<s>[INST] 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..
 [/INST] 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.
</s>