Finding roots of a Complex Polynomial 

Question
Given that x = 3 is a solution to $$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$ find the other two solutions. 


What I have attempted;
If $x = 3$ is a root then $(x-3)$ is a factor 
so $$ (x-3)(x^2 + Ax + B) = 0 $$
$$  x^3 + (A-3)x^2 + (B-3A)x - 3B = 0 $$ 
Equation Coefficients 
$$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$
$$ x^2 :−3+A=−7−3i $$
$$ ⇒ A=−4−3i  $$
$$\text{Constant: } -3B = -6(2+3i) $$
$$ ⇒ B=4+6i $$
$$ (x-3)(x^2 -(4+3i)x + 4+6i) = 0 $$
$$ x = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x = {4+3i\pm\sqrt{(4+3i)^2-4(4+6i)} \over 2} $$
$$ x = {4+3i\pm\sqrt{-9} \over 2} $$
$$ x = {4+3i±3i\over 2} $$
$$ x_1 = 3 $$ 
$$ x_2 = 2+3i $$
$$ x_3 = 2 $$
Hopefully this is correct , and if so I was wondering if there is an alternative method in solving this. 
I was thinking because we are given x = 3 is a root and (x-3) is a factor we can do polynomial division with complex numbers? Is it possible to do polynomial division with complex  numbers? If so can someone show how to get the same answer by using that method..
 A: 
Is it possible to do polynomial division with complex numbers? 

Yes, polynomial division works with complex numbers as well.

If so can someone show how to get the same answer by using that method..

A efficient method for polynomial evaluation and division is Horner's method.
Applied to your polynomial
$$
   f(x) = x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) 
$$
and $x_1 = 3$ it gives:
1     -7-3i    16+15i   -12-18i
       3      -12-9i     12+18i
-------------------------------
1     -4-3i     4+6i      0

From the Wikipedia article:

The entries in the third row are the sum of those in the first two. Each entry in the second row is the product of the x-value (3 in this example) with the third-row entry immediately to the left. The entries in the first row are the coefficients of the polynomial to be evaluated.
  ...
  As a consequence of the polynomial remainder theorem, the entries in the third row are the coefficients of the second-degree polynomial, the quotient of f(x) on division by  x-3 . 

It follows that $f(3) = 0$ and
$$
f(x) = (x-3)(x^2 -(4+3i)x + (4+6i))
$$
which is exactly what you got by equating coefficients.
If you know (or guess) another root then you can repeat the process.
Otherwise your method of solving the quadratic equation is fine.
A: If $a$ and $b$ are the other roots, then you can use Vieta's formulas for the sum and product of roots:
$$
3+a+b=7+3i,
\qquad
3ab=6(2+3i)
$$
and from those get a quadratic equation for $a$ and $b$.
A: Call the cubic $p(x).$ $\bullet$ Method 1: We have $p(3)=0.$ We have $$p(x)=p(x)-p(3)=(x-3)p'(3)+(x-3)^2p''(3)/2+(x-3)^3p'''(3)/6=$$ $$=[x-3]\cdot [p'(3)+(x-3)p''(3)/2+(x-3)^2p'''(3)/6].$$ For computation, let $x=y+3$ and find the solutions, in $y$, of $ p'(3)+y p''(3)/2+y^2 p'''(3)/6=0.$ $\bullet$  Method 2. For any cubic $p$ with leading term $A x^3$ we have $p(x)=(x-3) A x^2 +q(x)$ where $q$ is of degree $2$ or less, with leading term $B x^2$ ($B$ may be $0$). We have $q(x)=(x-3)B x+r(x)$ where $r(x)=Cx +D$ with constant $C,D.$ We have $r(x)= (x-3)C +E$ with constant $E.$ If $p(3)=0$ then $E=0,$ and $$p(x)=(x-3)A x^2 +q(x)=(x-3)A x^2+(x-3)B x+r(x)=$$ $$=(x-3)A x^2+(x-3)B x +(x-3)C=(x-3)(A x^2+B x +C).$$ In practice this method is fast and easy.
