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Question
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
t3 – 2t2 – 15t
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Solution
t3 – 2t2 – 15t
Taking t common, we get,
t(t2 – 2t – 15)
Splitting the middle term of the equation t2 – 2t – 15, we get,
t(t2 – 5t + 3t – 15)
Taking the common factors out, we get,
t(t(t – 5) + 3(t – 5)
On grouping, we get,
t(t + 3)(t – 5)
So, the zeroes are,
t = 0
t + 3 = 0
`\implies` t = – 3
t – 5 = 0
`\implies` t = 5
Therefore, zeroes are 0, 5 and – 3
Verification:
Sum of the zeroes = – (coefficient of x2) ÷ coefficient of x3
α + β + γ = `- b/a`
(0) + (– 3) + (5) = `- (-2)/1`
= 2
Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x3
αβ + βγ + αγ = `c/a`
(0)(– 3) + (– 3)(5) + (0)(5) = `-15/1`
= – 15
Product of all the zeroes = – (constant term) ÷ coefficient of x3
αβγ = `- d/a`
(0)(– 3)(5) = 0
= 0
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