The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.

You can make use of Horner's method. By further applying the formula $(a-b)(a+b) = a^2-b^2$ while sometimes factoring out common factors (-1 and 3 in this case), calculating the result becomes astonishingly nice:
$$\begin{array}{c|c|c|c|c|c|c|c|c|c|c|cc}
& 1 & 2 & -2 & -2 & 1 & 0 & 0 & 0 & 3 & 6 & 1\\
& + & + & + & + & + & + & + & + & + & + & + &\\
& 0 & \sqrt{2}-1 & 2-1 & -\sqrt{2}+1 & -1 & 0 & 0 & 0 & 0 & 3\sqrt{2}-3 & 3\\
\hline
x=\sqrt{2}-1 & 1 & \sqrt{2}+1 & -1 & -(\sqrt{2}+1) & 0 & 0 & 0 & 0 & 3 & 3(\sqrt{2}+1) & \boxed{\color{red}{4}}\\
\end{array}
$$
Short explanation:


*

*$1, 2, -2, \ldots, 1$ are the coefficient of the polynom from high to low, including zero ones!

*The third row (below the plus signs) always consists of the result in the last row of the last column multiplied by x. The value in the first column is 0.

*The fourth row is simply the sum of the first and third rows.
For example the first calculations of the scheme above would be as follows:


*

*$1 + 0 = 1$

*$(\sqrt{2}-1)*1 = \sqrt{2}-1$

*$2 + \sqrt{2}-1 = \sqrt{2}+1$

*$(\sqrt{2}+1)(\sqrt{2}-1) = 2 - 1 = 1$

*$\ldots$

