| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then solve related equation |
| Difficulty | Standard +0.3 Part (i) is a standard application of factor theorem to find a constant, followed by routine polynomial division and factorisation. Part (ii) requires substitution and algebraic manipulation but follows directly from part (i). This is a typical textbook exercise requiring multiple steps but no novel insight, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Substitute \(x = -2\) and equate to zero | M1 | |
| Obtain \(-8 + 4a - 28 + a + 1 = 0\) or equivalent and hence \(a = 7\) | A1 | |
| Attempt division by \((x+2)\) and reach partial quotient \(x^2 + kx\), where \(k\) is numeric, or use of identity or inspection or synthetic division | M1 | Synthetic division shown in mark scheme |
| Obtain quotient \(x^2 + 5x + 4\) soi | A1 | |
| Conclude with \((x+1)(x+2)(x+4)\) | A1 | |
| Total | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \((2x+1)(2x+2)(2x+4) = 3(x+1)(x+2)(x+4)\) | M1 | Following their complete factorised form |
| Obtain \(x = -1\) and \(x = -2\) | A1 | Calculator not permitted so necessary detail needed |
| Cancel common factors to obtain linear equation or factorise to find corresponding factor | M1 | |
| Obtain \(x = \frac{8}{5}\) or equivalent | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| State \((2x+1)(2x+2)(2x+4) = 3(x+1)(x+2)(x+4)\) or \((2x)^3 + 7(2x)^2 + 14(2x) + 8 = 3(x^3 + 7x^2 + 14x + 8)\) | M1 | Following their completed factorised form. Must see \(8x^3\) and \(28x^2\) if using second statement without bracketed terms in \(2x\) |
| Expand and simplify to obtain \(5x^3 + 7x^2 - 14x - 16 = 0\) | A1 | Must be equated to 0 for A1 |
| Attempt complete factorisation of cubic with leading term \(5x^3\) (may make use of synthetic division) | M1 | Synthetic division: \(-2 \mid 5 \quad 7 \quad -14 \quad -16\) with \(-10 \quad 6 \quad 16\) giving \(5 \quad -3 \quad -8 \quad 0\) |
| Obtain \((x+1)(x+2)(5x-8) = 0\) and conclude \(x = -1,\ x = -2,\ x = \frac{8}{5}\) | A1 | Calculator not permitted so necessary detail needed |
## Question 6(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Substitute $x = -2$ and equate to zero | M1 | |
| Obtain $-8 + 4a - 28 + a + 1 = 0$ or equivalent and hence $a = 7$ | A1 | |
| Attempt division by $(x+2)$ and reach partial quotient $x^2 + kx$, where $k$ is numeric, or use of identity or inspection or synthetic division | M1 | Synthetic division shown in mark scheme |
| Obtain quotient $x^2 + 5x + 4$ soi | A1 | |
| Conclude with $(x+1)(x+2)(x+4)$ | A1 | |
| **Total** | **5** | |
## Question 6(ii):
**Either method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State $(2x+1)(2x+2)(2x+4) = 3(x+1)(x+2)(x+4)$ | M1 | Following their complete factorised form |
| Obtain $x = -1$ and $x = -2$ | A1 | Calculator not permitted so necessary detail needed |
| Cancel common factors to obtain linear equation or factorise to find corresponding factor | M1 | |
| Obtain $x = \frac{8}{5}$ or equivalent | A1 | |
**Or method:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State $(2x+1)(2x+2)(2x+4) = 3(x+1)(x+2)(x+4)$ or $(2x)^3 + 7(2x)^2 + 14(2x) + 8 = 3(x^3 + 7x^2 + 14x + 8)$ | M1 | Following their completed factorised form. Must see $8x^3$ and $28x^2$ if using second statement without bracketed terms in $2x$ |
| Expand and simplify to obtain $5x^3 + 7x^2 - 14x - 16 = 0$ | A1 | Must be equated to 0 for A1 |
| Attempt complete factorisation of cubic with leading term $5x^3$ (may make use of synthetic division) | M1 | Synthetic division: $-2 \mid 5 \quad 7 \quad -14 \quad -16$ with $-10 \quad 6 \quad 16$ giving $5 \quad -3 \quad -8 \quad 0$ |
| Obtain $(x+1)(x+2)(5x-8) = 0$ and conclude $x = -1,\ x = -2,\ x = \frac{8}{5}$ | A1 | Calculator not permitted so necessary detail needed |
**Total: 4 marks**
---6 The cubic polynomial $\mathrm { f } ( x )$ is defined by
$$\mathrm { f } ( x ) = x ^ { 3 } + a x ^ { 2 } + 14 x + a + 1$$
where $a$ is a constant. It is given that ( $x + 2$ ) is a factor of $\mathrm { f } ( x )$.\\
(i) Use the factor theorem to find the value of $a$ and hence factorise $\mathrm { f } ( x )$ completely.\\
(ii) Hence, without using a calculator, solve the equation $\mathrm { f } ( 2 x ) = 3 \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{CAIE P2 2018 Q6 [9]}}