I am having trouble understanding how to solve polynomials for 0 when there are more then one fraction in the equation. Is there a trick I am missing?

Question

anonymous · Answer

Do you have an example of a question?

anonymous · Answer

I do, $$(z ^{2})/6-(2z)/2-3=0$$

bahrom7893 · Answer

Nice picture joemath!

anonymous · Answer

I know what do if there is only one fraction in the polynomial but I am totally lost with 2. I missed class :/

anonymous · Answer

lol thanks :P
for this problem:
$$\frac{z^2}{6}-\frac{2z}{2} - 3 = 0$$
what you want to do is look at the denominators, and think of one number, that when you multiply the equation by this number, will get rid of them. It turns out to be the LCM (least common multiple) of the denominators.

anonymous · Answer

So I pull out two and then leave the 3 under the first?

anonymous · Answer

if the equation I typed in my last response correct?

anonymous · Answer

er, not if, was.   was the equation correct.

anonymous · Answer

yes

anonymous · Answer

Alright, so the denominators are 2, and 6.  Their LCM turns out to also be 6.  So multiplying everything by  we get:
$$6*\frac{z^2}{6}+6*\frac{2z}{2}-6*3= 6*0 \Rightarrow z^2+3*2z-18 = 0 \Rightarrow z^2+6z-18 = 0$$
No More fractions :)

anonymous · Answer

Oh... Thanks.. I see what I was doing vs what needs to be done. Thanks again. Back to homework.