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


18
1