Answer: \(x=-1,-\dfrac12,3\); \(a=-5\), \(b=8\), \(c=-9\); \(p''(2)=2\gt 0\), so the stationary point at \(x=2\) is a minimum.

We start with the main method. Differentiate to find stationary points, then use the second derivative or the shape of the curve to classify them.

Since \(x-3\) and \(x+1\) are factors, the cubic has the form

\((x-3)(x+1)(2x+1).\)

Expanding confirms

\((x-3)(x+1)(2x+1)=2x^3-3x^2-8x-3.\)

Therefore the roots are

\(x=3,\quad x=-1,\quad x=-\frac12.\)

For part (b),

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

The stationary points occur at \(x=\frac43\) and \(x=2\), so \(\mathrm p'(x)\) has roots \(\frac43\) and \(2\).

Since the coefficient of \(x^2\) in \(\mathrm p'(x)\) is \(3\),

\(\mathrm p'(x)=3\left(x-\frac43\right)(x-2).\)

This expands to

\(\mathrm p'(x)=3x^2-10x+8.\)

Comparing with \(3x^2+2ax+b\),

\(2a=-10\) and \(b=8\).

So \(a=-5\) and \(b=8\).

The remainder when \(\mathrm p(x)\) is divided by \(x-1\) is \(\mathrm p(1)\). Hence

\(\mathrm p(1)=1+a+b+c=-5.\)

Substitute \(a=-5\) and \(b=8\):

\(1-5+8+c=-5.\)

Thus \(c=-9\).

Now

\(\mathrm p''(x)=6x+2a=6x-10.\)

At \(x=2\),

\(\mathrm p''(2)=12-10=2\gt 0.\)

Therefore the stationary point at \(x=2\) is a minimum.

This completes the solution and gives the required result.