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Question
Verify that 3, –2, 1 are the zeros of the cubic polynomial p(x) = (x3 – 2x2 – 5x + 6) and verify the relation between it zeros and coefficients.
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Solution
The given polynomial is p(x) = (x3 – 2x2 – 5x + 6)
∴ p(3) = (33 – 2 × 32 – 5 × 3 + 6)
= (27 – 18 – 15 + 6)
= 0
p(–2) = [(–23) – 2 × (–2)2 – 5 × (–2) + 6]
= (–8 – 8 + 10 + 6)
= 0
p(1) = (13 – 2 × 12 – 5 × 1 + 6)
= (1 – 2 – 5 + 6)
= 0
∴ 3, –2 and 1 are 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^3))`
(αβ + βγ + γα) = (–6 – 2 + 3)
= `(-5)/1`
= `("Coefficient of" x)/("Coefficient of" x^3)`
αβγ = {3 × (–2) × 1}
= `(-6)/1`
= `(-("Constant term"))/(("Coefficient of" x^3))`
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