Answer: \(a=6\), \(b=1\), \(c=6\).

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.

Since \(x+2\) is a factor, \(\mathrm p(-2)=0\):

\(-8a+4b+38+c=0.\)

When \(\mathrm p(x)\) is divided by \(x+1\), the remainder is \(\mathrm p(-1)\), so

\(-a+b+19+c=20.\)

Subtracting these equations eliminates \(c\):

\(-8a+4b+38+c)-(-a+b+19+c)=0-20\).

This gives \(-7a+3b+19=-20\), so \(7a-3b=39\), as required.

Now \(\mathrm p'(x)=3ax^2+2bx-19\). Since the remainder when \(\mathrm p'(x)\) is divided by \(x-1\) is \(1\),

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

Thus \(3a+2b-19=1\), so \(3a+2b=20\).

Solve

\(7a-3b=39,\quad 3a+2b=20.\)

This gives \(a=6\), \(b=1\).

Using \(-a+b+19+c=20\),

\(-6+1+19+c=20\), so \(c=6\).

This gives the required answer: \(a=6\), \(b=1\), \(c=6\).