| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.8 This is a standard Factor and Remainder Theorem question requiring straightforward application of p(2)=0 and p(-1)=27 to form two simultaneous equations, then factorising using the known factor. It's routine A-level algebra with no novel insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(2) = 0\): \(40 + 4a + 2b - 16 = 0 \Rightarrow 4a + 2b = -24 \Rightarrow 2a + b = -12\) | M1 A1 | Factor theorem applied correctly |
| \(p(-1) = 27\): \(-5 + a - b - 16 = 27 \Rightarrow a - b = 48\) | M1 A1 | Remainder theorem applied correctly |
| Solving: \(3a = 36 \Rightarrow a = 12\), \(b = -36\) | A1 | Both values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Since \(x = 2\) is a root (from part i), \((x-2)\) is a factor | M1 | Attempt polynomial division or inspection |
| \(p(x) = (x-2)(2x^2 + ax + b)\) leading to correct quadratic factor | M1 | |
| \(p(x) = (x-2)(2x^2+x-4)\) or equivalent complete factorisation | A1 |
# Question 5(i):
$p(x) = 5x^3 + ax^2 + bx - 16$
| $p(2) = 0$: $40 + 4a + 2b - 16 = 0 \Rightarrow 4a + 2b = -24 \Rightarrow 2a + b = -12$ | M1 A1 | Factor theorem applied correctly |
|---|---|---|
| $p(-1) = 27$: $-5 + a - b - 16 = 27 \Rightarrow a - b = 48$ | M1 A1 | Remainder theorem applied correctly |
| Solving: $3a = 36 \Rightarrow a = 12$, $b = -36$ | A1 | Both values correct |
I can see these are Cambridge A-Level Mathematics exam pages (9709/21/M/J/19). Let me extract the mark scheme content based on what I can derive from the questions shown.
## Question 5(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Since $x = 2$ is a root (from part i), $(x-2)$ is a factor | M1 | Attempt polynomial division or inspection |
| $p(x) = (x-2)(2x^2 + ax + b)$ leading to correct quadratic factor | M1 | |
| $p(x) = (x-2)(2x^2+x-4)$ or equivalent complete factorisation | A1 | |5 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 5 x ^ { 3 } + a x ^ { 2 } + b x - 16$$
where $a$ and $b$ are constants. It is given that $( x - 2 )$ is a factor of $\mathrm { p } ( x )$ and that the remainder is 27 when $\mathrm { p } ( x )$ is divided by $( x + 1 )$.\\
(i) Find the values of $a$ and $b$.\\
(ii) Hence factorise $\mathrm { p } ( x )$ completely.\\
\hfill \mbox{\textit{CAIE P2 2019 Q5 [8]}}