\(\mathrm{g}(x) = x^4 - 2x^3 + 3x - 5\) Find the remainder when \(\mathrm{g}(x)\) is divided by \((x - 3)\). (2) The remainder when \(\mathrm{g}(x)\) is div...

Part (a) is a direct application of the remainder theorem: the remainder on dividing \(\mathrm{g}(x)\) by \((x - a)\) is simply \(\mathrm{g}(a)\), so no division is needed. Part (b) extends the same idea to a quadratic divisor, and the extension rests on one structural fact: the remainder on dividing by a quadratic is of lower degree than the divisor, so it is at most linear, which is exactly why the question tells you it has the form \(ax + b\).

(a) Remainder when \(\mathrm{g}(x)\) is divided by \((x - 3)\) [2]

By the remainder theorem the remainder is \(\mathrm{g}(3)\) [M1]. Substituting into \(\mathrm{g}(x) = x^4 - 2x^3 + 3x - 5\):

\[\mathrm{g}(3) = 81 - 2(27) + 9 - 5 = 81 - 54 + 9 - 5 = 31 \qquad \textbf{[A1]}\]

The [M1] is for identifying \(x = 3\) as the value to substitute, so it survives an arithmetic slip; the [A1] is for the value \(31\). Note that \(\mathrm{g}(x)\) has no \(x^2\) term, so nothing is contributed there. The sign trap is the divisor: \((x - 3)\) means \(x = +3\). Long division would also reach \(31\) but wastes several minutes for the same two marks.

(b) Find \(a\) and \(b\), where the remainder on dividing by \(\left(x^2 - x - 6\right)\) is \(ax + b\) [4]

Factorise the divisor first, because that is what makes substitution possible:

\[x^2 - x - 6 = (x - 3)(x + 2)\]

The divisor is quadratic, so the remainder is at most linear, and the division statement is

\[\mathrm{g}(x) = (x - 3)(x + 2)\,\mathrm{Q}(x) + ax + b \qquad \textbf{[M1]}\]

where \(\mathrm{Q}(x)\) is the quotient. The [M1] is for setting up this identity. Its power is that substituting either root of the divisor kills the \(\mathrm{Q}(x)\) term entirely, whatever \(\mathrm{Q}(x)\) happens to be, so the quotient never has to be found.

Put \(x = 3\). The bracket \((x - 3)\) vanishes, so

\[3a + b = \mathrm{g}(3) = 31 \qquad \textbf{[A1]}\]

which reuses part (a) and is why the two parts sit together. Put \(x = -2\), which makes \((x + 2)\) vanish:

\[\mathrm{g}(-2) = 16 - 2(-8) + 3(-2) - 5 = 16 + 16 - 6 - 5 = 21, \quad\text{so } -2a + b = 21 \qquad \textbf{[A1]}\]

The evaluation of \(\mathrm{g}(-2)\) is where marks go astray: \((-2)^4 = +16\) and \((-2)^3 = -8\), so the term \(-2x^3\) becomes \(-2 \times (-8) = +16\), a double negative that turns positive. Getting it as \(-16\) gives \(\mathrm{g}(-2) = -11\) and wrecks the rest.

Now solve the pair. Subtracting the second equation from the first eliminates \(b\):

\[(3a + b) - (-2a + b) = 31 - 21 \quad\Rightarrow\quad 5a = 10 \quad\Rightarrow\quad a = 2\] \[b = 31 - 3a = 31 - 6 = 25 \qquad \textbf{[A1]}\]

So \(a = 2\) and \(b = 25\), and the remainder is \(2x + 25\). The final [A1] requires both constants. Each of the two accuracy marks before it is independent, one per substitution, so a slip in \(\mathrm{g}(-2)\) costs only that mark and the last one, while the method mark and the \(x = 3\) mark both stand.

The wrong turn this part is designed to catch is assuming the remainder is a constant, as it is for a linear divisor. Dividing by a quadratic can leave a linear remainder, which is why two unknowns and therefore two substitutions are needed. A second, subtler error is substituting \(x = 3\) and \(x = 2\), reading the factors from the coefficients rather than from \((x - 3)(x + 2)\).

Check: the remainder should reproduce both function values. At \(x = 3\), \(2(3) + 25 = 31 = \mathrm{g}(3)\), and at \(x = -2\), \(2(-2) + 25 = 21 = \mathrm{g}(-2)\). Both agree, which confirms \(a\) and \(b\) without carrying out the division. If more assurance is wanted, the full long division gives quotient \(x^2 - x + 5\), so \(\mathrm{g}(x) = \left(x^2 - x - 6\right)\left(x^2 - x + 5\right) + 2x + 25\), and multiplying out returns \(x^4 - 2x^3 + 3x - 5\) exactly.