I thought I had this down, but I was wrong.
$\displaystyle (4x^3 -17ix^2+5x+(7-3i))/x-3i$
Help?
Hello,
This would be the same as for real numbers
$\displaystyle f(x)=\frac{4x^3-17ix^2+5x+(7-3i)}{x-3i}$
$\displaystyle 4x^3=4x^2*x=4x^2*x+\underbrace{4x^2*(-3i)-4x^2*(-3i)}_{=0}=4x^2(x-3i)+12ix^2$
(I made appear the factor (x-3i))
$\displaystyle \begin{aligned} \implies f(x) &=\frac{4x^2(x-3i)+12ix^2-17ix^2+5x+(7-3i)}{x-3i} \\ \\
&=4x^2+\frac{-5ix^2+5x+(7-3i)}{x-3i} \end{aligned}$
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$\displaystyle -5ix^2=-5ix(x-3i)-15 i^2 x=-5ix(x-3i)+15x$
$\displaystyle \begin{aligned} \implies f(x) &=4x^2+\frac{-5ix(x-3i)+15x+5x+(7-3i)}{x-3i} \\ \\
&=4x^2-5ix+\frac{20x+(7-3i)}{x-3i} \end{aligned}$
---------------------
$\displaystyle 20x=20(x-3i)+60i$
$\displaystyle \begin{aligned} \implies f(x) &=4x^2-5ix+\frac{20(x-3i)+60i+7-3i}{x-3i} \\ \\
&=4x^2-5ix+20+\frac{7+57i}{x-3i} \end{aligned}$
Is it clear enough ? :x