If two zeros of the polynomial p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 are √2 and – √2, find its other two zeros.
It is given in the question that,
p (x) = 2x4 – 3x3 – 3x2 + 6x – 2 having zeros 
and - 
∴ The polynomial is (x + 
) (x - 
) = x2 – 2
Let us now divide p (x) by (x2 – 2)
2x4 – 3x3 – 3x2 + 6x – 2 = (x2 – 2) (2x2 – 3x + 1)
= (x2 – 2) [(2x2 – (2 + 1) x + 1]
= (x2 – 2) (2x2 – 2x – x + 1)
= (x2 – 2) [(2x (x – 1) – 1 (x – 1)]
= (x2 – 2) (2x – 1) (x – 1)
Hence, the other two zeros are 
and 1
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