Answer: \(a=2\), \(b=2\); \(x=-1,\frac12,2\).

Work through the given conditions step by step, keeping exact values until the final answer is required.

Since \(x=2\) is a root,

\(8a-12-6+b=0\), so \(8a+b=18\).

Since \(x=-1\) is a root,

\(-a-3+3+b=0\), so \(b=a\).

Therefore \(9a=18\), giving \(a=2\), and hence \(b=2\).

So \(\mathrm p(x)=2x^3-3x^2-3x+2\).

The known roots give factors \((x-2)\) and \((x+1)\). Dividing or factorising gives

\(2x^3-3x^2-3x+2=(x-2)(x+1)(2x-1)\).

Thus the solutions are \(x=2\), \(x=-1\), and \(x=\frac12\).

This gives the required answer: \(a=2\), \(b=2\); \(x=-1,\frac12,2\).