Since \((2x - 1)\) is a factor of \(p(x)\), substituting \(x = \frac{1}{2}\) into \(p(x)\) should yield zero:

\(p\left(\frac{1}{2}\right) = a\left(\frac{1}{2}\right)^3 - \left(\frac{1}{2}\right)^2 + 4\left(\frac{1}{2}\right) - a = 0\)

\(\Rightarrow \frac{a}{8} - \frac{1}{4} + 2 - a = 0\)

\(\Rightarrow \frac{a}{8} - a = -\frac{7}{4}\)

\(\Rightarrow \frac{a - 8a}{8} = -\frac{7}{4}\)

\(\Rightarrow -\frac{7a}{8} = -\frac{7}{4}\)

\(\Rightarrow 7a = 14\)

\(\Rightarrow a = 2\)

Substituting \(a = 2\) back into \(p(x)\):

\(p(x) = 2x^3 - x^2 + 4x - 2\)

We know \((2x - 1)\) is a factor, so divide \(p(x)\) by \((2x - 1)\) to find the other factor:

Using polynomial division or inspection, we find:

\(p(x) = (2x - 1)(x^2 + 2)\)