Answer: \(\mathrm{p}\left(\frac12\right)=0\); \(\mathrm{p}(x)=(2x-1)(x-4)(x+3)\); \(x=\frac12,4,-3\).

(a) Substitute \(x=\frac12\):

\(\mathrm{p}\left(\frac12\right) =2\left(\frac18\right)-3\left(\frac14\right)-23\left(\frac12\right)+12.\)

So

\(\mathrm{p}\left(\frac12\right) =\frac14-\frac34-\frac{23}{2}+12=0.\)

(b) Since \(\mathrm{p}\left(\frac12\right)=0\), \(2x-1\) is a factor.

Dividing \(2x^3-3x^2-23x+12\) by \(2x-1\) gives

\(x^2-x-12.\)

Therefore

\(\mathrm{p}(x)=(2x-1)(x^2-x-12).\)

Factorise the quadratic:

\(x^2-x-12=(x-4)(x+3).\)

Hence

\(\mathrm{p}(x)=(2x-1)(x-4)(x+3).\)

Solving \(\mathrm{p}(x)=0\),

\(x=\frac12,\qquad x=4,\qquad x=-3.\)