Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes 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) $3x^2-x-4.$

(i)  For finding the zeroes we have to equate the given polynomial to 0.

$x^2-2x-8=0.$

$\Rightarrow x^2+2x-4x-8=0.$

$\Rightarrow x(x+2)-4(x+2)=0.$

$$\begin{gathered} \Rightarrow (x+2)(x-4)=0. \\ \Rightarrow x=-2, x=4. \end{gathered}$$

Thus the zeros are $\alpha =-2, \beta =4.$

Sum of the zeros$=\alpha +\beta =-2+4=2.$

$\frac{-coefficient of x}{coefficient of x^2}=\frac{-(-2)}{1}=2.$

Thus, sum of the zeros$=\frac{-coefficient of x}{coefficient of x^2}$

Product of the zeros$=\alpha \beta =(-2)\times 4=-8.$

$\frac{cons\tan t term}{coefficient of x^2}=\frac{-8}{1}=-8.$

Thus, product of the zeros$=\frac{cons\tan t term}{coefficient of x^2}$

Hence, the relationship between the zeros and coefficient are verified.