Since two of the zeros are 

x = √(5/3) and   x = -√(5/3)

Then

x - √(5/3) = 0  and x + √(5/3) = 0

And so

x - √(5/3)  and x + √(5/3)  are factors of

the given polynomial.

Therefore, the product

[x - √(5/3)][x + √(5/3)]

is a factor of the given polynomial

Multiplying that out:

x² - 5/3

So we divide the given polynomial by that.

But we must put in placeholder 0x for the 
missing x term:

                           3x² +  6x + 3
x² + 0x - 5/3)3x^4 + 6x³ – 2x² – 10x – 5 
              3x^4 + 0x³ - 5x²
                     6x³ + 3x² - 10x
                     6x³ + 0x² - 10x
                           3x² +  0x - 5                
                           3x² +  0x - 5

So we have factored the given polynomial as

(x²-5/3)(3x²+6x+3)

Factoring a 3 out of the second factor,

(x²-5/3)3(x²+2x+1)

Multiplying the 3 into to first factor eliminates the
fraction, so the factorization is

(3x²-5)(x²+2x+1)

The other 2 zeros can be gotten by setting the factor

x²+2x+1 = 0

(x+1)(x+1) = 0

x+1 = 0;   x+1 = 0
  x = -1;    x = -1

The other zero is -1 of multiplicity 2.

Edwin