Answer: \(a=2\), \(b=12\), \(c=-9\).

Use the factor theorem and the remainder theorem to turn the given information into equations for the unknown constants. After finding the constants, factorise the polynomial if required.

\(\mathrm p(x)=ax^3+11x^2+bx+c\), so

\(\mathrm p'(x)=3ax^2+22x+b\).

Since \(\mathrm p'(0)=12\), we get \(b=12\).

Since \(x+3\) is a factor, \(\mathrm p(-3)=0\):

\(-27a+99-3b+c=0\).

Using \(b=12\), this becomes

\(-27a+63+c=0\).

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

\(a+11+b+c=16\).

Using \(b=12\),

\(a+c=-7\).

From \(-27a+63+c=0\), \(c=27a-63\). Substitute into \(a+c=-7\):

\(a+27a-63=-7\), so \(28a=56\), and \(a=2\).

Then \(c=-9\).

This gives the required answer: \(a=2\), \(b=12\), \(c=-9\).