Qualitative inspection of solutions to $x^{4}-2x+1=0$ Consider the following polynomial
$$
x^{4}-2x+1=0
$$
Is it possible to check if there is or there is not a solution in $x\in\left]0,1\right[$ without explicitly evaluating the expression? What other tests are there to qualitatively classify the solutions for this polynomial?
 A: Note that
$$
f'(x) = 4x^3-2 \quad \text{and} \quad f''(x) = 12x^2 > 0
$$
which means the function is always concave up.
Since $f(0)=1$ and $f(1)=0$, and the function is concave up, it has a root in $(0,1)$ only if there is a relative minimum in the interval, which happens iff $f'(x)=0$ for some $x$ in $(0,1)$. However,
$$
f'(x) = 0 \iff x^3 = 1/2 \iff x = 2^{-1/3} \in (0,1),
$$
so $f$ must have a root in $(0,1)$.

The actual graph is below:

A: Since $x=1$ works, 
$$
x^4 - 2x + 1 = (x-1)p(x)\quad [p \in \mathbb R[x]_3]. 
$$
Now
$$
x^4 - 2x +1 = x^2(x^2 -1) + (x-1)^2 = (x-1)(x-1+x^2(x+1)) = (x-1)(x^3 + x^2 + x - 1). 
$$
Then $p(x) = x^3 + x^2 + x-1$. Since
$$
p(1)= 2 >0, p(0) = -1 < 0,
$$
by intermediate value theorem, $p$ has a root in $(0,1)$, hence so does $x^4 - 2x + 1$. 
A: $f(x) = x^4-2x+1$.
There are two sign changes, so there are $0$ or $2$ positive roots.
$f(-x)$ has no sign changes. So there are no negative roots.
Since $f(1)=0$, there are two positive roots.
Also $f(0)=1$.
Since $f'(x) = 4x^3 - 2$ is positive for all $x \ge 1$ (because $f''(x) = 12x^2 \ge 0$ implies that $f'$ is increasing and $f'(1)=2 > 0$), then $f(x)$ is positive for all $x > 1$.
So the second root must be somewhere between $x=0$ and $x=1$.
