If two zeros of the polynomial p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2 are √2 and – √2, find its other two zeros.

It is given in the question that,

p (x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2 having zeros and -

∴ The polynomial is (x + ) (x - ) = x^{2} – 2

Let us now divide p (x) by (x^{2} – 2)

2x^{4} – 3x^{3} – 3x^{2} + 6x – 2 = (x^{2} – 2) (2x^{2} – 3x + 1)

= (x^{2} – 2) [(2x^{2} – (2 + 1) x + 1]

= (x^{2} – 2) (2x^{2} – 2x – x + 1)

= (x^{2} – 2) [(2x (x – 1) – 1 (x – 1)]

= (x^{2} – 2) (2x – 1) (x – 1)

Hence, the other two zeros are and 1

18