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

*

*Writing $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of two squares of quadratic forms

polynomials sum-of-squares-method matrices matrix-decomposition cholesky-decomposition
A: If $x$ is a real number, then $$x^4+x+1=\left(x^2-\frac12\right)^2+\left(x+\frac12\right)^2+\frac12>0\,.$$
A: 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.
A: $x^4$ 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.
A: 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.
A: $f'(x) = 4x^3 +1$.
To find extreme points $f'(x) = 4x^3 + 1=0 \implies x =-\sqrt[3]{\frac 14}$
$f''(x) = 12x^2$ and $f''(-\sqrt[3]{\frac 14})= 12(-\sqrt[3]{\frac 14}) > 0$ so this is a minimum.
So $f(x) \ge f(-\sqrt[3]{\frac 14})= \sqrt[3]{\frac 1{4^4}} -\sqrt[3]{\frac 14} + 1  > -\sqrt[3]{\frac 14}_1 >-1+1 = 0$ for all $x$
