Find the zeroes of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.
(i) \(x^2 - 2x - 8\)
(ii) \(4s^2 - 4s + 1\)
(iii) \(6x^2 - 3 - 7x\)
(iv) \(4u^2 + 8u\)
(v) \(t^2 - 15 \)
(vi) \(3x^2 - x - 4\)

General form of equation is \(ax^2 + bx + c\)
where, a = Coefficient of \(x^2\)
b = Coefficient of x
c = Constant term

Once we'll find the zeroes of the polynomial we can verify it using the following rules

Sum of zeroes = \({{-b}\over {a}}\)
Product of zeroes = \({{c}\over {a}}\)


(i) \(x^2 - 2x - 8\)
=\(x^2 - 4x + 2x - 8\)
= \(x(x - 4) + 2(x - 4)\)
= \((x + 2)(x - 4)\)

So, zeroes are -2 and 4
Now, Verification:

Sum of the zeroes = 2
= \({{-b}\over {a}}\) = \({{-(-2)}\over {1}}\) = 2
Product of the zeroes = -8
= \({{c}\over {a}}\) = \({{-8}\over {1}}\) = -8


(ii) \(4s^2 - 4s + 1\)
=\(4s^2 - 2s - 2s + 18\)
= \(2s(2s - 1) - 1(2s - 1)\)
= \((2s - 1)(2s - 1)\)

So, zeroes are \({{1}\over {2}}\) and \({{1}\over {2}}\)
Now, Verification:

Sum of the zeroes = 1
= \({{-b}\over {a}}\) = \({{-(-4)}\over {4}}\) = 1
Product of the zeroes = \({{1}\over {4}}\)
= \({{c}\over {a}}\) = \({{1}\over {4}}\)


(iii) \(6x^2 - 7x - 3\)
=\(6x^2 - 9x + 2x -3\)
= \(3x(2x - 3) + 1(2x - 3)\)
= \((3x + 1)(2x - 3)\)

So, zeroes are \({{-1}\over {3}}\) and \({{3}\over {2}}\)
Now, Verification:
Sum of the zeroes
= \({{-1}\over {3}} + {{3}\over {2}}\) = \({{7}\over {6}}\)
= \({{-b}\over {a}}\) = \({{-(-7)}\over {6}}\)
Product of the zeroes
= \({{-1}\over {3}} × {{3}\over {2}}\) = \({{-1}\over {2}}\)
= \({{c}\over {a}}\) = \({{-3}\over {6}}\) = \({{-1}\over {2}}\)


(iv) \(4u^2 + 8u\)
=\(4u(u + 2)\)
=\(4u = 0\)
=\(u = 0\)
=\(u = -2\)

So, zeroes are 0 and -2
Now, Verification:

Sum of the zeroes = -2
= \({{-b}\over {a}}\) = \({{-(8)}\over {4}}\) = -2
Product of the zeroes = 0
= \({{c}\over {a}}\) = \({{0}\over {4}}\) = 0


(v) \(t^2 - 15 \)
= \(t^2 - (\sqrt{15})^2\)
= \((t + \sqrt{15})(t -\sqrt{15})\)

So, zeroes are \(-\sqrt{15}\) and \(\sqrt{15}\)
Now, Verification:
Sum of the zeroes = 0
= \({{-b}\over {a}}\) = \({{-(0)}\over {1}}\) = 0
Product of the zeroes = -15
= \({{c}\over {a}}\) = \({{-15}\over {1}}\) = -15

(vi) \(3x^2 - x - 4\)
=\(3x^2 - 4x + 3x -4\)
= \(x(3x - 4) + 1(3x - 4)\)
= \((3x - 4)(x + 1)\)

So, zeroes are \({{4}\over {3}}\) and -1
Now, Verification:
Sum of the zeroes
= \({{4}\over {3}} + (-1)\) = \({{1}\over {3}}\)
= \({{-b}\over {a}}\) = \({{-(-1)}\over {3}}\) = \({{1}\over {3}}\)
Product of the zeroes
= \({{4}\over {3}} × (-1)\) = \({{-4}\over {3}}\)
= \({{c}\over {a}}\) = \({{-4}\over {3}}\)