Question

if √ 5/3 and -√ 5/3 are two zeroes of polynomial 3x^4 +6x^3-2x^2-10x-5, then it's other two zeroes are

Answer 1

Given x = root 5/3 and x = root -5/3.

           x - root 5/3 = 0 and x + root 5/3 = 0
 
           x^2 - 5/3 = 0.

Given polynomial 3x^4 + 6x^3 - 2x^2 - 10x - 5.

We should apply the division algorithm.

x^2 - 5/3) 3x^4 + 6x^3 - 2x^2 -10x - 5(3x^2 + 6x + 3

                3x^4             - 5x^2

               -----------------------------------------

                              6x^3  + 3x^2 - 10x - 5

                              6x^3              - 10x

              ---------------------------------------------

                                        3x^2             -5

                                       3x^2              -5

          -----------------------------------------------------
      
                                             0

           ------------------------------------------------------




Quotient is 3x^2 + 6x + 3

                   3x(x+1) + 3(x+1)

                   (3x+1)(x+1) = 0

                  3(x+1)(x+1) = 0

                    (x+1)(x+1) = 0

                     x = -1,-1.


      Hope this helps!     

Answer 2

Answer

x=1,12,√3,−√3
There appears to be error in posting the question.
As mentioned the two zeros are not √3,and√3.
I found that the two zeros must be √3,and−√3
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.
Given equation is 
f(x)=2x4−3x3−5x2+9x−3......(1)
Given that two of its zeros are x=√3,−√3
⇒(x+√3),(x−√3) are two factors of equation (1)
⇒(x+√3)(x−√3)=(x2−3)is a factor of the polynomial.

If we divide the equation (1) by the above quadratic by long division method we get another quadratic which is a factor of equation (1)
∴2x4−3x3−5x2+9x−3x2−3, we get dividend as 
2x2−3x+1

To find factors of second quadratic we use split the middle term method
2x2−2x−x+1, paring and taking out the common factors we get
2x(x−1)−(x−1)
⇒(x−1)(2x−1) 
Setting each factor =0, we obtain remaining two zeros as
x=1,12