| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.3 This is a standard two-condition factor/remainder theorem problem requiring straightforward substitution to form simultaneous equations, then routine factorization. While it involves multiple steps, the techniques are entirely procedural with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Substitute \(x = -2\) and equate to zero | \*M1 | |
| Substitute \(x = \frac{1}{2}\) and equate to \(40\) | \*M1 | |
| Obtain \(-8a + 4b - 64 = 0\) and \(\frac{1}{8}a + \frac{1}{4}b = \frac{23}{2}\) or equivalents | A1 | |
| Solve a pair of simultaneous equations for \(a\) or for \(b\) | DM1 | Needs at least one of the two previous M marks |
| Obtain \(a = 12\) and \(b = 40\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt division by \((x+2)\) or inspection at least as far as \(kx^2 + mx\) | M1 | |
| Obtain \(12x^2 + 16x + 5\) | A1 | |
| Conclude \((x+2)(2x+1)(6x+5)\) | A1 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Substitute $x = -2$ and equate to zero | \*M1 | |
| Substitute $x = \frac{1}{2}$ and equate to $40$ | \*M1 | |
| Obtain $-8a + 4b - 64 = 0$ and $\frac{1}{8}a + \frac{1}{4}b = \frac{23}{2}$ or equivalents | A1 | |
| Solve a pair of simultaneous equations for $a$ or for $b$ | DM1 | Needs at least one of the two previous M marks |
| Obtain $a = 12$ and $b = 40$ | A1 | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt division by $(x+2)$ or inspection at least as far as $kx^2 + mx$ | M1 | |
| Obtain $12x^2 + 16x + 5$ | A1 | |
| Conclude $(x+2)(2x+1)(6x+5)$ | A1 | |
---5 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = a x ^ { 3 } + b x ^ { 2 } + 37 x + 10$$
where $a$ and $b$ are constants. It is given that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$. It is also given that the remainder is 40 when $\mathrm { p } ( x )$ is divided by ( $2 x - 1$ ).\\
(i) Find the values of $a$ and $b$.\\
(ii) When $a$ and $b$ have these values, factorise $\mathrm { p } ( x )$ completely.\\
\hfill \mbox{\textit{CAIE P2 2017 Q5 [8]}}