| Exam Board | Edexcel |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with parameter in coefficient |
| Difficulty | Standard +0.3 This is a straightforward application of the Factor Theorem requiring substitution, algebraic manipulation, and solving a quadratic. The multi-part structure guides students through each step, and the techniques (factor theorem, factorising quadratics, solving cubics after finding one factor) are all standard A-level pure maths content with no novel insight required. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((f(a)=)\ 4a^3+5a^2-10a+4a=0 \Rightarrow a(...)=0\) | M1 | Attempts \(f(a)=0\) leading to equation in \(a\) only, attempts to factor out \(a\). |
| \(a(4a^2+5a-6)=0\) | A1* | Achieves given answer with no errors including brackets. Minimum acceptable: \(4a^3+5a^2-10a+4a=0 \Rightarrow a(4a^2+5a-6)=0\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = \dfrac{3}{4}\) | B1 | Deduces \(a=\frac{3}{4}\) only. May be implied by resultant cubic. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4x^3+5x^2-10x+4\times\frac{3}{4}=3 \Rightarrow 4x^3+5x^2-10x\ (=0)\) | M1 | Substitutes \(a=\frac{3}{4}\) (must be positive) into \(f(x)\), sets \(f(x)=3\), collects terms. |
| \(x=0\) | B1 | |
| \(x = \dfrac{-5\pm\sqrt{185}}{8}\) | A1 | These values only. Or exact equivalent. Withhold if fraction line not under both terms. |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(f(a)=)\ 4a^3+5a^2-10a+4a=0 \Rightarrow a(...)=0$ | M1 | Attempts $f(a)=0$ leading to equation in $a$ only, attempts to factor out $a$. |
| $a(4a^2+5a-6)=0$ | A1* | Achieves given answer with no errors including brackets. Minimum acceptable: $4a^3+5a^2-10a+4a=0 \Rightarrow a(4a^2+5a-6)=0$. |
## Question 2(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = \dfrac{3}{4}$ | B1 | Deduces $a=\frac{3}{4}$ only. May be implied by resultant cubic. |
## Question 2(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4x^3+5x^2-10x+4\times\frac{3}{4}=3 \Rightarrow 4x^3+5x^2-10x\ (=0)$ | M1 | Substitutes $a=\frac{3}{4}$ (must be positive) into $f(x)$, sets $f(x)=3$, collects terms. |
| $x=0$ | B1 | |
| $x = \dfrac{-5\pm\sqrt{185}}{8}$ | A1 | These values only. Or exact equivalent. Withhold if fraction line not under both terms. |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
$$f ( x ) = 4 x ^ { 3 } + 5 x ^ { 2 } - 10 x + 4 a \quad x \in \mathbb { R }$$
where $a$ is a positive constant.\\
Given ( $x - a$ ) is a factor of $\mathrm { f } ( x )$,\\
(a) show that
$$a \left( 4 a ^ { 2 } + 5 a - 6 \right) = 0$$
(b) Hence\\
(i) find the value of $a$\\
(ii) use algebra to find the exact solutions of the equation
$$f ( x ) = 3$$
\hfill \mbox{\textit{Edexcel Paper 1 2023 Q2 [6]}}