<s>[INST] 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. 
 [/INST] Ill give a guide line we have $x=-2-\sqrt{3}i$ now bring two on lhs and square both sides. So we get $x^2+4x+4=-3$ so $x^2+4x+7=0$ ..(1).now just divide the given expression by this qyadratic so we have . $\frac{2x^4+5x^3+7x^2-x+41}{x^2+4x+7}$ . I hope now you know this type of division . If not look up on net. Sowe get the answer as 6 on division. Now we have $2x^4+5x^3+7x^2-x+41=(x^2+4x+7)(2x^3-3x+5)+6$ but according to eqn(1) we have $x^2+4x+7=0$ . Thus the value of $2x^4+5x^3+7x^-x+41=6$ . Hope now you can get ideas for other problems. Look up on the net for division of polynomials and you will know how did i divide the quartic with a quadratic. For first take value as $x=3+2i$ and for second we have $z=-{5}+2.2i$ so we have $z=-{5}+4i$. Suppose if we have $x=a+bi$ the best trick is to bring the  iota  independent term together with x here 'a' and squaring both the sides . This removes the iota thus dqn becomes $x^2-a^2=-b^2$ note $i^2=-1$ . Then we divide the given polynomial with this quadratic and get the answer. As a referrence take the example of my answer. Thats what a general solution can be like.
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