## 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 Group Title

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

2. lgbasallote Group Title

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 Group Title

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

4. apoorvk Group Title

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 Group Title

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

6. nilankshi Group Title

yes that is right..

7. nilankshi Group Title

root is on both..

8. apoorvk Group Title

$\mathbb{yes}$

9. Lumenaire Group Title

and is it an x^3 in the equation?

10. Lumenaire Group Title

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

11. nilankshi Group Title

yes it is in equion

12. lgbasallote Group Title

$$\large x = \sqrt{\frac{5}{3}}$$ $$\large x^2 = \frac{5}{3}$$ $$\large 3x^2 = 5$$ $$\large 3x^2 - 5 = 0$$ 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 Group Title

something doesnt feel right...

14. nilankshi Group Title

yaa..

15. lgbasallote Group Title

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

16. nilankshi Group Title

its ok..

17. apoorvk Group Title

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

18. TranceNova Group Title

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

19. phi Group Title

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 Group Title

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