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.
 A: 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).
A: You do not need to compute everything. From $x = \sqrt{2}-1$, we get the following:$$p(x) = x^2 + 2x -1  = 0$$Now perform long division on the given polynomial and get this form:$$f(x) = Q(x)p(x) + R(x)$$Now we know that $p(\sqrt{2}-1) = 0$ so we have$$f(\sqrt{2}-1) = R(\sqrt{2}-1)$$I will leave the computational work to you, but it should be easy because the remainder turns out to be a constant, so you don't even need to plug anything in.
A: $$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1\\=x(x((x-1)(x+1)(x(x+2)-1)x^4+3)+6)+1$$
$\Longrightarrow f(\sqrt 2 -1)=\boxed{\color{red}{4}}$
A: 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:


*

*$1 + 0 = 1$

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

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

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

*$\ldots$

