Edexcel FP1 2015 June — Question 1 5 marks

Exam BoardEdexcel
ModuleFP1 (Further Pure Mathematics 1)
Year2015
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeGiven factor, find all roots
DifficultyModerate -0.5 This is a straightforward application of the factor theorem requiring polynomial division (or inspection) to find the quadratic factor, then solving the resulting quadratic. While it's Further Maths content, it's a routine algorithmic procedure with no conceptual challenges—slightly easier than average since the method is explicitly guided and the arithmetic is manageable.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

1. $$f ( x ) = 9 x ^ { 3 } - 33 x ^ { 2 } - 55 x - 25$$ Given that \(x = 5\) is a solution of the equation \(\mathrm { f } ( x ) = 0\), use an algebraic method to solve \(\mathrm { f } ( x ) = 0\) completely.
(5)

Question 1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((x-5)\) is a factor, so \(f(x) = (x-5)(9x^2\ldots)\)M1 Uses \((x-5)\) as factor and begins division to obtain quadratic with \(9x^2\); award if no working but quadratic factor completely correct
\(f(x) = (x-5)(9x^2+12x+5)\)A1 \(9x^2+12x+5\) completely correct
Solve \(9x^2+12x+5=0\) to give \(x=\)M1 Solves their quadratic by usual rules; award if one complex root correct with no working; award for \((9x^2+\ldots\) incorrectly factorised to \((3x+p)(3x+q)\) where \(
\(x = -\frac{2}{3}-\frac{1}{3}i,\ -\frac{2}{3}+\frac{1}{3}i\) or \(-\frac{2}{3}\pm\frac{1}{3}i\) or \(\frac{-2\pm i}{3}\) (and 5)A1cao A1ft (5) A1: One correct complex root; accept exact equivalent form, single fraction and \(\pm\). A1ft: Conjugate of their first complex root
# Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(x-5)$ is a factor, so $f(x) = (x-5)(9x^2\ldots)$ | M1 | Uses $(x-5)$ as factor and begins division to obtain quadratic with $9x^2$; award if no working but quadratic factor completely correct |
| $f(x) = (x-5)(9x^2+12x+5)$ | A1 | $9x^2+12x+5$ completely correct |
| Solve $9x^2+12x+5=0$ to give $x=$ | M1 | Solves their quadratic by usual rules; award if one complex root correct with no working; award for $(9x^2+\ldots$ incorrectly factorised to $(3x+p)(3x+q)$ where $|pq|=5$ |
| $x = -\frac{2}{3}-\frac{1}{3}i,\ -\frac{2}{3}+\frac{1}{3}i$ or $-\frac{2}{3}\pm\frac{1}{3}i$ or $\frac{-2\pm i}{3}$ (and 5) | A1cao A1ft (5) | A1: One correct complex root; accept exact equivalent form, single fraction and $\pm$. A1ft: Conjugate of their first complex root |

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1.

$$f ( x ) = 9 x ^ { 3 } - 33 x ^ { 2 } - 55 x - 25$$

Given that $x = 5$ is a solution of the equation $\mathrm { f } ( x ) = 0$, use an algebraic method to solve $\mathrm { f } ( x ) = 0$ completely.\\
(5)\\

\hfill \mbox{\textit{Edexcel FP1 2015 Q1 [5]}}