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Question
Find a cubic polynomial whose zeroes are `1/2, 1 and -3.`
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Solution
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
`x^3 – (a + b + c)x^2 + (ab + bc + ca)x – abc` .............(1)
Let a = `1/2, b=14 and c=-3`
Substituting the values in (1), we get
`x^3-(1/2+1-3)x^2+(1/2-3-3/2)x-(-3/2)`
⇒ `x^3-(-3/2)x^2-4x+3/2`
`⇒ 2x^3+3x^2-8x+3`
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