Vtyjkyuk

Consider the following polynomial

$$
x^4-2x+1=0
$$

Is it possible to check if there is or there is not a solution in $xinleft]0,1right[$ without explicitly evaluating the expression? What other tests are there to qualitatively classify the solutions for this polynomial?

Note that
$$
f'(x) = 4x^3-2 quad textand quad f''(x) = 12x^2 > 0
$$
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:

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 $x^4 - 2x + 1$.

$f(x) = x^4-2x+1$.

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 $x=1$.