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 \(3{a^3} + {a^2} + 5a + 7 = 0\).