Answer: \(a=-13\), \(b=1\); remainder \(0\); \(\mathrm{p}(x)=(x-2)(2x-1)(3x+1)\).

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

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

Therefore

\(6(2)^3+a(2)^2+b(2)+2=0,\)

so

\(48+4a+2b+2=0.\)

This gives

\(2a+b=-25.\)

Also \(\mathrm{p}(0)=2\), so \(\mathrm{p}(1)=-2\mathrm{p}(0)=-4\). Hence

\(6+a+b+2=-4,\)

which gives

\(a+b=-12.\)

Solving

\(2a+b=-25,\qquad a+b=-12,\)

gives

\(a=-13,\qquad b=1.\)

So

\(\mathrm{p}(x)=6x^3-13x^2+x+2.\)

The remainder when dividing by \(2x-1\) is \(\mathrm{p}\left(\frac12\right)\):

\(\mathrm{p}\left(\frac12\right) =6\left(\frac18\right)-13\left(\frac14\right)+\frac12+2 =0.\)

For the factorisation, since \(x-2\) and \(2x-1\) are factors, the remaining factor is \(3x+1\). Therefore

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