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(i) \(x^2 - 2x - 8\)

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

(iii) \(6x^2 - 3 - 7x\)

(iv) \(4u^2 + 8u\)

(v) \(t^2 - 15 \)

(vi) \(3x^2 - x - 4\)

Answer :

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

= \({{4}\over {3}} × (-1)\) = \({{-4}\over {3}}\)

= \({{c}\over {a}}\) = \({{-4}\over {3}}\)

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