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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question