How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?

Let $$p(x)=x^4+x+1$$ be a polynomial in $$\mathbb{R}[x]$$. How can I prove that $$p$$ doesn't have real roots?

My attempt:

From calculus, I know that

$$\lim_{x \to \pm\infty} p(x) = \infty\,.$$

Then, if it has real roots, then we have two or four real roots. I know that if $$\gcd \left( p(x), p'(x) \right) = 1\,,$$ then the roots is simple. Is there another hint?

• "Then, if it has real roots we have two or four real roots." -- Minor correction on this point, but this isn't necessarily true. For instance, $f(x)=(x-1)^2$ has only one root, $x=1$, because it "sits" on the $x$-axis. $f$ might only have real roots, but they're also not distinct ... so I guess it's a matter of convention/preference as to how you count them. It might be worth being clear on such details due to such ambiguities. – Eevee Trainer Aug 11 at 1:35
• They are "usually" counted with multiplicity though: a lot of the theorems (e.g. the corollary of the FTA that states that an nth degree polynomial has n roots and Bezout's theorem) are stated using multiplicity in order to simplify their statements. IMO, that should be the default interpretation. – NickD Aug 11 at 16:45

Consider three cases . . .

• If $$x\ge 0$$ then $$x^4+x+1\ge 1$$.$$\\[4pt]$$
• If $$-1 < x < 0$$ then $$x^4+x+1 > x^4 + (-1) + 1 > 0$$.$$\\[4pt]$$
• If $$x\le -1$$ then $$x^4+x+1 \ge x^2+x+1 = \Bigr(x+\frac{1}{2}\Bigr)^2+\frac{3}{4}$$.

Thus in all three cases, $$x^4+x+1$$ is positive.

It follows that $$x^4+x+1$$ has no real roots.

• if $x \leq -1$ then $x^4 \geq |x|,$ so $x^4 + x \geq 0$ – Will Jagy Aug 11 at 2:38
• Yes, that's a nice way to dispatch that case. – quasi Aug 11 at 2:46

If $$x$$ is a real number, then $$x^4+x+1=\left(x^2-\frac12\right)^2+\left(x+\frac12\right)^2+\frac12>0\,.$$

Hint: $$p'(x)=4x^3+1$$ has one root $$c$$ (to see this, remark that $$p"(x)\geq 0$$), if $$x if $$x>c, p'(c>0$$, show that $$p(c)>0$$. The function decreases from $$-\infty$$ to $$c$$ and increases from $$c$$ to $$+\infty$$.

• check your $p'$ calculation. Leading coefficient is...? – user24142 Aug 11 at 1:23
• I know that $p'$ has one real root, it is given by $\frac{-1}{\sqrt[3]{4}}$, but honestly I don't know how it solves my question.. – Joãonani Aug 11 at 1:25
• Show that $p(root)>0$ – Tsemo Aristide Aug 11 at 1:27
• It's easy, $\frac{-1}{\sqrt[3]{4}}< 1$ then $(\frac{-1}{\sqrt[3]{4}})^4 + \frac{-1}{\sqrt[3]{4}} +1$ is positive...But how can I conclude that? – Joãonani Aug 11 at 1:28
• in your sum, you have $1$, a positive term and a negative term inferior to $1$ – Tsemo Aristide Aug 11 at 1:30

$$p := x^4 + x + 1 = \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}^\top \begin{bmatrix} 1 & 0 & -t\\ 0 & 2 t & 0.5\\ -t & 0.5 & 1\end{bmatrix} \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}$$

where $$t \in \Bbb R$$. Using Sylvester's criterion, we learn that the (symmetric) matrix above is positive semidefinite for $$t = 0.5$$. Using the Cholesky decomposition,

$$\begin{bmatrix} 1 & 0 & -0.5\\ 0 & 1 & 0.5\\ -0.5 & 0.5 & 1\end{bmatrix} = {\rm L} {\rm L}^\top$$

where

$${\rm L} = \begin{bmatrix} \color{blue}{1} & 0 & 0\\ 0 & \color{magenta}{1} & 0\\ \color{blue}{-\frac{1}{2}} & \color{magenta}{\frac{1}{2}} & \color{red}{\frac{\sqrt{2}}{2}}\end{bmatrix}$$

and, thus,

$$p = \left( \color{blue}{x^2 - \frac12} \right)^2 + \left( \color{magenta}{x + \frac12} \right)^2 + \left( \color{red}{\frac{\sqrt{2}}{2}} \right)^2 > 0$$

which is the exact same sum of squares (SOS) decomposition in this answer.

SymPy code

>>> from sympy import *
>>> t = symbols('t', real=True)
>>> M = Matrix([[ 1,   0,  -t],
[ 0, 2*t, 1/2],
[-t, 1/2,   1]])
>>> L = M.subs(t,1/2).cholesky()
>>> L
Matrix([
[   1,   0,         0],
[   0,   1,         0],
[-1/2, 1/2, sqrt(2)/2]])


Related

$$x^4$$ is positive except for $$x = 0$$. $$x + 1$$ is positive for $$x > -1$$. So the only possibility for the polynomial (the sum of those 2 parts) to be negative is for $$x \le -1$$.

But the polynomial is positive at $$x = -1$$ and $$x^4$$ grows much faster than $$x + 1$$ for $$|x| > 1$$ so the polynomial is positive everywhere.

By studying $$p'$$ you easily see that $$p$$ is a decreasing function on $$]-\infty, -1/\sqrt[3]{4}]$$ and an increasing function on $$[-1/\sqrt[3]{4},\infty[$$. From that, you now that the minimal value $$p(x)$$ is $$p(-1/\sqrt[3]{4})$$ which is positive. Then $$p$$ has no real root.