| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2019 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Find constants using remainder theorem |
| Difficulty | Moderate -0.8 This is a straightforward application of the remainder theorem requiring routine substitution and algebraic manipulation. Part (a) is immediate from the form, part (b) involves one substitution and simple algebra, and part (c) is standard factorisation once the constant is known. All steps are procedural with no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(-35\) | B1 | \(-35\) stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f\left(\pm\frac{2}{3}\right)=0 \Rightarrow \left(\frac{2}{3}-3\right)\left[3\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)+a\right]-35=0\) | M1 | Attempts to set \(f\left(\pm\frac{2}{3}\right)=0\); condone bracket errors; "=0" may be implied |
| \(-\frac{7}{3}(2+a)=35 \Rightarrow a=-17\) | A1* | Achieves given answer with no incorrect work and at least one correct intermediate line; "=0" must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x)=(x-3)(3x^2+x-17)-35=3x^3-8x^2-20x+16\) | B1 | Multiplies out to achieve \(f(x)=3x^3-8x^2-20x+16\); may be implied by \((3x-2)(x^2-2x-8)\) |
| \(=(3x-2)(x^2-2x-8)\) | M1, A1 | M1 attempts to divide/factorise out \((3x-2)\); A1 correct quadratic factor |
| \(=(3x-2)(x+2)(x-4)\) | M1, A1 | M1 factorises quadratic of form \(Ax^2+Bx+C\) where \((Ax^2+Bx+C)=(dx+e)(fx+g)\), \( |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $-35$ | B1 | $-35$ stated |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f\left(\pm\frac{2}{3}\right)=0 \Rightarrow \left(\frac{2}{3}-3\right)\left[3\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)+a\right]-35=0$ | M1 | Attempts to set $f\left(\pm\frac{2}{3}\right)=0$; condone bracket errors; "=0" may be implied |
| $-\frac{7}{3}(2+a)=35 \Rightarrow a=-17$ | A1* | Achieves given answer with no incorrect work and at least one correct intermediate line; "=0" must be seen |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x)=(x-3)(3x^2+x-17)-35=3x^3-8x^2-20x+16$ | B1 | Multiplies out to achieve $f(x)=3x^3-8x^2-20x+16$; may be implied by $(3x-2)(x^2-2x-8)$ |
| $=(3x-2)(x^2-2x-8)$ | M1, A1 | M1 attempts to divide/factorise out $(3x-2)$; A1 correct quadratic factor |
| $=(3x-2)(x+2)(x-4)$ | M1, A1 | M1 factorises quadratic of form $Ax^2+Bx+C$ where $(Ax^2+Bx+C)=(dx+e)(fx+g)$, $|A|=|d\times f|$ or $|C|=|e\times g|$; A1 fully correct on one line following B1M1A1M1 |
---4. $\mathrm { f } ( x ) = ( x - 3 ) \left( 3 x ^ { 2 } + x + a \right) - 35$ where $a$ is a constant
\begin{enumerate}[label=(\alph*)]
\item State the remainder when $\mathrm { f } ( x )$ is divided by $( x - 3 )$.
Given $( 3 x - 2 )$ is a factor of $\mathrm { f } ( x )$,
\item show that $a = - 17$
\item Using algebra and showing each step of your working, fully factorise $\mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2019 Q4 [8]}}