1. ## factorise with i

im stuck on how to factor out the i.

$\displaystyle 108+36\sqrt{3i}+12i\sqrt{3i}+36i$

2. Originally Posted by skystar
im stuck on how to factor out the i.

$\displaystyle 108+36\sqrt{3i}+12i\sqrt{3i}+36i$
Try rewriting the expression. I'll give you a jump start:

$\displaystyle 108+36\sqrt{3}\sqrt{i}+12i\sqrt{3}\sqrt{i}+36i$.

My next hint is that $\displaystyle \sqrt{i}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$.

I'm not quite sure that you can indeed factor out an i, but at least you can simplify the expression and get it in the form $\displaystyle a+bi$. Hope this helped you out a little bit!

3. Hello, skystar!

Why do you want to factor out an $\displaystyle i$ ?

Factor: .$\displaystyle 108+36\sqrt{3i}+12i\sqrt{3i}+36i$

We have: .$\displaystyle 108 + 36\sqrt{3i} + 36i + 12i\sqrt{3i}$

Factor out 12: .$\displaystyle 12\,\bigg[9 + 3\sqrt{3i} + 3i + i\sqrt{3i}\bigg]$

Factor by grouping: .$\displaystyle 12\,\bigg[3(3+\sqrt{3i}) + i(3+\sqrt{3i})\bigg]$

. . Therefore: .$\displaystyle 12\,(3+\sqrt{3i})\,(3 + i)$