Find all the zeros of (x^{4} + x^{3} ‒ 23x^{2} ‒ 3x + 60), if it is given that two of its zeros are √3 and ‒√3.

Let us assume f (x) = x^{4} + x^{3} ‒ 23x^{2} ‒ 3x + 60

As √3 and –√3 are the zeros of the given polynomial therefore each one of (x - √3) and (x - √3) is a factor of f (x)

Consequently, (x – √3) (x + √3) = (x^{2} – 3) is a factor of f(x)

Now, on dividing f (x) by (x^{2} – 3) we get:

f (x) = 0

(x^{2} + x – 20) (x^{2} – 3) = 0

(x^{2} + 5x – 4x – 20) (x^{2} – 3)

[x (x + 5) – 4 (x + 5)] (x^{2} – 3)

(x – 4) (x + 5) (x – ) (x + ) = 0

∴ x = 4 or x = - 5 or x = or x = -

Hence, all the zeros of the given polynomial are , -, 4 and - 5

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