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Question
Verify that 5, –2 and `1/3` are the zeroes of the cubic polynomial p(x) = (3x3 – 10x2 – 27x + 10) and verify the relation between its zeros and coefficients.
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Solution
p(x) = (3x3 – 10x2 – 27x + 10)
p(5) = (3 × 53 – 10 × 52 – 27 × 5 + 10)
= (375 – 250 – 135 + 10)
= 0
p(–2) = [3 × (–23) – 10 × (–22) – 27 × (–2) + 10]
= (–24 – 40 + 54 + 10)
= 0
`p(1/3) = {3 xx (1/3)^3 - 10 xx (1/3)^2 - 27 xx 1/3 + 10}`
= `(3 xx 1/27 - 10 xx 1/9 - 9 + 10)`
= `(1/9 - 10/9 + 1)`
= `((1 - 10 - 9)/9)`
= `(0 / 9)`
= 0
∴ 5, –2 and `1/3` are the zeroes of p(x).
Let α = 5, β = –2 and γ = `1/3`.
Then we have:
`(α + β + γ) = (5 - 2 + 1/3)`
= `10/3`
= `(-("Coefficient of" x^2))/(("Coefficient of" x^3))`
`(αβ + βγ + γα) = (-10 - 2/3 + 5/3)`
= `(-27)/3`
= `("Coefficient of" x)/("Coefficient of"" x^3)`
`αβγ ={5 xx (-2) xx 1/3}`
= `(-10)/3`
= `(-("Constant term"))/(("Coefficient of" x^3))`
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