# Thread: Synthetic Division with imaginary numbers!

1. ## Synthetic Division with imaginary numbers!

I thought I had this down, but I was wrong.

$(4x^3 -17ix^2+5x+(7-3i))/x-3i$

Help?

2. Hello,

Originally Posted by mankvill
I thought I had this down, but I was wrong.

$(4x^3 -17ix^2+5x+(7-3i))/x-3i$

Help?
This would be the same as for real numbers

$f(x)=\frac{4x^3-17ix^2+5x+(7-3i)}{x-3i}$

$4x^3=4x^2*x=4x^2*x+\underbrace{4x^2*(-3i)-4x^2*(-3i)}_{=0}=4x^2(x-3i)+12ix^2$
(I made appear the factor (x-3i))

\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}

---------------------

$-5ix^2=-5ix(x-3i)-15 i^2 x=-5ix(x-3i)+15x$

\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}

---------------------

$20x=20(x-3i)+60i$

\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

3. ...I think so. It's still kind of confusing, but I'll work on it.

Thanks!