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

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

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.

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 $+\infty$.