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

1. lgbasallote Group Title

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

2. vishal_kothari Group Title
3. nubeer Group Title

yes its in caculus.

4. mathmate Group Title

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

5. nubeer Group Title

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 Group Title

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

7. nubeer Group Title

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

8. Mr.Math Group Title

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 Group Title

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

10. Mr.Math Group Title

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

11. nubeer Group Title

so this last term will give us three roots?

12. Mr.Math Group Title

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

13. nubeer Group Title

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 Group Title

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. arijit.mech Group Title

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

16. nubeer Group Title

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

17. Mr.Math Group Title

You're welcome!