Edexcel FP1 2009 January — Question 1 5 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2009
SessionJanuary
Marks5
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Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeGiven factor, find all roots
DifficultyModerate -0.8 This is a straightforward application of the factor theorem with the factor already given. Students divide by (x-3) using polynomial division or inspection, then solve the resulting quadratic—a routine multi-step procedure requiring no problem-solving insight, making it easier than average.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem
1. $$f ( x ) = 2 x ^ { 3 } - 8 x ^ { 2 } + 7 x - 3$$ Given that \(x = 3\) is a solution of the equation \(\mathrm { f } ( x ) = 0\), solve \(\mathrm { f } ( x ) = 0\) completely.
\(x - 3\) is a factor
AnswerMarks
\(f(x) = (x-3)(2x^2 - 2x + 1)\)B1
Attempt to solve quadratic i.e. \(x = \frac{2 \pm \sqrt{4-8}}{4}\)M1
\(x = \frac{1 \pm i}{2}\)A1
M1
A1
[5]
$x - 3$ is a factor

$f(x) = (x-3)(2x^2 - 2x + 1)$ | B1

Attempt to solve quadratic i.e. $x = \frac{2 \pm \sqrt{4-8}}{4}$ | M1

$x = \frac{1 \pm i}{2}$ | A1

M1

A1

[5]
1.

$$f ( x ) = 2 x ^ { 3 } - 8 x ^ { 2 } + 7 x - 3$$

Given that $x = 3$ is a solution of the equation $\mathrm { f } ( x ) = 0$, solve $\mathrm { f } ( x ) = 0$ completely.\\

\hfill \mbox{\textit{Edexcel FP1 2009 Q1 [5]}}
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Q1 5 Q2 7 Q3 4 Q4 5 Q5 9 Q6 5 Q7 6 Q8 10 Q9 10 Q10 14