Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
$x^2-2x-8$

Answer 1

$$\begin{gathered} x^2-2x-8=0 \\ \Rightarrow x^2-4x+2x-8=0 \\ \Rightarrow x\left(x-4\right)+2\left(x-4\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x-4\right)\left(x+2\right)=0 \\ \Rightarrow \left(x-4\right)=0\mathrm{or}\left(x+2\right)=0 \\ \Rightarrow x=4\mathrm{or}x=-2 \end{gathered}$$

Sum of zeroes = $4+\left(-2\right)=2=\frac{2}{1}=\frac{-\left(\mathrm{coefficient}\mathrm{of}x\right)}{\mathrm{coefficient}\mathrm{of}{x}^2}$ −4+(−3)=−71=−(coefficient of x)(coefficient of x2)
Product of zeroes = $4\left(-2\right)=-8=\frac{-8}{1}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2}$ (−4)(−3)=121=constant termcoefficient of x2