Show that the two zeros are 3 and \( - 6\).

Find the value of \(q\) and of \(r\).

recognition that the \(x\)-coordinate of the vertex is \( - 1.5\) (seen anywhere)     (M1)

eg\(\,\,\,\,\,\)axis of symmetry is \( - 1.5\), sketch, \(f'( - 1.5) = 0\)

correct working to find the zeroes     A1

eg\(\,\,\,\,\,\)\( - 1.5 \pm 4.5\)

\(x =  - 6\) and \(x = 3\)     AG     N0

[2 marks]

METHOD 1 (using factors)

attempt to write factors     (M1)

eg\(\,\,\,\,\,\)\((x - 6)(x + 3)\)

correct factors     A1

eg\(\,\,\,\,\,\)\((x - 3)(x + 6)\)

\(q = 3,{\text{ }}r =  - 18\)    A1A1     N3

METHOD 2 (using derivative or vertex)

valid approach to find \(q\)     (M1)

eg\(\,\,\,\,\,\)\(f'( - 1.5) = 0,{\text{ }} - \frac{q}{{2a}} =  - 1.5\)

\(q = 3\)    A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( - 6)^2} + 3( - 6) + r = 0\)

\(r =  - 18\)    A1

\(q = 3,{\text{ }}r =  - 18\)    N3

METHOD 3 (solving simultaneously)

valid approach setting up system of two equations     (M1)

eg\(\,\,\,\,\,\)\(9 + 3q + r = 0,{\text{ }}36 - 6q + r = 0\)

one correct value

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  - 18\)     A1

correct substitution     A1

eg\(\,\,\,\,\,\)\({3^2} + 3(3) + r = 0,{\text{ }}{( - 6)^2} + 3( - 6) + r = 0,{\text{ }}{3^2} + 3q - 18 = 0,{\text{ }}36 - 6q - 18 = 0\)

second correct value     A1

eg\(\,\,\,\,\,\)\(q = 3,{\text{ }}r =  - 18\)

\(q = 3,{\text{ }}r =  - 18\)    N3

[4 marks]

As a ‘show that’ question, part a) required a candidate to independently find the answers. Again, too many candidates used the given answers (of 3 and \( - 6\)) to show that the two zeros were 3 and \( - 6\) (a circular argument). Those who were able to recognize that the \(x\)-coordinate of the vertex is \( - 1.5\) tended to then use the given answers and work backwards thus scoring no further marks in part a).

Answers to part b) were more successful with a good variety of methods used and correct solutions seen.