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 \(\frac{1}{5}\).