| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial identity or expansion |
| Difficulty | Moderate -0.8 Part (a) is a straightforward algebraic identity requiring expansion and coefficient matching to find b=-1, c=1. Part (b) follows immediately from (a) showing K=(n+1)(n²-n+1), proving it's composite. This is a routine application of factorization with minimal problem-solving required, easier than average A-level questions. |
| Spec | 1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(b=-1,\ c=1\) | B1 | or \((n+1)(n^2-n+1)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(n+1\) (or \(n^2-n+1\)) is a factor of \(K\) | B1 | Stated. Allow \(x\) instead of \(n\). NOT \(n^3+1\) can be expressed as \((n+1)(n^2-n+1)\) |
| \(n>2\) so \(n+1>1\) or \(n+1>3\) or \(n+1\neq 1\); (\(n^2-n+1\) is a factor of \(K\) and \(n^2-n+1>1\) or \(\neq 1\)) | B1 | Must see \(n>2\). Allow omission of this step |
| Assume these factors are equal. Let \(n^2-n+1=n+1\Rightarrow n^2-2n=0\) | M1 | |
| \(n=0\) or \(2\) | A1 | |
| \(n>2\) so both invalid; hence 2 distinct factors. Ignore attempted proofs that either factor \(\neq K\) | A1 | Conclusion stated, from correct working seen. Dep at least B1M1 and correct reasoning given. SC: \((n+1)>1\) or \(n+1>3\) (or \(n^2-n+1>1\)) B1; \((n+1)\) & \((n^2-n+1)\) are factors of \(K\) B1 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $b=-1,\ c=1$ | **B1** | or $(n+1)(n^2-n+1)$ |
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## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $n+1$ (or $n^2-n+1$) is a factor of $K$ | **B1** | Stated. Allow $x$ instead of $n$. NOT $n^3+1$ can be expressed as $(n+1)(n^2-n+1)$ |
| $n>2$ so $n+1>1$ or $n+1>3$ or $n+1\neq 1$; ($n^2-n+1$ is a factor of $K$ and $n^2-n+1>1$ or $\neq 1$) | **B1** | Must see $n>2$. Allow omission of this step |
| Assume these factors are equal. Let $n^2-n+1=n+1\Rightarrow n^2-2n=0$ | **M1** | |
| $n=0$ or $2$ | **A1** | |
| $n>2$ so both invalid; hence 2 distinct factors. Ignore attempted proofs that either factor $\neq K$ | **A1** | Conclusion stated, from correct working seen. Dep at least B1M1 and correct reasoning given. SC: $(n+1)>1$ or $n+1>3$ (or $n^2-n+1>1$) B1; $(n+1)$ & $(n^2-n+1)$ are factors of $K$ B1 |
---8 The number $K$ is defined by $K = n ^ { 3 } + 1$, where $n$ is an integer greater than 2 .
\begin{enumerate}[label=(\alph*)]
\item Given that $n ^ { 3 } + 1 \equiv ( n + 1 ) \left( n ^ { 2 } + b n + c \right)$, find the constants $b$ and $c$.
\item Prove that $K$ has at least two distinct factors other than 1 and $K$.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2021 Q8 [6]}}