Two zeroes are √3 and -√3,

Therefore (x -√3)(x - (-√3) = (x - √3)(x + √3) is a factor of p(x).

Let us divide p(x) by (x - √3)(x + √3) = (x² - 3)

⇒ (2x⁴ - 3x³ - 5x² + 9x - 3) = (x² - 3)(2x² - 3x + 1)

= (x - √3)(x + √3)(2x² - 2x - x + 1)

= (x - √3)(x + √3)(2x(x - 1) - 1(x - 1))

= (x - √3)(x + √3)(2x - 1)(x - 1)

Hence,

2x - 1 = 0 or x - 1 = 0

or x = 1

Hence, other two zeroes are or 1.