Edexcel C2 Specimen — Question 7 12 marks

Exam BoardEdexcel
ModuleC2 (Core Mathematics 2)
SessionSpecimen
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeShow equation reduces to polynomial
DifficultyStandard +0.3 This is a structured multi-part question that guides students through standard C2 techniques: applying factor theorem (routine), factorizing a cubic and solving (standard), and manipulating logarithms to reach a given polynomial (methodical but not demanding). Part (c) requires careful log manipulation but the target equation is given, and part (d) is trivial once (b) and (c) are complete. Slightly above average due to the multi-step nature and logarithm work, but all techniques are standard C2 fare with significant scaffolding.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b

7. (a) Use the factor theorem to show that \(( x + 1 )\) is a factor of \(x ^ { 3 } - x ^ { 2 } - 10 x - 8\).
(b) Find all the solutions of the equation \(x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0\).
(c) Prove that the value of \(x\) that satisfies $$2 \log _ { 2 } x + \log _ { 2 } ( x - 1 ) = 1 + \log _ { 2 } ( 5 x + 4 )$$ is a solution of the equation $$x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0$$ (d) State, with a reason, the value of \(x\) that satisfies equation (I).

Question 7:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(f(-1) = -1 - 1 + 10 - 8\)M1 \(f(+1)\) or \(f(-1)\)
\(= 0\) so \((x+1)\) is a factorA1 \(= 0\) and comment
(2)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x^3 - x^2 = 2(5x+4)\)M1 Out of logs
i.e. \(x^3 - x^2 - 10x - 8 = 0\)M1 A1 cso
\(x = -1,\ -2,\ 4\)A2(1,0)
(4)
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\log_2 x^2 + \log_2(x-1) = 1 + \log_2(5x+4)\)M1 Use of \(\log x^n\)
\(\log_2\left(\frac{x^2(x-1)}{5x+4}\right) = 1\)M1 Use of \(\log a + \log b\)
Part (d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x = 4\), since \(x < 0\) is not valid in logsB1, B1
(2)
# Question 7:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(-1) = -1 - 1 + 10 - 8$ | M1 | $f(+1)$ or $f(-1)$ |
| $= 0$ so $(x+1)$ is a factor | A1 | $= 0$ and comment |
| | **(2)** | |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x^3 - x^2 = 2(5x+4)$ | M1 | Out of logs |
| i.e. $x^3 - x^2 - 10x - 8 = 0$ | M1 | A1 cso |
| $x = -1,\ -2,\ 4$ | A2(1,0) | |
| | **(4)** | |

## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_2 x^2 + \log_2(x-1) = 1 + \log_2(5x+4)$ | M1 | Use of $\log x^n$ |
| $\log_2\left(\frac{x^2(x-1)}{5x+4}\right) = 1$ | M1 | Use of $\log a + \log b$ |

## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 4$, since $x < 0$ is not valid in logs | B1, B1 | |
| | **(2)** | |

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7. (a) Use the factor theorem to show that $( x + 1 )$ is a factor of $x ^ { 3 } - x ^ { 2 } - 10 x - 8$.\\
(b) Find all the solutions of the equation $x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0$.\\
(c) Prove that the value of $x$ that satisfies

$$2 \log _ { 2 } x + \log _ { 2 } ( x - 1 ) = 1 + \log _ { 2 } ( 5 x + 4 )$$

is a solution of the equation

$$x ^ { 3 } - x ^ { 2 } - 10 x - 8 = 0$$

(d) State, with a reason, the value of $x$ that satisfies equation (I).\\

\hfill \mbox{\textit{Edexcel C2  Q7 [12]}}