Correct option is (D) \(10x^2 -x-3\)
Here, the zeroes are \(\frac{3}{5}\) and \(\frac{-1}{2}\).
Let, \(\alpha=\frac{3}{5}\) and \(\beta=-\frac{1}{2}\)
Sum of the zeroes \((\alpha+\beta)=\frac{3}{5}+\left(-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{1}{2}\)
\(=\frac{6-5}{10} \)
\(=\frac{1}{10}\)
Product of the zeroes \((\alpha \beta)=\frac{3}{5} \times\left(-\frac{1}{2}\right)\)
\(=-\frac{3}{10}\)
The polynomial will be \(x^2-(\alpha+\beta) x+\alpha \beta=0\)
\( \Rightarrow x^2-\frac{1}{10} x+\left(-\frac{3}{10}\right)=0\)
\(\Rightarrow \frac{x^2}1-\frac{x}{10}-\frac{3}{10}=0\)
\(\Rightarrow \frac{10 x^2-x-3}{10}=0\)
\(\Rightarrow10 x^2-x-3=0\)
