| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | One factor, one non-zero remainder |
| Difficulty | Moderate -0.3 This is a straightforward application of the factor and remainder theorems to find two unknowns, followed by routine factorization and a simple substitution. The question requires standard techniques (substituting x=-2 and x=-1, solving simultaneous equations, factorizing a cubic with one known factor) with no novel insight needed. It's slightly easier than average due to its mechanical nature, though the multi-part structure and final substitution prevent it from being trivial. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x=-2\) and equate to zero | \*M1 | |
| Substitute \(x=-1\) and equate to \(-11\) | \*M1 | |
| Obtain \(4a-2b-68=0\) and \(a-b-26=-11\) or equivalents | A1 | |
| Solve a pair of relevant simultaneous linear equations to find \(a\) or \(b\) | DM1 | Dependent on at least one M1 mark |
| Obtain \(a=19\) and \(b=4\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Divide by \(x+2\) at least as far as the \(x\) term | M1 | or equivalent (inspection, …) |
| Obtain \((x+2)^2(6x-5)\) | A1 | OE |
| Replace (or imply replacement of) \(x\) by \(3x\) in factorised form | M1 | |
| Obtain \(-\dfrac{2}{3}\) and \(\dfrac{5}{18}\) | A1 | and no others |
| Total: 4 |
## Question 5(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x=-2$ and equate to zero | \*M1 | |
| Substitute $x=-1$ and equate to $-11$ | \*M1 | |
| Obtain $4a-2b-68=0$ and $a-b-26=-11$ or equivalents | A1 | |
| Solve a pair of relevant simultaneous linear equations to find $a$ or $b$ | DM1 | Dependent on at least one M1 mark |
| Obtain $a=19$ and $b=4$ | A1 | |
| **Total: 5** | | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Divide by $x+2$ at least as far as the $x$ term | M1 | or equivalent (inspection, …) |
| Obtain $(x+2)^2(6x-5)$ | A1 | OE |
| Replace (or imply replacement of) $x$ by $3x$ in factorised form | M1 | |
| Obtain $-\dfrac{2}{3}$ and $\dfrac{5}{18}$ | A1 | and no others |
| **Total: 4** | | |5 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 6 x ^ { 3 } + a x ^ { 2 } + b x - 20$$
where $a$ and $b$ are constants. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ and that the remainder is - 11 when $\mathrm { p } ( x )$ is divided by $( x + 1 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $a$ and $b$.
\item Hence factorise $\mathrm { p } ( x )$, and determine the exact roots of the equation $\mathrm { p } ( 3 x ) = 0$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q5 [9]}}