Consider $p(x) = 3{x^3} + ax + 5a,\;\;\;a \in \mathbb{R}$.

The polynomial $p(x)$ leaves a remainder of $- 7$ when divided by $(x - a)$.

Show that only one value of $a$ satisfies the above condition and state its value.

using $p(a) =  - 7$ to obtain $3{a^3} + {a^2} + 5a + 7 = 0$     M1A1

$(a + 1)(3{a^3} - 2a + 7) = 0$     (M1)(A1)

Note:     Award M1 for a cubic graph with correct shape and A1 for clearly showing that the above cubic crosses the horizontal axis at $( - 1,{\text{ }}0)$ only.

$a =  - 1$     A1

EITHER

showing that $3{a^2} - 2a + 7 = 0$ has no real (two complex) solutions for $a$     R1

OR

showing that $3{a^3} + {a^2} + 5a + 7 = 0$ has one real (and two complex) solutions for $a$     R1

Note:     Award R1 for solutions that make specific reference to an appropriate graph.

[6 marks]

A large number of candidates, either by graphical (mostly) or algebraic or via use of a GDC solver, were able to readily obtain $a =  - 1$. Most candidates who were awarded full marks however, made specific reference to an appropriate graph. Only a small percentage of candidates used the discriminant to justify that only one value of $a$ satisfied the required condition. A number of candidates erroneously obtained $3{a^3} + {a^2} + 5a - 7 = 0$ or equivalent rather than $3{a^3} + {a^2} + 5a + 7 = 0$.