| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constant then solve inequality or further work |
| Difficulty | Moderate -0.3 This is a straightforward application of the factor theorem requiring routine algebraic manipulation. Part (a) uses substitution to find k, part (b) involves polynomial division and factorisation (standard techniques), and part (c) combines root identification with solving a modulus equation. While multi-step, each component is a textbook exercise with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02l Modulus function: notation, relations, equations and inequalities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x = 3\), equate to zero and attempt solution | M1 | Condone \(-\frac{51}{3}\) |
| Obtain \(k = -17\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Divide by \(x - 3\) at least as far as \(2x^2 + mx\) | M1 | |
| Obtain \(2x^2 + 9x + 10\) | A1 | |
| Obtain \((x-3)(2x+5)(x+2)\) | A1 | SC – no attempt at division (or equivalent) and only correct fully factorised form shown: award B1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Solve linear equation \(4y - 5 = 3\) to obtain \(y = 2\) | B1 FT | Following any positive root from *their* factorised \(p(x)\) |
| Attempt solution of linear equation \(4y - 5 = -3\) or equivalent | M1 | With RHS of equation being \(-\)(*their* positive root) |
| Obtain \(y = \frac{1}{2}\) | A1 | |
| Alternative: State or imply \((4y-5)^2 = 3^2\) | B1 FT | Following any positive root from *their* factorised \(p(x)\) |
| Attempt solution of 3-term quadratic equation \((4y-5)^2 = 3^2\) | M1 | With RHS of equation involving *their* positive root |
| Obtain \(\frac{1}{2}\) and 2 and no other solutions | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = 3$, equate to zero and attempt solution | M1 | Condone $-\frac{51}{3}$ |
| Obtain $k = -17$ | A1 | |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Divide by $x - 3$ at least as far as $2x^2 + mx$ | M1 | |
| Obtain $2x^2 + 9x + 10$ | A1 | |
| Obtain $(x-3)(2x+5)(x+2)$ | A1 | SC – no attempt at division (or equivalent) and only correct fully factorised form shown: award B1 only |
## Question 4(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Solve linear equation $4y - 5 = 3$ to obtain $y = 2$ | B1 FT | Following any positive root from *their* factorised $p(x)$ |
| Attempt solution of linear equation $4y - 5 = -3$ or equivalent | M1 | With RHS of equation being $-$(*their* positive root) |
| Obtain $y = \frac{1}{2}$ | A1 | |
| **Alternative:** State or imply $(4y-5)^2 = 3^2$ | B1 FT | Following any positive root from *their* factorised $p(x)$ |
| Attempt solution of 3-term quadratic equation $(4y-5)^2 = 3^2$ | M1 | With RHS of equation involving *their* positive root |
| Obtain $\frac{1}{2}$ and 2 and no other solutions | A1 | |4 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 2 x ^ { 3 } + 3 x ^ { 2 } + k x - 30$$
where $k$ is a constant. It is given that $( x - 3 )$ is a factor of $\mathrm { p } ( x )$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Hence find the quotient when $\mathrm { p } ( x )$ is divided by ( $x - 3$ ) and factorise $\mathrm { p } ( x )$ completely.
\item It is given that $a$ is one of the roots of the equation $\mathrm { p } ( x ) = 0$.
Given also that the equation $| 4 y - 5 | = a$ is satisfied by two real values of $y$, find these two values of $y$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P2 2023 Q4 [8]}}