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Question
Find all the zeroes of (x4 + x3 – 23x2 – 3x + 60), if it is given that two of its zeroes are `sqrt(3)` and `-sqrt(3)`.
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Solution
Let f(x) = x4 + x3 – 23x2 – 3x + 60
Since `sqrt(3)` and `-sqrt(3)` are the zeroes of f(x), it follows that each one of `(x - sqrt(3))` and `(x + sqrt(3))` is a factor of f(x).
Consequently, `(x - sqrt(3)) (x + sqrt(3)) = (x^2 - 3)` is a factor of f(x).
On dividing f(x) by (x2 – 3), we get:
`x^2 - 3")"overline(x^4 + x^3 - 23x^2 - 3x + 60)"("x^2 + x - 20`
x4 – 3x2
– +
x3 – 20x2 – 3x + 60
x3 – 3x
– +
–20x2 + 60
–20x2 + 60
+ –
x
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 - sqrt(3)) (x + sqrt(3)) = 0`
⇒ x = 4 or x = –5 or x = `sqrt(3)` or x = `-sqrt(3)`
Hence, all the zeroes are `sqrt(3), -sqrt(3), 4` and –5.
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