Find the zeroes of the quadratic polynomial 7y2- 11/3 y- 2/3 and verify the relationship between the zeroes and the coefficients.

The given polynomial is\(f(y)=7y^2-\dfrac{11}{3}y-\dfrac{2}{3}\)\(\Rightarrow f\left(y\right)=\frac13\left(21y^2-11y-2\right)\\\Rightarrow f\left(y\right)=\frac13\left(21y^2-14y+3y-2\right)\\\Rightarrow f\left(y\right)=\frac13\left\{7y\left(3y-2\right)+1\left(3y-2\right)\right\}\\\Rightarrow f\left(y\right)=\frac13\left(3y-2\right)\left(7y+1\right)\)Therefore, the value of \(f\left(y\right)=7y^2-\frac{11}3y-\frac23\) is zero when \(3y-2=0\) or \(7y+1=0\) i.e., when \(y=\frac23\) or \(y=-\frac17\).\(\therefore\) The zeros of the given polynomial are \(\frac23\) and \(-\frac17\).Sum of zeros \(=\frac23+\left(-\frac17\right)=\frac23-\frac17=\frac{14-3}{21}=\frac{11}{21}=-\frac{\left(-{\displaystyle\frac{11}3}\right)}7=-\frac{\mathrm{Coefficient}\;\mathrm{of}\;y}{\mathrm{Coefficient}\;\mathrm{of}\;y^2}\) and Product of zeroes \(\frac23\times\left(-\frac17\right)=\frac2{21}=-\frac{\left(- {\displaystyle\frac23}\right)}7=\frac{\mathrm{Coefficient}\;term}{\mathrm{Coefficient}\;\mathrm{of}\;y^2}\)