| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two unknowns with show-that step |
| Difficulty | Moderate -0.8 This is a straightforward application of the Remainder Theorem requiring students to substitute x=1 and x=2 into the polynomial, then solve two simultaneous linear equations. The 'show that' step in part (a) is routine verification, and part (b) involves basic algebra with no conceptual challenges beyond standard P2 content. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| VIIIV SIHI NI JIIIM ION OC | VIIV SIHI NI JINAM ION OC | VEYV SIHI NI JULIM ION OO |
| Answer | Marks |
|---|---|
| Attempting \(f(1)\) or \(f(-1)\) | M1 |
| \(f(1) = 1 + 1 + 2 + a + b = 7\) or \(4 + a + b = 7\) \(\Rightarrow a + b = 3\) (as required) AG | A1* |
| Answer | Marks |
|---|---|
| Attempting \(f(-2)\) or \(f(2)\) | M1 |
| \(f(-2) = 16 - 8 + 8 - 2a + b = -8\) {\(\Rightarrow -2a + b = -24\)} | A1 |
| Solving both equations simultaneously to get as far as \(a = \ldots\) or \(b = \ldots\) | dM1 |
| Any one of \(a = 9\) or \(b = -6\) | A1 |
| Both \(a = 9\) and \(b = -6\) | A1 |
# Question 1
## 1(a)
$f(x) = x^4 + x^3 + 2x^2 + ax + b$
Attempting $f(1)$ or $f(-1)$ | M1
$f(1) = 1 + 1 + 2 + a + b = 7$ or $4 + a + b = 7$ $\Rightarrow a + b = 3$ (as required) AG | A1*
**Marks: 2**
### Notes for 1(a)
M1: For attempting either $f(1)$ or $f(-1)$.
A1: For applying $f(1)$, setting the result equal to 7, and manipulating this correctly to give the result given on the paper as $a + b = 3$. Note that the answer is given in part (a).
**Alternative**
M1: For long division by $(x - 1)$ to give a remainder in $a$ and $b$ which is independent of $x$.
A1: Or remainder $= b + a + 4 = 7$ leading to the correct result of $a + b = 3$ (answer given).
## 1(b)
Attempting $f(-2)$ or $f(2)$ | M1
$f(-2) = 16 - 8 + 8 - 2a + b = -8$ {$\Rightarrow -2a + b = -24$} | A1
Solving both equations simultaneously to get as far as $a = \ldots$ or $b = \ldots$ | dM1
Any one of $a = 9$ or $b = -6$ | A1
Both $a = 9$ and $b = -6$ | A1
**Marks: 5**
### Notes for 1(b)
M1: Attempting either $f(-2)$ or $f(2)$.
A1: Correct underlined equation in $a$ and $b$; e.g. $16 - 8 + 8 - 2a + b = -8$ or equivalent, e.g. $-2a + b = -24$.
dM1: An attempt to eliminate one variable from 2 linear simultaneous equations in $a$ and $b$. Note that this mark is dependent upon the award of the first method mark.
A1: Any one of $a = 9$ or $b = -6$.
A1: Both $a = 9$ and $b = -6$ and a correct solution only.
**Alternative**
M1: For long division by $(x + 2)$ to give a remainder in $a$ and $b$ which is independent of $x$.
A1: For remainder $= b - 2(a - 8) = -8$ $\Rightarrow -2a + b = -24$.
Then dM1A1A1 are applied in the same way as before.
**Total Marks: 7**1.
$$\mathrm { f } ( x ) = x ^ { 4 } + x ^ { 3 } + 2 x ^ { 2 } + a x + b ,$$
where $a$ and $b$ are constants.\\
When $\mathrm { f } ( x )$ is divided by ( $x - 1$ ), the remainder is 7
\begin{enumerate}[label=(\alph*)]
\item Show that $a + b = 3$
When $\mathrm { f } ( x )$ is divided by ( $x + 2$ ), the remainder is - 8
\item Find the value of $a$ and the value of $b$\\
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VIIIV SIHI NI JIIIM ION OC & VIIV SIHI NI JINAM ION OC & VEYV SIHI NI JULIM ION OO \\
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\hfill \mbox{\textit{Edexcel P2 2018 Q1 [7]}}