Obtain all other zeros of (x^{4} + 4x^{3} ‒ 2x^{2} ‒ 20x ‒15) if two of its zeros are √5 and – √5.

Let us assume f (x) = x^{4} + 4x^{3} ‒ 2x^{2} ‒ 20x - 15

As (x-√5) and (x-√5) are the zeros of the given polynomial therefore each one of (x-√5) and (x + √5) is a factor of f (x)

Consequently, (x-√5) (x + √5) = (x^{2} – 5) is a factor of f (x)

Now, on dividing f (x) by (x^{2} – 5) we get:

f (x) = 0

x^{4} + 4x^{3} – 7x^{2} – 20x – 15 = 0

(x^{2} – 5) (x^{2} + 4x + 3) = 0

(x -√5) (x + √5) (x + 1) (x + 3) = 0

∴ x = √5 or x = - √5 or x = - 1 or x = - 3

Hence, all the zeros of the given polynomial are √5, -√5, - 1 and - 3

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