First find the factors of \(\displaystyle{\left({x}^{{2}}+{2}{x}-{3}\right)}.\)

Compare the \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}\) and \(\displaystyle{x}^{{2}}+{2}{x}-{3}\), implies that \(a = 1, b = 2\), and \(c = -3\).

The ac-method of factoring quadratics: To find factors of quadratic \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}\), Find two numbers whose sum is b and product is \(a\cdot c\).

Here \(a\cdot c = 1\cdot (-3) = -3 \)and \(b = 2\).

To find factors of quadratic \(\displaystyle{x}^{{2}}+{2}{x}-{3}\), find two numbers whose sum is 2 and the product is -3.

Such numbers are 3 and -1.

The factors of \(\displaystyle{\left({x}^{{2}}+{2}{x}-{3}\right)}\) are \((x + 3)(x - 1).\)

The given function \(\displaystyle{f{{\left({x}\right)}}}={\left({2}{x}−{1}\right)}\cdot{\left({x}^{{2}}+{2}{x}−{3}\right)}\) becomes

\(\displaystyle{f{{\left({x}\right)}}}={\left({2}{x}−{1}\right)}·{\left({x}+{3}\right)}·{\left({x}-{1}\right)}\)

Set \(f(x) = 0\), implies that

\(\displaystyle{\left({2}{x}−{1}\right)}\cdot{\left({x}+{3}\right)}\cdot{\left({x}-{1}\right)}={0}\)

By using zero product property,

\(\displaystyle{\left({2}{x}−{1}\right)}={0},{\left({x}+{3}\right)}={0},{\left({x}-{1}\right)}={0}\)

Implies that,

\(2x = 1, x = -3, x = 1\)

That is, the roots of the given function are,

\(\displaystyle{x}=\frac{{1}}{{2}},{x}=-{3},{x}={1}.\)