Question

find the zeroes of the polynomial and verify the relationship between the zeroes and the coefficients.

6x^2+x-2

Answer 1

$$\begin{gathered} \mathrm{Let}p\left(x\right)=6x^2+x-2,\mathrm{be}\mathrm{the}\mathrm{given}\mathrm{polynomial} \\ \mathrm{Zeroes}\mathrm{of}p\left(x\right)\mathrm{is}\mathrm{given}\mathrm{by}, \\ p\left(x\right)=0 \\ \Rightarrow 6x^2+x-2=0 \\ \Rightarrow 6x^2+4x-3x-2=0 \\ \Rightarrow 2x\left(3x+2\right)-1\left(3x+2\right)=0 \\ \Rightarrow \left(2x-1\right)\left(3x+2\right)=0 \\ \Rightarrow 2x-1=0\mathrm{or}3x+2=0 \\ \Rightarrow x=\frac{1}{2}\mathrm{or}x=-\frac{2}{3} \\ \mathrm{So},\mathrm{zeroes}\mathrm{of}\mathrm{given}\mathrm{polynomial}\mathrm{are}:- \\ \alpha =\frac{1}{2}\mathrm{and}\beta =-\frac{2}{3} \\ \mathrm{Now},\mathrm{sum}\mathrm{of}\mathrm{zeroes}=\alpha +\beta =\frac{1}{2}-\frac{2}{3}=-\frac{1}{6}=-\frac{\left(\mathrm{coefficient}\mathrm{of}x\right)}{\mathrm{coefficient}\mathrm{of}{x}^2} \\ \mathrm{Product}\mathrm{of}\mathrm{zeroes}=\alpha \beta =\frac{1}{2}\times \left(-\frac{2}{3}\right)=-\frac{1}{3}=\frac{\mathrm{constant}\mathrm{term}}{\mathrm{coefficient}\mathrm{of}{x}^2} \end{gathered}$$