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Question
Verify that 5, -2 and 13 are the zeroes of the cubic polynomial `p(x) = (3x^3 – 10x^2 – 27x + 10)` and verify the relation between its zeroes and coefficients.
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Solution
p(x) = `(3x^3 – 10x^2 – 27x + 10)`
`p(5) = (3 × 5^3 – 10 × 5^2 – 27 × 5 + 10) = (375 – 250 – 135 + 10) = 0`
`p(–2) = [3 × (–2^3) – 10 × (–2^2) – 27 × (–2) + 10] = (–24 – 40 + 54 + 10) = 0 `
`p(1/3)={3xx(1/3)^3-10(1/3)^2-27xx1/3+10}=(3xx1/27-10xx1/9-9+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` ={5xx(-2)xx1/3}=-10/3=-("Constant term")/(("Coefficient of "x^3))`
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