Answer: \(m=6\), \(n=-10\), so \(x-2\) is a factor.

Since \(3x-1\) is a factor, \(x=\frac13\) is a root. Therefore

\(\mathrm p\left(\frac13\right)=0.\)

This gives

\(\frac{m}{27}-\frac{29}{9}+13+n=0. \tag{1}\)

The remainder when divided by \(x-1\) is \(6\), so

\(\mathrm p(1)=6.\)

Thus

\(m-29+39+n=6,\)

or

\(m+n=-4. \tag{2}\)

Solving (1) and (2) gives

\(m=6,\qquad n=-10.\)

Now substitute \(x=2\):

\(\mathrm p(2)=6(2)^3-29(2)^2+39(2)-10.\)

So

\(\mathrm p(2)=48-116+78-10=0.\)

Since \(\mathrm p(2)=0\), \(x-2\) is a factor of \(\mathrm p(x)\).