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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:
5t2 + 12t + 7
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Solution
5t2 + 12t + 7
Splitting the middle term, we get,
5t2 + 5t + 7t + 7
Taking the common factors out, we get,
5t(t + 1) + 7(t + 1)
On grouping, we get,
(t + 1)(5t + 7)
So, the zeroes are,
t + 1 = 0
`\implies` y = –1
5t + 7 = 0
`\implies` 5t = –7
`\implies` t = `-7/5`
Therefore, zeroes are `(-7/5)` and –1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = `- b/a`
`(-1) + (-7/5) = - (12)/5`
= `-12/5 = -12/5`
Product of the zeroes = constant term ÷ coefficient of x2
αβ = `c/a`
`(-1)(-7/5) = 7/5`
`7/5 = 7/5`
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