Let us assume f (x) = x⁴ + x³ ‒ 34x² ‒ 4x + 120

As 2 and – 2 are the zeros of the given polynomial therefore each one of (x - 2) and (x + 2) is a factor of f (x)

Consequently, (x – 3) (x + 3) = (x² – 4) is a factor of f (x)

Now, on dividing f (x) by (x² – 4) we get:

f (x) = 0

(x² + x – 30) (x² – 4) = 0

(x² + 6x – 5x – 30) (x – 2) (x + 2)

[x (x + 6) – 5 (x + 6)] (x – 2) (x + 2)

(x – 5) (x + 6) (x – 2) (x + 2) = 0

∴ x = 5 or x = - 6 or x = 2 or x = - 2

Hence, all the zeros of the given polynomial are 2, -2, 5 and -6