A polynomial $p(x)$ with real coefficients is of degree five. The equation $p(x) = 0$ has a complex root 2 + i. The graph of $y = p(x)$ has the x-axis as a tangent at (2, 0) and intersects the coordinate axes at (−1, 0) and (0, 4).

Find $p(x)$ in factorised form with real coefficients.

other root is 2 – i     (A1)

a quadratic factor is therefore $(x - 2 + i)(x - 2 - i)$     (M1)

$= {x^2} - 4x + 5$     A1

x + 1 is a factor     A1

${(x - 2)^2}$ is a factor     A1

$p(x) = a(x + 1){(x - 2)^2}({x^2} - 4x + 5)$     (M1)

$p(0) = 4 \Rightarrow a = \frac{1}{5}$     A1

$p(x) = \frac{1}{5}(x + 1){(x - 2)^2}({x^2} - 4x + 5)$

[7 marks]

Whilst most candidates knew that another root was $2 - {\text{i}}$ , much fewer were able to find the quadratic factor. Surprisingly few candidates knew that $\left( {x - 2} \right)$ must be a repeated factor and less surprisingly many did not recognise that the whole expression needed to be multiplied by $\frac{1}{5}$.