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प्रश्न
Find the zeroes of the following polynomials by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials:
4x2 – 3x – 1
योग
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उत्तर
4x2 – 3x – 1
Splitting the middle term, we get,
4x2 – 4x + 1x – 1
Taking the common factors out, we get,
4x(x – 1) + 1(x – 1)
On grouping, we get,
(4x + 1)(x – 1)
So, the zeroes are,
4x + 1 = 0
`\implies` 4x = – 1
`\implies` x = `(-1/4)`
(x – 1) = 0
`\implies` x = 1
Therefore, zeroes are `(-1/4)` and 1
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = `– b/a`
`1 - 1/4 = - (- 3)/4 = 3/4`
Product of the zeroes = constant term ÷ coefficient of x2
αβ = `c/a`
`1(-1/4) = - 1/4`
`-1/4 = -1/4`
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