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

Mathematics
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I assume you mean: 3x^4+6x^3-2x^2-10x-5
the roots aree \(\large \sqrt{\frac{5}{3}}\) and \(\large -\sqrt{\frac{5}{3}}\) right? or is the root applied on 5 only?
is there a square root sign at all? I understood that 5/3 and -5/3 are the roots

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Other answers:

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.
he/she said root 5/3 and root -5/3..she already said zeroes so i assume root->radical
yes that is right..
root is on both..
\[\mathbb{yes}\]
and is it an x^3 in the equation?
(6x^2) should be 6x^3
yes it is in equion
\(\large x = \sqrt{\frac{5}{3}}\) \(\large x^2 = \frac{5}{3}\) \(\large 3x^2 = 5\) \(\large 3x^2 - 5 = 0\) 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
something doesnt feel right...
yaa..
hmm idk how to do this sorry... @apoorvk may know though
its ok..
Sorry, Trance. :) and Nilakshi. I 'll work on the solution now
Thanks :) @Rohangrr if you want to talk about it, please make a new question.
  • 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 \( 3(x^2-\frac{5}{3})(x^2+2x+1) =0\) \( (3x^2-5) (x^2+2x+1)=0\)
Hmm, I got so bored, Phi already posted it. :/ Good job, Phi

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