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.

Just compute the value for each power $$x=\sqrt{2}-1$$ $$x^2=x \times x=3-2 \sqrt{2}$$ $$x^3=x\times x^2=5 \sqrt{2}-7$$ $$x^6=x^3 \times x^3=99-70 \sqrt{2}$$ $$x^7=x\times x^6=169 \sqrt{2}-239$$ $$x^9=x^7\times x^2=985 \sqrt{2}-1393$$ $$x^{10}=x^7\times x^3=3363-2378 \sqrt{2}$$ With this kind of problem, just work in a systematic manner.
Happy New Year !!
Edit
Parth Kohli's answer provides another interesting way to compute the different powers of $x$ (his/her solution is much more elegant and faster than my brute force based method).
Let us consider that the value for which we need to compute the polynomial is one of the roots of the quadratic $x^2+2x-1=0$. So $$x^2=-2x+1$$ $$x^3=-2x^2+x=-2(-2x+1)+x=5x-2$$ $$x^4=5x^2-2x=5(-2x+1)-2x=-12x+5$$ $$x^5=-12x^2+5x=-12(-2x+1)+5x=29x-12$$ $$x^6=29x^2-12x=29(-2x+1)-12x=-70x+29$$ $$x^7=-70x^2+29x=-70(-2x+1)+29x=169x-70$$ $$x^8=169x^2-70x=169(-2x+1)-70x=-408x+169$$ $$x^9=-408x^2+169x=-408(-2x+1)+169x=985x-408$$ $$x^{10}=985x^2-408x=985(-2x+1)-408x=-2378x+985$$ Please notice the nice pattern in the numbers. Compute now the expression and get the final result (and notice that you will not use at all the fact that $x=\sqrt 2-1$; this is explained in Parth Kohli's answer by the constant remainder).
