Obtain all zeros of the polynomial

2y^3-4y-y^2+2,
if two of its zeroes are 2 and - 2

The zeros of this polynomial are not 2 and - 2.

Write a polynomial in this form:

( 2 y³ - y² ) + ( - 4 y + 2 )

Now factorize the polynomial.

( 2 y³ - y² ) + ( - 4 y + 2 ) =

y² ( 2 y - 1 ) - 2 ( 2 y - 1 ) =

( 2 y - 1 ) ( y² - 2 )

2 y³ - 4 y - y² + 2 = 0

is same as:

( 2 y - 1 ) + ( y² - 2 ) = 0

The solutions are:

2 y - 1 = 0

2 y = 1

y = 1 / 2

and

y² - 2 = 0

y² = 2

y = ± √2