\(\mathrm{f}(x) = 3x^3 + 2x^2 - 5x + 7\) Find the quotient and the remainder when \(\mathrm{f}(x)\) is divided by \((x + 2)\). (3) Hence write down the valu...

Part (a) asks for a division, so it must be carried out; part (b) then rewards a candidate who recognises that the remainder theorem makes a second calculation unnecessary. The word "Hence" in part (b) is the instruction to reuse part (a) rather than to substitute afresh.

(a) The quotient and remainder on division by \(\left(x + 2\right)\) [3 marks]

Long division of polynomials works exactly like long division of numbers: at each stage divide the leading term of what is left by the leading term of the divisor, multiply back, and subtract. The divisor is \(x + 2\) and the dividend is \(\mathrm{f}(x) = 3x^{3} + 2x^{2} - 5x + 7\).

1. \(3x^{3} \div x = 3x^{2}\). Multiplying back, \(3x^{2}\left(x + 2\right) = 3x^{3} + 6x^{2}\). Subtracting from \(3x^{3} + 2x^{2} - 5x + 7\) leaves \(-4x^{2} - 5x + 7\), since \(2x^{2} - 6x^{2} = -4x^{2}\). M1
2. \(-4x^{2} \div x = -4x\). Multiplying back, \(-4x\left(x + 2\right) = -4x^{2} - 8x\). Subtracting leaves \(3x + 7\), since \(-5x - \left(-8x\right) = 3x\). M1
3. \(3x \div x = 3\). Multiplying back, \(3\left(x + 2\right) = 3x + 6\). Subtracting leaves \(1\), and since \(1\) has lower degree than \(x + 2\) the division stops.

\[\text{Quotient } = 3x^{2} - 4x + 3, \qquad \text{remainder } = 1 \qquad \textbf{A1}\]

The two M marks are for the division process, so they survive an arithmetic slip inside a subtraction; the single A1 requires both the quotient and the remainder to be completely correct, and it does not. The subtraction at each stage is where marks are lost, because subtracting a negative term changes its sign: at the second stage \(-5x - \left(-8x\right)\) is \(+3x\), not \(-13x\).

Verify by multiplying out, which takes only a moment and catches every slip:

\[\left(x + 2\right)\left(3x^{2} - 4x + 3\right) + 1 = 3x^{3} - 4x^{2} + 3x + 6x^{2} - 8x + 6 + 1 = 3x^{3} + 2x^{2} - 5x + 7 = \mathrm{f}(x).\]

Note also that the divisor is \(x + 2\), so the number used in a synthetic-division layout is \(-2\), not \(+2\). Using \(+2\) is the single most common route to a wrong remainder here.

(b) The value of \(\mathrm{f}(-2)\) [1 mark]

The remainder theorem says that when a polynomial \(\mathrm{f}(x)\) is divided by \(\left(x - a\right)\), the remainder equals \(\mathrm{f}(a)\). Writing the divisor as \(x - \left(-2\right)\) identifies \(a = -2\), so the remainder found in part (a) is precisely \(\mathrm{f}(-2)\):

\[\mathrm{f}(-2) = 1 \qquad \textbf{B1}\]

This is a B mark: a single independent mark for the correct value, with no method credit, so the answer must simply be right. The reason the theorem holds is visible in the identity above. Since

\[\mathrm{f}(x) = \left(x + 2\right)\left(3x^{2} - 4x + 3\right) + 1,\]

putting \(x = -2\) makes the bracket \(\left(x + 2\right)\) zero, which annihilates the whole quotient term and leaves only the remainder.

Direct substitution confirms it: \(\mathrm{f}(-2) = 3\left(-8\right) + 2\left(4\right) - 5\left(-2\right) + 7 = -24 + 8 + 10 + 7 = 1\). That calculation is a legitimate check but not what the part is testing, and a candidate who does it without ever linking the answer to the remainder has missed the point of the word "Hence". The sign errors to guard against in the check are \(3\left(-2\right)^{3} = -24\) rather than \(+24\), and \(-5 \times \left(-2\right) = +10\).

Examination takeaway. Convert the divisor into the form \(x - a\) before using the remainder theorem, so that \(x + 2\) gives \(a = -2\). When a question asks first for a division and then for a value of the function, the remainder is the answer, and the marks reward spotting that rather than repeating the work.