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Find all the zeros of (x4 + x3 ‒ 23x2 ‒ 3x + 60), if it is given that two of its zeros are √3 and ‒√3.
Let us assume f (x) = x4 + x3 ‒ 23x2 ‒ 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) = (x2 – 3) is a factor of f(x)
Now, on dividing f (x) by (x2 – 3) we get:
f (x) = 0
(x2 + x – 20) (x2 – 3) = 0
(x2 + 5x – 4x – 20) (x2 – 3)
[x (x + 5) – 4 (x + 5)] (x2 – 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