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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?

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    $\begingroup$ "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. $\endgroup$ – Eevee Trainer Aug 11 at 1:35
  • $\begingroup$ 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. $\endgroup$ – NickD Aug 11 at 16:45
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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.

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    $\begingroup$ if $x \leq -1$ then $x^4 \geq |x|,$ so $x^4 + x \geq 0$ $\endgroup$ – Will Jagy Aug 11 at 2:38
  • $\begingroup$ Yes, that's a nice way to dispatch that case. $\endgroup$ – quasi Aug 11 at 2:46
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If $x$ is a real number, then $$x^4+x+1=\left(x^2-\frac12\right)^2+\left(x+\frac12\right)^2+\frac12>0\,.$$

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

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    $\begingroup$ check your $p'$ calculation. Leading coefficient is...? $\endgroup$ – user24142 Aug 11 at 1:23
  • $\begingroup$ 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.. $\endgroup$ – Joãonani Aug 11 at 1:25
  • $\begingroup$ Show that $p(root)>0$ $\endgroup$ – Tsemo Aristide Aug 11 at 1:27
  • $\begingroup$ 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? $\endgroup$ – Joãonani Aug 11 at 1:28
  • $\begingroup$ in your sum, you have $1$, a positive term and a negative term inferior to $1$ $\endgroup$ – Tsemo Aristide Aug 11 at 1:30
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$$p := x^4 + x + 1 = \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}^\top \begin{bmatrix} 1 & 0 & -t\\ 0 & 2 t & 0.5\\ -t & 0.5 & 1\end{bmatrix} \begin{bmatrix} x^2\\ x\\ 1\end{bmatrix}$$

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

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$x^4$ 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.

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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.

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