Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

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

Answer 6

yes that is right..

Answer 7

root is on both..

Answer 8

\[\mathbb{yes}\]

Answer 9

and is it an x^3 in the equation?

Answer 10

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

Answer 11

yes it is in equion

Answer 12

\(\large x = \sqrt{\frac{5}{3}}\)

\(\large x^2 = \frac{5}{3}\)

\(\large 3x^2 = 5\)

\(\large 3x^2 - 5 = 0\)

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

Answer 13

something doesnt feel right...

Answer 14

yaa..

Answer 15

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

Answer 16

its ok..

Answer 17

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

Answer 18

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

Answer 19

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
\( 3(x^2-\frac{5}{3})(x^2+2x+1) =0\)
\( (3x^2-5) (x^2+2x+1)=0\)

Answer 20

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