Answer: \(a=6,\ b=-3\), \(\mathrm p(x)=(x-2)(6x^2+3x+3)\), one real solution only.

Since \(x-2\) is a factor,

\(\mathrm p(2)=0.\)

So

\(8a-36+2b-6=0,\)

which gives

\(8a+2b=42. \tag{1}\)

The remainder when divided by \(x-3\) is \(66\), so

\(\mathrm p(3)=66.\)

Hence

\(27a-81+3b-6=66,\)

so

\(27a+3b=153. \tag{2}\)

Divide (1) by \(2\) and (2) by \(3\):

\(4a+b=21,\qquad 9a+b=51.\)

Subtracting gives

\(5a=30,\)

so

\(a=6.\)

Then

\(4(6)+b=21,\)

so

\(b=-3.\)

Thus

\(\mathrm p(x)=6x^3-9x^2-3x-6.\)

Since \(x-2\) is a factor, divide by \(x-2\):

\(\mathrm p(x)=(x-2)(6x^2+3x+3).\)

For the quadratic factor, the discriminant is

\(3^2-4(6)(3)=9-72=-63.\)

This is negative, so \(6x^2+3x+3=0\) has no real solutions. Therefore the only real solution of \(\mathrm p(x)=0\) is \(x=2\).