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Question
Verify that 3, -2, 1 are the zeros of the cubic polynomial `p(x) = (x^3 – 2x2 – 5x + 6)` and verify the relation between it zeros and coefficients.
Solution
The given polynomial is `p(x) = (x^3 – 2x^2 – 5x + 6)`
`∴ p(3) = (3^3 – 2 × 3^2 – 5 × 3 + 6) = (27 – 18 – 15 + 6) = 0`
`p(-2) = [ (– 2^3) – 2 × (– 2)^2 – 5 × (– 2) + 6] = (–8 –8 + 10 + 6) = 0`
`p(1) = (1^3 – 2 × 1^2 – 5 × 1 + 6) = ( 1 – 2 – 5 + 6) = 0`
∴ 3, –2 and 1are the zeroes of p(x),
Let 𝛼 = 3, 𝛽 = –2 and γ = 1. Then we have:
(𝛼 + 𝛽 + γ) = (3 – 2 + 1) = 2 = `(-("Coefficient of" x^2))/(("Coefficient of" x^2))`
(𝛼𝛽 + 𝛽γ + γ𝛼) = (–6 –2 + 3) = `−5/1 = ("Coefficient of x")/("Coefficient of x"^2)`
𝛼𝛽γ = { 3 × (-2) × 1}=`(-6)/1= -(("Constant term"))/(("Coefficient of" x^3))`
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