\(\mathrm{f}(x) = 2x^3 + ax^2 + bx + 12\), where \(a\) and \(b\) are constants. \((x + 1)\) is a factor of \(\mathrm{f}(x)\), and the remainder when \(\math...

Two theorems drive this question. The remainder theorem says the remainder when \(\mathrm{f}(x)\) is divided by \((x - a)\) is \(\mathrm{f}(a)\), and the factor theorem is its special case: \((x - a)\) is a factor exactly when \(\mathrm{f}(a) = 0\). Substituting a single number is enormously quicker than long division, and each piece of information given converts into one equation in \(a\) and \(b\). Two unknowns need two equations, which is exactly what parts (a) and (b) supply.

(a) Show that \(a - b = -10\) [2]

Since \((x + 1)\) is a factor, write it as \((x - (-1))\), so the value to substitute is \(x = -1\):

\[\mathrm{f}(-1) = 0 \qquad \textbf{[M1]}\] \[2(-1)^3 + a(-1)^2 + b(-1) + 12 = -2 + a - b + 12 = 0 \quad\Rightarrow\quad a - b = -10 \qquad \textbf{[A1]}\]

The [M1] is for setting \(\mathrm{f}(-1) = 0\), the [A1] for the printed result. The sign of the substituted value is the trap: the factor \((x + 1)\) corresponds to \(x = -1\), not \(x = 1\). Then \((-1)^3 = -1\) while \((-1)^2 = +1\), so the \(a\) term stays positive and the \(b\) term turns negative. Since the target is printed, the derivation carries all the credit.

(b) A second equation, and hence \(a\) and \(b\) [4]

The remainder on division by \((x - 2)\) is \(\mathrm{f}(2)\), and it is given as \(-6\):

\[\mathrm{f}(2) = -6 \qquad \textbf{[M1]}\] \[2(8) + a(4) + b(2) + 12 = -6 \quad\Rightarrow\quad 4a + 2b = -34 \quad\Rightarrow\quad 2a + b = -17 \qquad \textbf{[A1]}\]

Here \(16 + 12 = 28\) must be taken across to give \(-6 - 28 = -34\); arriving at \(-34\) rather than \(+22\) is the sign check to make. Now solve the pair simultaneously. Adding removes \(b\) at once:

\[(2a + b) + (a - b) = -17 + (-10) \quad\Rightarrow\quad 3a = -27 \quad\Rightarrow\quad a = -9 \qquad \textbf{[A1]}\] \[b = a + 10 = -9 + 10 = 1 \qquad \textbf{[A1]}\]

The [M1] is for using the remainder theorem, and the three accuracy marks are for the second equation, for \(a\), and for \(b\), so a slip in one value does not automatically cost the other. Adding is the right operation because the \(b\) coefficients are \(+1\) and \(-1\); subtracting instead leaves both unknowns in play.

(c) Factorise \(\mathrm{f}(x)\) completely [4]

With the constants found, \(\mathrm{f}(x) = 2x^3 - 9x^2 + x + 12\). Part (a) already guarantees that \((x + 1)\) is a factor, so divide it out, by long division, by inspection, or by comparing coefficients:

\[\mathrm{f}(x) = (x + 1)\left(2x^2 - 11x + 12\right) \qquad \textbf{[M1 A1]}\]

The [M1] is for a correct division method, the [A1] for the quadratic factor. Checking by inspection is fast: the leading term must be \(2x^3\), so the quadratic starts \(2x^2\), and the constant term must be \(12\), so the quadratic ends \(+12\).

Now factorise the quadratic. Look for two numbers multiplying to \(2 \times 12 = 24\) and adding to \(-11\), namely \(-3\) and \(-8\):

\[2x^2 - 11x + 12 = (2x - 3)(x - 4) \qquad \textbf{[M1]}\] \[\mathrm{f}(x) = (x + 1)(2x - 3)(x - 4) \qquad \textbf{[A1]}\]

The word "completely" means all three linear factors must appear; stopping at \((x + 1)\left(2x^2 - 11x + 12\right)\) forfeits the last two marks.

(d) Hence solve \(\mathrm{f}(x) = 0\) [1]

A product is zero when one of its factors is zero, so

\[x = -1, \qquad x = \frac{3}{2}, \qquad x = 4 \qquad \textbf{[B1]}\]

The single [B1] needs all three roots. The one most often mishandled comes from \(2x - 3 = 0\), which gives \(x = \tfrac{3}{2}\); reading the root as \(3\) or as \(-\tfrac{3}{2}\) loses the mark. Since this part is worth one mark for a full solution set, there is no partial credit.

Check: expand the factorisation, or test the given conditions. \(\mathrm{f}(-1) = -2 - 9 - 1 + 12 = 0\), confirming the factor, and \(\mathrm{f}(2) = 16 - 36 + 2 + 12 = -6\), confirming the remainder. Both original statements hold with \(a = -9\) and \(b = 1\), so the values are right and the factorisation built on them is sound.