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Question
Find all the zeroes of `(2x^4 – 3x^3 – 5x2 + 9x – 3)`, it is being given that two of its zeroes are `sqrt3 and –sqrt3`.
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Solution
The given polynomial is f(x) = `2x^4 – 3x^3 – 5x^2 + 9x – 3`
Since √3 and –√3 are the zeroes of f(x), it follows that each one of `(x – sqrt3) `and `(x + sqrt3)`is a factor of f(x).
Consequently, `(x – sqrt3) (x + sqrt3)` = (x2 – 3) is a factor of f(x).
On dividing f(x) by (x2 – 3), we get:
`f(x) = 0`
`⇒ 2x^4 – 3x^3 – 5x2 + 9x – 3 = 0`
`⇒ (x^2 – 3) (2x^2– 3x + 1) = 0`
`⇒ (x^2 – 3) (2x2– 2x – x + 1) = 0`
`⇒ (x – sqrt3) (x + sqrt3) (2x – 1) (x – 1) = 0`
`⇒ x = sqrt3 or x = -sqrt3 or x = 12 or x = 1`
Hence, all the zeroes are `sqrt3, -sqrt3`, 12 and 1.
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