Number of real roots of $f(x) = (x^6) + 2(x^4) + (x^2) - 2(x) + 1$

Question

The question is to find the number of real roots of the polynomial -

$f(x) = (x^6) + 2(x^4) + (x^2) - 2(x) + 1$

I used the Descartes rule of signs. Using it, it is clear that there are maximum number of 2 positive real roots and no possible negative root. But after that I was stuck.

Then I changed the approach and calculated it's derivative. Using Descartes rule of signs, I determined that it's derivative has maximum of 1 positive real root and no imaginary root. It means graph is non-monotonic. So, I tried to find the minimum value of function by keeping derivative zero. But I stuck yet again.

Answer 1

Note
$x^6 \ge 0$ with equality only for $x=0$;
$2x^4 \ge 0$ with equality only for $x=0$;
$x^2-2x+1 = (x-1)^2 \ge 0$ with equality only for $x=1$.
Adding, we conclude $f(x) \ge 0$ with no equality.

Answer 2

I'm not sure but the last three terms we can factor them \begin{equation*} x^2-2x+1=(x-1)^2 \end{equation*} Hence, $f(x)=x^6+2x^4+(x-1)^2\geq 0$. Since each term is quadratic, we have $f(x)=0$ if and only if \begin{equation*} x^6=2x^4=(x-1)^2=0 \end{equation*} The above implies that $0=x=1$ and therefore, $f(x)$ has no real roots.