| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Show equation reduces to polynomial |
| Difficulty | Standard +0.3 This is a straightforward multi-part question combining factor theorem (routine substitution and factorization) with logarithm manipulation (standard laws) and domain analysis. Part (a) requires simple substitution and factoring; part (b) uses basic log laws to reach a polynomial; part (c) connects back via domain restrictions. All techniques are standard A-level fare with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(-6)=(-6)^3+a(-6)^2-a(-6)+48\) | M1 | Substitutes \(x=-6\) |
| \(=-216+36a+6a+48=0 \Rightarrow 42a=168 \Rightarrow a=4\)* | A1* | Correct value (starred — part of proof) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(x)=(x+6)(x^2-2x+8)\) | M1, A1 | Correct factorisation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\log_2(x+2)^2+\log_2 x - \log_2(x-6)=3\) or \(2\log_2(x+2)+\log_2\left(\frac{x}{x-6}\right)=3\) | M1 | Correct use of log laws to combine terms |
| \(\log_2\left(\frac{x(x+2)^2}{x-6}\right)=3\) \(\left[\text{or } \log_2(x(x+2)^2)=\log_2(8(x-6))\right]\) | M1 | Single log form |
| \(\frac{x(x+2)^2}{x-6}=2^3\) \(\left\{\text{i.e. } \log_2 a=3 \Rightarrow a=2^3 \text{ or } 8\right\}\) | B1 | Correct removal of log |
| \(x^3+4x^3-4x+48=0\)* | A1* | Correct cubic (starred) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((x+6)(x^2-2x+8)=0\) — Reason 1: \(\log_2 x\) is not defined when \(x=-6\); or \(\log_2(x-6)\) not defined when \(x=-6\); or \(x=-6\) but \(\log_2 x\) only defined for \(x>0\) | B1 | At least one valid reason |
| Reason 2: \(b^2-4ac=-28<0\), so \((x^2-2x+8)=0\) has no real roots | B1 | Both reasons required for second mark |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Applies \(f(-6)\) | M1 | |
| Applies \(f(-6) = 0\) to show that \(a = 4\) | A1* | Fully justified |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces \((x+6)\) is a factor and finds quadratic factor by equating coefficients or long division | M1 | |
| \((x+6)(x^2 - 2x + 8)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Evidence of applying a correct law of logarithms | M1 | |
| Uses correct laws to give \(\log_2(h(x)) = k\) or \(\log_2(g(x)) = \log_2(h(x))\) | M1 | |
| \(\log_2 a = 3 \Rightarrow a = 2^3\) or \(8\) | B1 | |
| Correctly proves \(x^3 + 4x^3 - 4x + 48 = 0\) with no errors | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| See scheme | B1 | |
| See scheme | B1 |
## Question 5:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(-6)=(-6)^3+a(-6)^2-a(-6)+48$ | M1 | Substitutes $x=-6$ |
| $=-216+36a+6a+48=0 \Rightarrow 42a=168 \Rightarrow a=4$* | A1* | Correct value (starred — part of proof) |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x)=(x+6)(x^2-2x+8)$ | M1, A1 | Correct factorisation |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\log_2(x+2)^2+\log_2 x - \log_2(x-6)=3$ or $2\log_2(x+2)+\log_2\left(\frac{x}{x-6}\right)=3$ | M1 | Correct use of log laws to combine terms |
| $\log_2\left(\frac{x(x+2)^2}{x-6}\right)=3$ $\left[\text{or } \log_2(x(x+2)^2)=\log_2(8(x-6))\right]$ | M1 | Single log form |
| $\frac{x(x+2)^2}{x-6}=2^3$ $\left\{\text{i.e. } \log_2 a=3 \Rightarrow a=2^3 \text{ or } 8\right\}$ | B1 | Correct removal of log |
| $x^3+4x^3-4x+48=0$* | A1* | Correct cubic (starred) |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(x+6)(x^2-2x+8)=0$ — Reason 1: $\log_2 x$ is not defined when $x=-6$; or $\log_2(x-6)$ not defined when $x=-6$; or $x=-6$ but $\log_2 x$ only defined for $x>0$ | B1 | At least one valid reason |
| Reason 2: $b^2-4ac=-28<0$, so $(x^2-2x+8)=0$ has no real roots | B1 | Both reasons required for second mark |
## Question 5:
### Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Applies $f(-6)$ | M1 | |
| Applies $f(-6) = 0$ to show that $a = 4$ | A1* | Fully justified |
### Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces $(x+6)$ is a factor and finds quadratic factor by equating coefficients or long division | M1 | |
| $(x+6)(x^2 - 2x + 8)$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Evidence of applying a correct law of logarithms | M1 | |
| Uses correct laws to give $\log_2(h(x)) = k$ or $\log_2(g(x)) = \log_2(h(x))$ | M1 | |
| $\log_2 a = 3 \Rightarrow a = 2^3$ or $8$ | B1 | |
| Correctly proves $x^3 + 4x^3 - 4x + 48 = 0$ with no errors | A1* | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| See scheme | B1 | |
| See scheme | B1 | |
---\begin{enumerate}
\item $\mathrm { f } ( x ) = x ^ { 3 } + a x ^ { 2 } - a x + 48$, where $a$ is a constant
\end{enumerate}
Given that $\mathrm { f } ( - 6 ) = 0$\\
(a) (i) show that $a = 4$\\
(ii) express $\mathrm { f } ( x )$ as a product of two algebraic factors.
Given that $2 \log _ { 2 } ( x + 2 ) + \log _ { 2 } x - \log _ { 2 } ( x - 6 ) = 3$\\
(b) show that $x ^ { 3 } + 4 x ^ { 2 } - 4 x + 48 = 0$\\
(c) hence explain why
$$2 \log _ { 2 } ( x + 2 ) + \log _ { 2 } x - \log _ { 2 } ( x - 6 ) = 3$$
has no real roots.
\hfill \mbox{\textit{Edexcel Paper 1 Q5 [10]}}