| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Standard +0.3 This is a standard AS-level Factor Theorem question with routine algebraic steps. Part (a) requires simple substitution, (b) involves factorizing a quadratic and checking the discriminant, (c) uses graph transformation reasoning, and (d) applies horizontal translation. All parts follow predictable patterns with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(f(4)=2\times4^3-13\times4^2+8\times4+48\) | M1 | Do not accept \(f(4)=0\) without sight of embedded values; alternatively attempts division by \((x-4)\) |
| \(f(4)=0 \Rightarrow (x-4)\) is a factor | A1 | Correct reason with conclusion; must state \(f(4)=0\) hence factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x^3-13x^2+8x+48=(x-4)(2x^2\ldots x-12)\) | M1 | Attempts quadratic factor by inspection (correct first and last terms) or division |
| \(=(x-4)(2x^2-5x-12)\) | A1 | |
| Attempts to factorise quadratic factor or solve quadratic equation | dM1 | Dependent on previous M1 |
| \(f(x)=(x-4)^2(2x+3) \Rightarrow f(x)=0\) has only two roots, 4 and \(-1.5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Deduces either three roots or deduces \(f(x)\) is moved down two units | M1 | |
| States three roots, as when \(f(x)\) is moved down two units there will be three points of intersection with the \(x\)-axis | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sight of \(k=\pm4, \pm\frac{3}{2}\) | M1 | |
| \(k=4, -\frac{3}{2}\) | A1ft |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct quadratic factor \(\left(2x^2 - 5x - 12\right)\) visible in correct place | A1 | For division award for sight of this "in the correct place". Don't need to see it paired with \((x-4)\) |
| Correct attempt to solve or factorise \(\left(2x^2 - 5x - 12\right)\) including use of formula. Apply usual rules \((2x^2-5x-12)=(ax+b)(cx+d)\) where \(ac=\pm2\) and \(bd=\pm12\) | dM1 | Allow candidate to move from \((x-4)(2x^2-5x-12)\) to \((x-4)^2(2x+3)\) |
| \(f(x)=(x-4)(2x+3)(x-4)\) or \(f(x)=(x-4)^2(2x+3)\) or \(f(x)=2(x-4)^2\left(x+\frac{3}{2}\right)\) followed by valid explanation of only two roots | A1 | Via factorisation. Roots must be correct: \(x=4\) and \(-\frac{3}{2}\) only. Must show understanding between roots and factors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| States there are 3 roots or states it is a solution of \(f(x)=2\) or \(f(x)-2=0\) | M1 | Valid deduction |
| States three roots, as \(f(x)\) is moved down by two units giving three points of intersection with \(x\)-axis. Or: states three roots as it is where \(f(x)=2\) | A1 | Fully explains |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Sight of \(\pm4\) and \(\pm\frac{3}{2}\) | M1 | Follow through on \(\pm\) their roots |
| \(k=4, -\frac{3}{2}\) | A1ft | Follow through on their roots. Accept \(4, -\frac{3}{2}\) but not \(x=4, -\frac{3}{2}\) |
## Question 11(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $f(4)=2\times4^3-13\times4^2+8\times4+48$ | M1 | Do not accept $f(4)=0$ without sight of embedded values; alternatively attempts division by $(x-4)$ |
| $f(4)=0 \Rightarrow (x-4)$ is a factor | A1 | Correct reason with conclusion; must state $f(4)=0$ hence factor |
## Question 11(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^3-13x^2+8x+48=(x-4)(2x^2\ldots x-12)$ | M1 | Attempts quadratic factor by inspection (correct first and last terms) or division |
| $=(x-4)(2x^2-5x-12)$ | A1 | |
| Attempts to factorise quadratic factor or solve quadratic equation | dM1 | Dependent on previous M1 |
| $f(x)=(x-4)^2(2x+3) \Rightarrow f(x)=0$ has only two roots, 4 and $-1.5$ | A1 | |
## Question 11(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Deduces either three roots or deduces $f(x)$ is moved down two units | M1 | |
| States three roots, as when $f(x)$ is moved down two units there will be three points of intersection with the $x$-axis | A1 | |
## Question 11(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sight of $k=\pm4, \pm\frac{3}{2}$ | M1 | |
| $k=4, -\frac{3}{2}$ | A1ft | |
# Question 11 (continued):
## Part (b) continued:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct quadratic factor $\left(2x^2 - 5x - 12\right)$ visible in correct place | A1 | For division award for sight of this "in the correct place". Don't need to see it paired with $(x-4)$ |
| Correct attempt to solve or factorise $\left(2x^2 - 5x - 12\right)$ including use of formula. Apply usual rules $(2x^2-5x-12)=(ax+b)(cx+d)$ where $ac=\pm2$ and $bd=\pm12$ | dM1 | Allow candidate to move from $(x-4)(2x^2-5x-12)$ to $(x-4)^2(2x+3)$ |
| $f(x)=(x-4)(2x+3)(x-4)$ or $f(x)=(x-4)^2(2x+3)$ or $f(x)=2(x-4)^2\left(x+\frac{3}{2}\right)$ followed by valid explanation of only two roots | A1 | Via factorisation. Roots must be correct: $x=4$ and $-\frac{3}{2}$ only. Must show understanding between roots and factors |
**Via solving:** Factorises to $(x-4)(2x^2-5x-12)$ and solves $2x^2-5x-12=0 \Rightarrow x=4, -\frac{3}{2}$, followed by explanation that roots are $4, 4, -\frac{3}{2}$ so only two distinct roots.
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| States there are 3 roots **or** states it is a solution of $f(x)=2$ or $f(x)-2=0$ | M1 | Valid deduction |
| States three roots, as $f(x)$ is moved down by **two** units giving three points of intersection with $x$-axis. Or: states three roots as it is where $f(x)=2$ | A1 | Fully explains |
## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Sight of $\pm4$ **and** $\pm\frac{3}{2}$ | M1 | Follow through on $\pm$ their roots |
| $k=4, -\frac{3}{2}$ | A1ft | Follow through on their roots. Accept $4, -\frac{3}{2}$ but not $x=4, -\frac{3}{2}$ |
---11.
$$f ( x ) = 2 x ^ { 3 } - 13 x ^ { 2 } + 8 x + 48$$
\begin{enumerate}[label=(\alph*)]
\item Prove that $( x - 4 )$ is a factor of $\mathrm { f } ( x )$.
\item Hence, using algebra, show that the equation $\mathrm { f } ( x ) = 0$ has only two distinct roots.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{deba6a2b-1821-4110-bde8-bde18a5f9be9-24_727_1059_566_504}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve with equation $y = \mathrm { f } ( x )$.
\item Deduce, giving reasons for your answer, the number of real roots of the equation
$$2 x ^ { 3 } - 13 x ^ { 2 } + 8 x + 46 = 0$$
Given that $k$ is a constant and the curve with equation $y = \mathrm { f } ( x + k )$ passes through the origin, (d) find the two possible values of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2019 Q11 [10]}}