A general approach for this type of questions I've come across multiple questions like these. $\left(\iota=\sqrt{-1}\right)$

If $f\left(x\right)=x^4-4x^3+4x^2+8x+44$, find the value of $f\left(3+2\iota\right)$. 
  If $f\left(z\right)=z^4+9z^3+35z^2-z+4$, find $f\left(-5+2\sqrt{-4}\right)$. 
  Find the value of $2x^4+5x^3+7x^2-x+41$, when $x=-2-\sqrt{3}\iota$. 

...and so on. The answers to these are real numbers. And the methods given to solve these include random manipulations of $x=\text{<the given complex number>}$ until we reach the given $f\left(x\right)$ or a term which $f\left(x\right)$ can be rewritten in terms of, so that we get the remainder. But all these approaches are completely random, and I see no logical approach. Can someone explain me exactly how would you solve a question of this format?

EDIT : Here's what I mean by random manipulations. Here's the solution to the third one. 
$x+2=-\sqrt{3}\iota\Rightarrow x^2+4x+7=0$
Therefore, $2x^4+5x^3+7x^2-x+41$  $=\left(x^2+4x+7\right)\left(2x^2-3x+5\right)+6$$=0\times\left(2x^2-3x+5\right)+6=6$
Now how are we supposed to observe that the given quartic could be factorized? We have not been taught any method to factorise degree four polynomials, unless one root is known, in which case I can reduce it into a product of a linear and a cubic. Further if a root of the cubic is visible. 

Unless there is some noticeable  relation between f and x(for example if $ f'(x)=0$ ) or some simplifying expression for f  (for example if $f(x)=1+x+x^2+x^3 +x^5$ and $x\ne 1$), or some special property of x (for example if $x^3=i$), all you can do is compute the complex arithmetic.There is considerable computer-science theory on efficient ways to compute polynomials. 
