OCR MEI C1 — Question 4 3 marks

Exam BoardOCR MEI
ModuleC1 (Core Mathematics 1)
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFactorise polynomial completely
DifficultyStandard +0.3 This is a straightforward C1 question requiring factorisation by taking out common factors, then recognising the product of three consecutive integers. The divisibility proof follows immediately from the factorisation. Slightly above average difficulty due to the proof element, but the reasoning is direct once factorised.
Spec1.01a Proof: structure of mathematical proof and logical steps1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

Factorise \(n^3 + 3n^2 + 2n\). Hence prove that, when \(n\) is a positive integer, \(n^3 + 3n^2 + 2n\) is always divisible by 6. [3]

Question 4:
AnswerMarks
4n (n + 1)(n + 2)
argument from general consecutive
numbers leading to:
at least one must be even
AnswerMarks
[exactly] one must be multiple of 3M1
A1
AnswerMarks
A1condone division by n and then
(n + 1)(n + 2) seen, or separate factors
shown after factor theorem used;
or divisible by 2;
if M0:
allow SC1 for showing given
AnswerMarks
expression always evenignore ‘ = 0’;
an induction approach using the factors may also be
used eg by those doing paper FP1 as well;
A0 for just substituting numbers for n and stating
results;
allow SC2 for a correct induction approach using the
original cubic (SC1 for each of showing even and
showing divisible by 3)
Question 4:
4 | n (n + 1)(n + 2)
argument from general consecutive
numbers leading to:
at least one must be even
[exactly] one must be multiple of 3 | M1
A1
A1 | condone division by n and then
(n + 1)(n + 2) seen, or separate factors
shown after factor theorem used;
or divisible by 2;
if M0:
allow SC1 for showing given
expression always even | ignore ‘ = 0’;
an induction approach using the factors may also be
used eg by those doing paper FP1 as well;
A0 for just substituting numbers for n and stating
results;
allow SC2 for a correct induction approach using the
original cubic (SC1 for each of showing even and
showing divisible by 3)
Factorise $n^3 + 3n^2 + 2n$. Hence prove that, when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by 6. [3]

\hfill \mbox{\textit{OCR MEI C1  Q4 [3]}}