## anonymous 4 years ago obtain all the zeroes of 3x^4+6x^2-2x^2-10x-5,if two zeroes are root5/3 and -root 5/3

1. anonymous

I assume you mean: 3x^4+6x^3-2x^2-10x-5

2. anonymous

the roots aree $$\large \sqrt{\frac{5}{3}}$$ and $$\large -\sqrt{\frac{5}{3}}$$ right? or is the root applied on 5 only?

3. anonymous

is there a square root sign at all? I understood that 5/3 and -5/3 are the roots

4. anonymous

divide the whole system by its two factors that are given (or try synthetic division) After two whole divisions by each factor, you 'll be left with a quadratic as the quotient, factorising which you'll get you the other two roots.

5. anonymous

he/she said root 5/3 and root -5/3..she already said zeroes so i assume root->radical

6. anonymous

yes that is right..

7. anonymous

root is on both..

8. anonymous

$\mathbb{yes}$

9. anonymous

and is it an x^3 in the equation?

10. anonymous

(6x^2) should be 6x^3

11. anonymous

yes it is in equion

12. anonymous

$$\large x = \sqrt{\frac{5}{3}}$$ $$\large x^2 = \frac{5}{3}$$ $$\large 3x^2 = 5$$ $$\large 3x^2 - 5 = 0$$ as for the other zero... $$\large x = -\sqrt{\frac{5}{3}}$$ $$\large x^2 = \frac{5}{3}$$ $$\large 3x^2 = 5$$ $$\large 3x^2 - 5 = 0$$ so two of the factors would be (3x^2 - 5)^2

13. anonymous

something doesnt feel right...

14. anonymous

yaa..

15. anonymous

hmm idk how to do this sorry... @apoorvk may know though

16. anonymous

its ok..

17. anonymous

Sorry, Trance. :) and Nilakshi. I 'll work on the solution now

18. TranceNova

Thanks :) @Rohangrr if you want to talk about it, please make a new question.

19. phi

I would multiply out $(x-\sqrt{5/3})(x+\sqrt{5/3}) = x^2-\frac{5}{3}$ we can write this as $$3x^2-5$$ (why? see below) divide this into the original equation to get $$x^2+2x+1$$ now factor this to get the remaining roots. * we are looking for $$(x^2-\frac{5}{3}) (x^2+2x+1) =0$$ if we multiply both sides by 3 we get $$3(x^2-\frac{5}{3})(x^2+2x+1) =0$$ $$(3x^2-5) (x^2+2x+1)=0$$

20. anonymous

Hmm, I got so bored, Phi already posted it. :/ Good job, Phi