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.