OCR MEI Paper 1 2018 June — Question 1 3 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
Year2018
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFactor condition (zero remainder)
DifficultyEasy -1.8 This is a straightforward application of the factor theorem requiring only substitution of x=2 into the polynomial and verification that it equals zero. It's a single-step 'show that' question with no problem-solving or algebraic manipulation needed, making it significantly easier than average A-level questions.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

1 Show that ( \(x - 2\) ) is a factor of \(3 x ^ { 3 } - 8 x ^ { 2 } + 3 x + 2\).

Question 1:
EITHER method:
AnswerMarks Guidance
AnswerMarks Guidance
\(f(2) = 3\times2^3 - 8\times2^2 + 3\times2 + 2 = 24-32+6+2=0\)M1, A1 Function notation need not be used; zero must be seen
Therefore by the factor theorem \((x-2)\) is a factorE1 [3] Reason required
OR method:
AnswerMarks Guidance
AnswerMarks Guidance
\(f(x) = (x-2)(3x^2-2x-1)\)M1, A1 Using algebraic division as far as \(3x^2\); correct quotient
No remainder so \((x-2)\) is a factorE1 [3] Reason required
## Question 1:

**EITHER method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(2) = 3\times2^3 - 8\times2^2 + 3\times2 + 2 = 24-32+6+2=0$ | M1, A1 | Function notation need not be used; zero must be seen |
| Therefore by the factor theorem $(x-2)$ is a factor | E1 [3] | Reason required |

**OR method:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = (x-2)(3x^2-2x-1)$ | M1, A1 | Using algebraic division as far as $3x^2$; correct quotient |
| No remainder so $(x-2)$ is a factor | E1 [3] | Reason required |

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1 Show that ( $x - 2$ ) is a factor of $3 x ^ { 3 } - 8 x ^ { 2 } + 3 x + 2$.

\hfill \mbox{\textit{OCR MEI Paper 1 2018 Q1 [3]}}