| Exam Board | CAIE |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants using remainder theorem |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem requiring substitution of x=1/2 to find 'a', then verification by substitution at x=3, followed by routine factorization and solving a quadratic. All steps are standard textbook procedures with no novel insight required, making it easier than average but not trivial due to the multi-part nature and algebraic manipulation involved. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Substitute \(x = \frac{1}{2}\) and equate to 10 | M1 | |
| Obtain answer \(a = -16\) | A1 | |
| Either show that \(f(3) = 0\) or divide by \((x-3)\) obtaining a remainder of zero | B1 | [3] |
| (ii) At any stage state that \(x = 3\) is a solution | B1 | |
| Attempt division by \((x-3)\) reaching a partial quotient of \(4x^2 + kx\) | M1 | |
| Obtain quadratic factor \(4x^2 - 4x - 3\) | A1 | |
| Obtain solutions \(x = \frac{3}{2}\) and \(x = -\frac{1}{2}\) | A1 | |
| S.C. M1A1√ if value of '\(a\)' incorrect | [4] |
**(i)** Substitute $x = \frac{1}{2}$ and equate to 10 | M1 |
Obtain answer $a = -16$ | A1 |
Either show that $f(3) = 0$ or divide by $(x-3)$ obtaining a remainder of zero | B1 | [3]
**(ii)** At any stage state that $x = 3$ is a solution | B1 |
Attempt division by $(x-3)$ reaching a partial quotient of $4x^2 + kx$ | M1 |
Obtain quadratic factor $4x^2 - 4x - 3$ | A1 |
Obtain solutions $x = \frac{3}{2}$ and $x = -\frac{1}{2}$ | A1 |
S.C. M1A1√ if value of '$a$' incorrect | [4]5 The polynomial $4 x ^ { 3 } + a x ^ { 2 } + 9 x + 9$, where $a$ is a constant, is denoted by $\mathrm { p } ( x )$. It is given that when $\mathrm { p } ( x )$ is divided by $( 2 x - 1 )$ the remainder is 10 .\\
(i) Find the value of $a$ and hence verify that ( $x - 3$ ) is a factor of $\mathrm { p } ( x )$.\\
(ii) When $a$ has this value, solve the equation $\mathrm { p } ( x ) = 0$.
\hfill \mbox{\textit{CAIE P2 2011 Q5 [7]}}