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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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 $+\infty$.

Answer 4

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

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.

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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]])

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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

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polynomials sum-of-squares-method matrices matrix-decomposition cholesky-decomposition

Answer 5

Another way would be to use AM-GM, like so: $$x^4+1=x^4+\frac13+\frac13+\frac13\geqslant 4\sqrt[4]{\frac{x^4}{3^3}}\geqslant |x|\geqslant -x$$ $\implies x^4+x+1>0$ as the inequalities cannot be simultaneously equal.

Answer 6

Yet another way:

$$ p(x) p(-x) \;=\; x^8 + 2 x^4 - x^2 + 1 \;=\; x^8 + x^4 + \left(x^2 - \frac{1}{2}\right)^2 + \frac{3}{4} \;\gt\; 0 $$

Answer 7

$x^4$ 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.

Answer 8

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.

Answer 9

$f'(x) = 4x^3 +1$.

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 $x$

Answer 10

Inconsistency of discriminants

Let $r$ be any such real root. Then $x^4+x+1=r^2x^2+x+1=0$ at that root, and so the latter quadratic polynomial has that same real root $r$.

It follows that the discriminant of the quadratic must satisfy

$1-4r^2\ge0\implies|r|\le\frac12.$

So the proposed real root must be no greater than $\frac12$ in absolute value, but this is contradicted by the fact that $1>|x^4|+|x|$ for all of $\{x:|x|\le\frac12\}$.

As an aside, the contradiction step also rules out complex conjugate roots with absolute value $\le\frac12$.

Answer 11

Another way

1. for $x> -1$

$$x^4+x+1>x^4\ge 0$$

2. for $x\le -1$ by $y=-\frac1{x}$ with $0<y\le 1$

$$x^4+x+1=\frac{y^4-y^3+1}{y^4}$$

and

$$y^4-y^3+1 \ge y^4-y^2+1\ge \frac34 $$

Answer 12

Note that

$1)$ If $x\ge 0$, then $p(x)\ge 1\implies p(x)$ does not have any non negative root.

$2)$ For $x\ne \pm 1, 0$, A.M $\ge G.M$ gives $p(x)=x^4+x+1\gt 2x^2+x$

In view of $1)$ suppose that $x<0$.

Note that $2x^2+x\le 0\implies 2x+1\ge 0\implies x\ge-1/2$

It follows that $x^4+x+1\ge x^4+1/2>1/2\implies p(x)\ne 0$ for $x<0: 2x^2+x\le 0$.

And if $2x^2+x>0$, then by $2)$, it follows that $p(x)>0\implies p(x)\ne 0$.

It follows that $p$ has no real solutions.