1. **Problem Statement:** Find the zeroes of the quadratic polynomial $x^2 - 2x - 8$ and verify the relationship between the zeroes and the coefficients. 2. **Formula and Important Rules:** - For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes $\alpha + \beta = -\frac{b}{a}$. - The product of zeroes $\alpha \beta = \frac{c}{a}$. - Zeroes are values of $x$ for which the polynomial equals zero. 3. **Finding Zeroes:** - Factorize $x^2 - 2x - 8$. - Find two numbers whose product is $-8$ and sum is $-2$. - These numbers are $-4$ and $2$ because $-4 \times 2 = -8$ and $-4 + 2 = -2$. - So, $x^2 - 2x - 8 = (x - 4)(x + 2)$. - Zeroes are $x = 4$ and $x = -2$. 4. **Verification:** - Sum of zeroes $= 4 + (-2) = 2$. - Coefficients: $a = 1$, $b = -2$, $c = -8$. - $-\frac{b}{a} = -\frac{-2}{1} = 2$ matches sum of zeroes. - Product of zeroes $= 4 \times (-2) = -8$. - $\frac{c}{a} = \frac{-8}{1} = -8$ matches product of zeroes. 5. **Conclusion:** The zeroes $4$ and $-2$ satisfy the relationships with coefficients. **Final answer:** Zeroes are $4$ and $-2$.