(i) x^2 - 2x - 8 = 0
x^2 - 4x + 2x - 8= 0
x(x - 4) + 2( x - 4) = 0
(x+2) (x - 4) = 0
x = - 2 , x= 4
Sum of zeros = - b/a = - (-2)/ 1 = 2
Product of zeros = c/a = - 8/1 = - 8.

(ii) 4s^2 - 4s + 1 = 0
4s^2 - 2s - 2s + 1 = 0
2s ( 2s - 1) - 1 ( 2s - 1) = 0
( 2s - 1) ( 2s - 1) = 0
2s - 1= 0
2s = 1
s = 1/2
Sum of zeros = - b/a = - ( - 4) / 4 = 1
Product of zeros = c/a = 1/4


( iii) 6x^2 - 3 - 7x = 0
6x^2 - 7x - 3 = 0
6x^2 - 9x + 2x - 3 = 0
3x ( 2x - 3) + 1( 2x - 3) = 0
( 3x +1) ( 2x - 3) = 0
3x = - 1
x = - 1/3
2x = 3
2x = 3
x = 3/2
Sum of zeros = - b/a = - (-7)/6 = 7/6
Product of zeros = c/ a = - 3/6 = - 1/2

1) Finding the zeros of the quadratic polynomial:
To find the zeros of a quadratic polynomial, we need to solve the equation when the polynomial is equal to zero.

The given quadratic polynomial is: x^2 - 2x - 8.

Setting it equal to zero, we have: x^2 - 2x - 8 = 0.

To solve this quadratic equation, we can factorize it or use the quadratic formula. In this case, let's factorize it:

(x - 4)(x + 2) = 0.

Now, we have two factors that can be individually set to zero:

x - 4 = 0 or x + 2 = 0.

Solving each equation, we find:

x = 4 or x = -2.

Therefore, the zeros of the quadratic polynomial x^2 - 2x - 8 are x = 4 and x = -2.

2) Verifying the relationship between the zeros and the coefficients:
The relationship between the zeros and the coefficients of a quadratic polynomial can be explained using Vieta's formulas.

For a quadratic polynomial in the form ax^2 + bx + c = 0, the sum of the zeros is equal to -b/a, and the product of the zeros is equal to c/a.

Let's apply this to the given quadratic polynomial: 4s^2 - 4s + 1.

The sum of the zeros is given by: -(-4)/4 = 1.

The product of the zeros is given by: 1/4.

Therefore, the relationship between the zeros and the coefficients for the quadratic polynomial 4s^2 - 4s + 1 is as follows:
- The sum of the zeros is 1.
- The product of the zeros is 1/4.

3) Verifying the relationship between the zeros and the coefficients:
Now, let's analyze the quadratic polynomial 6x^2 - 3x - 7x.

Setting it equal to zero, we have: 6x^2 - 10x = 0.

We can factor out the common factor of 2x from both terms:

2x(3x - 5) = 0.

Now, we have two factors that can be individually set to zero:

2x = 0 or 3x - 5 = 0.

Solving each equation, we find:

x = 0 or x = 5/3.

Therefore, the zeros of the quadratic polynomial 6x^2 - 3x - 7x are x = 0 and x = 5/3.

Now, let's verify the relationship between the zeros and the coefficients using Vieta's formulas:

The sum of the zeros is given by: -(0)/6 = 0.

The product of the zeros is given by: (0)(5/3)/6 = 0.

Therefore, the relationship between the zeros and the coefficients for the quadratic polynomial 6x^2 - 3x - 7x is as follows:
- The sum of the zeros is 0.
- The product of the zeros is 0.

In conclusion, we have found the zeros of the given quadratic polynomials and verified the relationship between the zeros and the coefficients using