unique real - integer polynomial

Question

If $ f(x) = x^{10} + 2x^9 - 2x^8 - 2x^7 + x^6 + 3x^2 + 6x + 2014 $ so can anyone here proof $f(\sqrt[2]{2} -1) = 2017$

Please do it with hands not by computer help or calculator help

Answer 1

This is my thought:

$$f(x)=x^6(x^4+2x^3-2x^2-2x+1)+3x(x+2)+2014=x^6h(x)+g(x)+2014$$ where the definitions of the polynomials $h$ and $g$ are obvious.

Now observe that $g(\sqrt{2}-1)=3$ and so $g(x)+2014=2017$.

Finally, you can explicitly see that $h(x)$ is divisible by the linear polynomial $x-\sqrt{2}+1$ and so $h(x)=(x-\sqrt{2}+1)h'(x)$, i.e. $h(\sqrt{2}-1)=0$.

Answer 2

Note that $\sqrt2-1$ is a zero of $x^2+2x-1$. Now observe that the original polynomial, with the more convenient constant term $-3$ replacing $2014$, is divisible by this quadratic.

Answer 3

$ ( \sqrt{2} -1 )^2=2+1-2 \sqrt{2}=3-2 \sqrt{2}$
$( \sqrt{2} -1 )^6 = (3-2 \sqrt{2})^3 = 27 - 54 \sqrt{2} + 72 - 16 \sqrt{2} = 99 - 70 \sqrt{2}$
$( \sqrt{2} -1 )^7 = (99-70 \sqrt{2})(\sqrt{2}-1)=169 \sqrt{2} - 239$
$( \sqrt{2} -1 )^8 = (169 \sqrt{2} -239)(\sqrt{2}-1) = 577 - 408 \sqrt{2}$
Using same method we find, $( \sqrt{2} -1 )^9 = 985 \sqrt{2}-1393$ and $( \sqrt{2} -1 )^{10} = 3363-2378 \sqrt{2}$.

Therefore, $f(\sqrt{2} - 1)= 3363-2378 \sqrt{2} +2(985 \sqrt{2}-1393) -2(577 - 408 \sqrt{2}) -2(169 \sqrt{2} - 239) + 99 - 70 \sqrt{2} + 3(3-2 \sqrt{2}) + 6 (\sqrt{2}-1) + 2014 $

And by simplifying it, you will get, $$f(\sqrt{2}-1) = 2017$$

Answer 4

Using Horner's Method. It may not look pretty, but notice that all the calculations are easy, and the numbers never grow very large, and you don't have to notice in advance any special features of the polynomial.

$x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+2014$ $=2014+x(6+x(3+x^4(1+x(-2+x(-2+x(2+x))))))$

$x=\sqrt2-1$, $\quad2+x=\sqrt2+1$, $\quad x(2+x)=(\sqrt2-1)(\sqrt2+1)=1$,

$-2+x(2+x)=-1$, $\quad x(-2+x(2+x))=1-\sqrt2$, $\quad-2+x(-2+x(2+x))=-1-\sqrt2$,

$x(-2+x(-2+x(2+x)))=-1$, $\quad1+x(-2+x(-2+x(2+x)))=0$,

$x^4(1+x(-2+x(-2+x(2+x))))=0$, $\quad3+x^4(1+x(-2+x(-2+x(2+x))))=3$,

$x(3+x^4(1+x(-2+x(-2+x(2+x)))))=3\sqrt2-3$,

$6+x(3+x^4(1+x(-2+x(-2+x(2+x)))))=3(\sqrt2+1)$,

$x(6+x(3+x^4(1+x(-2+x(-2+x(2+x))))))=3$, $\quad f(\sqrt2-1)=2014+3=2017$.