How can I do a polynomial long division with complex numbers?

Question

So I have been trying to solve following equation since yesterday, could someone tell me what I am missing or doing wrong? I would be very grateful.

x^2 + (-2-6i)x - 11 + 2i : (x-(3+4i))=

So I tried to do this the following way:´

    x^2 + (-2-6i)x - 11 + 2i : (x-(3+4i))= x`+ (1+2i)
 -  x^2 -(3+ 4i)x
    -------------
        (1+2i)x -11 `
   -    (1+2i)x + 10i -5
    ----------------------
                10i -6 +2i 
                  = 12i - 6 

Answer 1

It is not correct. Note that\begin{align}x^2+(-2-6i)x-11+2i-x\times\bigl(x-(3+4i)\bigr)&=(1-2i)x-11+2i\\&=(1-2i)\bigl(x-(3+4i)\bigr)\end{align}and that therefore$$x^2+(-2-6i)x-11+2i-\bigl(x-(1-2i)\bigr)\times\bigl(x-(3+4i)\bigr)=0.$$