, and at least zero assuming all the odd exponents greater than or equal to three leaving cancel out but the linear term $bx^n$ and the c are kept, but I don't know if they do for real roots.

Answer 1

Yes, there are trinomials of arbitrarily large degree whose roots are all real, e.g., $$x^{n + 4} - (r^2 + s^2) x^{n + 2} + r^2 s^2 x^n = (x + s) (x + r) x^n (x - r) (x - s) , \quad r, s \in \Bbb R .$$

We can say more: Descartes' Rule of Signs asserts that the number of positive roots of a real polynomial (counting multiplicity) is at most the number of sign changes of the its coefficients. So, if $p(x)$ is a trinomial, there are at most $2$ sign changes, hence it has at most $2$ positive roots. Likewise $p(-x)$ has at most $2$ positive roots, hence $p(x)$ has at most $2$ negative roots, so a trinomial has at most $4$ nonzero real roots.

We've thus shown:

Proposition If $p(x)$ is a (nonzero) trinomial whose roots are all real, it must have the form $$p(x) = x^m q(x)$$ for some trinomial $q(x)$ (whose roots are all real) of degree $\operatorname{deg} g \leq 4$ and satisfying $q(0) \neq 0$.

It's a straightforward exercise to describe all such trinomials $q$, hence all of the trinomials whose roots are all real.

In any case, we can use the theorem to analyze directly the specific families of trinomials in the question statement. For example, for the family $$p(x) = a x^{2 n} + b x^n + c :$$

• If $c \neq 0$, then $p(x) = q(x)$, so $2 n = \deg q \leq 4$, hence $n = 2$, that is, $$p(x) = a x^4 + b x^2 + c .$$ Since $p$ is even, its roots are $\pm r, \pm s \in \Bbb R \setminus \{0\}$, hence $$p(x) = a x^4 - a (r^2 + s^2) x^2 + a r^2 s^2 ,$$ which recovers the example at the beginning of this answer. Equivalently all four roots are real if $b^2 \geq 4 a c \geq 0$ and $\operatorname{sign}(b) = -\operatorname{sign}(a)$.
• If $c = 0$ but $b \neq 0$, then $p(x) = x^n (a x^n + b)$. Descartes' Rule of Signs gives that $q(x) = a x^n + b$ has at most $2$ real roots, so the only possibility is $n = 2$. Again $q$ is even, so $q(x) = a (x + r) (x - r)$ for some $r \in \Bbb R \setminus \{0\}$: $$p(x) = a x^4 - a r^2 x^2 .$$
• If $b = c = 0$ but $a \neq 0$, then $p(x) = a x^{2 n}$, and all of the roots of this polynomial are of course real for all $n \geq 0$: $$p(x) = a x^{2 n}, \qquad n \geq 2.$$

The other three classes are easier.

In any case the proposition immediately implies the following:

Corollary If $p(x)$ is a nonzero trinomial whose roots are real and distinct, then $\operatorname{deg} p \leq 5$.