| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants using remainder theorem |
| Difficulty | Moderate -0.3 This is a straightforward application of the factor and remainder theorems with routine algebraic manipulation. Part (i) requires substituting x=-1 and showing the result is zero regardless of a (simple algebra), part (ii) uses remainder theorem with x=2 to find a single constant (one equation, one unknown), and part (iii) involves factorising a polynomial in x² which is slightly less routine but still standard technique. The question requires multiple steps but each is a direct application of well-practiced methods with no novel insight needed. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x = -1\) and simplify | M1 | Allow attempt at long division, must get down to a remainder. Allow M1 if at least 2 numerical values of \(a\) are used. May equate to \((x+1)(Ax^2 + Bx + C) + R\). Allow M1 if they get as far as finding \(R\) |
| Obtain \(-4 + a - a + 4 = 0\) and conclude appropriately | A1 | Must have a conclusion – allow 'hence shown', or made a statement of intent at the start of the question |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(x = 2\) and equate to \(-42\) and attempt to solve | M1 | May equate to \((x-2)(Ax^2 + Bx + C)\), must have a complete method to get as far as \(a = \ldots\) to obtain M1 |
| Obtain \(a = -13\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Divide \(p(x)\) with their \(a\) at least as far as \(4x^2 + kx\) | M1 | |
| Obtain \(4x^2 - 17x + 4\) | A1 | |
| Obtain \((x+1)(4x-1)(x-4)\) or equivalent if \(x^2\) already involved | A1 | If \((x+1)(4x-1)(x-4)\) seen with no evidence of long division then allow the marks |
| Obtain \((x^2+1)(2x-1)(2x+1)(x-2)(x+2)\) | A1 |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = -1$ and simplify | M1 | Allow attempt at long division, must get down to a remainder. Allow M1 if at least 2 numerical values of $a$ are used. May equate to $(x+1)(Ax^2 + Bx + C) + R$. Allow M1 if they get as far as finding $R$ |
| Obtain $-4 + a - a + 4 = 0$ and conclude appropriately | A1 | Must have a conclusion – allow 'hence shown', or made a statement of intent at the start of the question |
## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $x = 2$ and equate to $-42$ and attempt to solve | M1 | May equate to $(x-2)(Ax^2 + Bx + C)$, must have a complete method to get as far as $a = \ldots$ to obtain M1 |
| Obtain $a = -13$ | A1 | |
## Question 4(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Divide $p(x)$ with their $a$ at least as far as $4x^2 + kx$ | M1 | |
| Obtain $4x^2 - 17x + 4$ | A1 | |
| Obtain $(x+1)(4x-1)(x-4)$ or equivalent if $x^2$ already involved | A1 | If $(x+1)(4x-1)(x-4)$ seen with no evidence of long division then allow the marks |
| Obtain $(x^2+1)(2x-1)(2x+1)(x-2)(x+2)$ | A1 | |
---4 The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 4 x ^ { 3 } + a x ^ { 2 } + a x + 4$$
where $a$ is a constant.\\
(i) Use the factor theorem to show that ( $x + 1$ ) is a factor of $\mathrm { p } ( x )$ for all values of $a$.\\
(ii) Given that the remainder is - 42 when $\mathrm { p } ( x )$ is divided by ( $x - 2$ ), find the value of $a$.\\
(iii) When $a$ has the value found in part (ii), factorise $\mathrm { p } \left( x ^ { 2 } \right)$ completely.
\hfill \mbox{\textit{CAIE P2 2016 Q4 [8]}}