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

1. Lumenaire

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

2. lgbasallote

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

3. Lumenaire

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

4. apoorvk

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. lgbasallote

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

6. nilankshi

yes that is right..

7. nilankshi

root is on both..

8. apoorvk

$\mathbb{yes}$

9. Lumenaire

and is it an x^3 in the equation?

10. Lumenaire

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

11. nilankshi

yes it is in equion

12. lgbasallote

$$\large x = \sqrt{\frac{5}{3}}$$ $$\large x^2 = \frac{5}{3}$$ $$\large 3x^2 = 5$$ $$\large 3x^2 - 5 = 0$$ 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. lgbasallote

something doesnt feel right...

14. nilankshi

yaa..

15. lgbasallote

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

16. nilankshi

its ok..

17. apoorvk

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. apoorvk

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