## nubeer 5 years ago find all roots of z^5+2z^3+6z^2+12 =0 using demoivre theroem anyone plz?

1. lgbasallote

the demoivre theorem i am familiar with is the one in Calculus. Is that the one you're pertaining to?

2. vishal_kothari
3. Nubeer

yes its in caculus.

4. mathmate

You'll see why when you factor the left-hand side of the equation to (z^2+2)*(z^3+6) = 0

5. Nubeer

ahh yes i think this is the factor this helps alot.. can u tell me how u got that... dont tell me its again hit and trial

6. Mr.Math

$(z^5+2z^3)+(6z^2+12)=z^3(z^2+2)+6(z^2+2)=(z^3+6)(z^2+2).$

7. Nubeer

well thanks .. this way is also a good one.. can u tell me any website whichhave questions like these with solutions.

8. Mr.Math

This was just a simple factorization, You can search factorization techniques. You still need to solve $$(z^3+6)(z^2+2)=0$$, which implies $$z^2+2=0$$ or $$z^3+6=0$$.

9. Mr.Math

$$z^2+2=0$$ is very easy to solve and you it gives the roots $$z=\pm \sqrt{2}i$$.

10. Mr.Math

You can use the demoivre theorem to solve $$z^3+6$$, (i.e find the third roots of $$-6$$).

11. Nubeer

so this last term will give us three roots?

12. Mr.Math

Yep, The notes vishal kothari gave are very good. You should probably read them.

13. Nubeer

ya i read them all.. they are helpful.. but still was kinda confused of how getting these roots.. but thank you.. do u know any website for more questions about this? with solutions?

14. Mr.Math

I don't, but you can search the internet for solutions of polynomials over complex numbers. I would look for you, but I have to leave now.

15. anonymous

put z=e^ix =cosx+isinx then proceed

16. Nubeer

@Mr.Math . thank you very much have a nice day :)

17. Mr.Math

You're welcome!