Answer:
A. \((x-1)(2x+3)\)
Step-by-step explanation:
We are given the quadratic equation \(2x^2+x-3\).
We will find the roots of the equation quadratic \(2x^2+x-3=0\).
The roots of the quadratic equation \(ax^2+bx+c=0\) is given by \(x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\).
On comparing the equations, we get,
a= 2, b= 1, c= -3.
Substituting the values gives us the roots are,
\(x=\frac{-1\pm \sqrt{1^2-4\times 2\times (-3)}}{2\times 2}\)
i.e. \(x=\frac{-1\pm \sqrt{1+24}}{4}\)
i.e. \(x=\frac{-1\pm \sqrt{25}}{4}\)
i.e. \(x=\frac{-1\pm 5}{4}\)
i.e. \(x=\frac{-1+5}{4}\) and \(x=\frac{-1-5}{4}\)
i.e. \(x=\frac{4}{4}\) and \(x=\frac{-6}{4}\)
i.e. x = 1 and \(x=\frac{-3}{2}\)
Thus, the factors of the quadratic equation are \((x-1)\) and \((2x+3)[/tex].
Hence, the factored form is [tex](x-1)(2x+3)\).
