Just put the value of x (sounds easy)

$x-5=√3$ Squaring $\Rightarrow x^2-10x+22=0…(1)$ The required expression is $S=x^{11}-10x^{10}+22x^9+x^7-10x^6+22x^5+x^2-10x+41$ $=x^{9}(x^2-10x+22)+x^5(x^2-10x+22)+(x^2-10x+22)+19$ Now putting the value of $(1))$ $\Rightarrow S=0+0+0+19=\boxed{19}$

$f(x) = x^{11} - 10x^{10} + 22x^9 + x^7 - 10x^6 + 22x^5 + x^2 - 10x + 22x + 41$

Let $g(x) = x^2 - 10x +22$ . then we have:

$f(x) = x^9g(x) + x^5g(x) + g(x) + 19 = (x^9+x^5+1)g(x) + 19$

We note that the roots of $g(x)$ are $5 \pm \sqrt{3}$ . Therefore, $g(\sqrt{3}+5) = 0$ .

Therefore, $f(\sqrt{3}+5) = 0 + 19 = \boxed{19}$

If you use synthetic division, it turns out to be really easy.

Just factorize the equation given in the question as done in the following line { x }^{ 9 }({ x }^{ 2 }-10x+22)+{ x }^{ 5 }({ x }^{ 2 }-10x+22)+{ x }^{ 2 }-10x+41 now just check the first two brackets ,as they are same so take them as common and solve the equation in the brackets u will find that it has (\sqrt { 3 } \pm 5) as its solutions hence it becomes zero when we find f(\sqrt { 3 } +5 ) and the left equation can be easily solved.