Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients: $x^{2}-2x-8$.

(N/A) Given polynomial: $p(x) = x^{2}-2x-8$.
To find the zeroes,we set $p(x) = 0$:
$x^{2}-2x-8 = 0$
$x^{2}-4x+2x-8 = 0$
$x(x-4)+2(x-4) = 0$
$(x-4)(x+2) = 0$
Thus,the zeroes are $x = 4$ and $x = -2$.
Verification:
Comparing $x^{2}-2x-8$ with $ax^{2}+bx+c$,we get $a=1, b=-2, c=-8$.
Sum of zeroes $= 4 + (-2) = 2$.
Formula: $-b/a = -(-2)/1 = 2$.
Since $2 = 2$,the sum of zeroes is verified.
Product of zeroes $= 4 \times (-2) = -8$.
Formula: $c/a = -8/1 = -8$.
Since $-8 = -8$,the product of zeroes is verified.