Show that this polynomial is positive

1. $x\mapsto Q(5, x)$ is positive. There is numerous ways to prove that, more or less algorithmically.

2. Assume that there exists a $k$ such that $x\mapsto Q(k,x)$ is not positive, and consider $L$ be the infimum of all real numbers $k \geqslant 5$ such that $x\mapsto Q(k, x)$ is not positive. I claim that $Q(L,x)$ has a double real root, this is equivalent to say that $L$ is a root of the discriminant of $Q$ w.r.t. the variable $x$.

3. The discriminant of $Q$ with respect to $x$ is (computed with Maple) : $$ \operatorname{disc}_x Q = 92162779488452608 k^{16} (k-4)^4 $$

4. The real roots of this discriminant are 0 and 4, thus $L$ cannot be finite, and hence the claim.

Proofs

• Point 1 can be proved using Sturm sequences, and you can compute these sequences using Maple :

  sturm(subs(k=5, Q),x,-infinity, infinity);
        0
  

This means that $Q(5, x)$ has no real roots. Since the leading coefficient is positive, the polynomial $Q(5, x)$ is positive whenever $x\in \mathbb R$.

• To prove point 2 consider $x_0$ a real numbers such that $Q(L, x)$ is minimal. We have $Q(L, x_0) = 0$, because if $Q(L, x_0) > 0$ then $x\mapsto Q(L+\epsilon, x)$ is positive for $\epsilon > 0$ small enough, which contradict the definition of $L$. If $Q(L, x_0) < 0$, then $Q(L-\epsilon, x_0)$ is stille negative for $\epsilon > 0$ small enough, which is again a contradiction. So $x_0$ is a root of $Q(L,x)$ and of $\partial_x Q(L,x)$, thus it is a double root. The discriminant thing is usual.

• The point 3 is Maple :

  factor(discrim(Q, x));

You can also use quantifier elimination algorithms. This gives you precisely for which $k$ the polynomial $Q(k,x)$ is positive. For example Mathematica implements such an algorithm :

In[1] := Resolve[ForAll[{x}, Q > 0]]
Out[1] := k < 0 || k > 4

I insist that you can trust this result, it is an exact algorithm not a numerical heuristic.