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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:
3x2 + 4x – 4
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Solution
3x2 + 4x – 4
Splitting the middle term, we get,
3x2 + 6x – 2x – 4
Taking the common factors out, we get,
3x(x + 2) – 2(x + 2)
On grouping, we get,
(x + 2)(3x – 2)
So, the zeroes are,
x + 2 = 0
`\implies` x = – 2
3x – 2 = 0
`\implies` 3x = 2
`\implies` x = `2/3`
Therefore, zeroes are `(2/3)` and – 2
Verification:
Sum of the zeroes = – (coefficient of x) ÷ coefficient of x2
α + β = `– b/a`
`-2 + (2/3) = - (4)/3`
= `- 4/3 = -4/3`
Product of the zeroes = constant term ÷ coefficient of x2
αβ = `c/a`
Product of the zeroes = `(-2) (2/3) = - 4/3`
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Case Study -1

The figure given alongside shows the path of a diver, when she takes a jump from the diving board. Clearly it is a parabola.
Annie was standing on a diving board, 48 feet above the water level. She took a dive into the pool. Her height (in feet) above the water level at any time ‘t’ in seconds is given by the polynomial h(t) such that h(t) = -16t2 + 8t + k.
The zeroes of the polynomial r(t) = -12t2 + (k - 3)t + 48 are negative of each other. Then k is ______.
The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.




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