| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Known polynomial, verify then factorise |
| Difficulty | Moderate -0.8 This is a straightforward application of the factor theorem requiring substitution of x=-4 to verify the factor, followed by polynomial division and factorising a quadratic. All steps are routine C2 techniques with no problem-solving insight needed, making it easier than average but not trivial since it requires accurate algebraic manipulation across multiple steps. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Attempt to find \(f(-4)\) or \(f(4)\) | M1 | |
| \(f(-4) = 2(-4)^3 - 3(-4)^2 - 39(-4) + 20 = -128 - 48 + 156 + 20 = 0\), so \((x+4)\) is a factor | A1 | A1 requires zero and a simple conclusion (even a tick or Q.E.D.), or preamble e.g. "If \(f(-4)=0\), \((x+4)\) is a factor". Long division scores no marks in (a); factor theorem required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2x^3 - 3x^2 - 39x + 20 = (x+4)(2x^2 - 11x + 5)\) | M1 A1 | First M: use of \((x+4)\) to obtain \((2x^2 + ax + b)\), \(a \neq 0\), \(b \neq 0\), even with remainder |
| \((2x-1)(x-5)\) | M1 A1cso | Second M: attempt to factorise three-term quadratic. Usual rule: \((kx^2+ax+b)=(px+c)(qx+d)\) where \( |
| or \(\left(x - \frac{1}{2}\right)(2x-10)\) or equivalent | Losing a factor of 2: \((x+4)\left(x-\frac{1}{2}\right)(x-5)\) scores M1 A1 M1 A0. Answer only one sign wrong e.g. \((x+4)(2x-1)(x+5)\) scores M1 A1 M1 A0 |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Attempt to find $f(-4)$ or $f(4)$ | M1 | |
| $f(-4) = 2(-4)^3 - 3(-4)^2 - 39(-4) + 20 = -128 - 48 + 156 + 20 = 0$, so $(x+4)$ is a factor | A1 | A1 requires zero and a simple conclusion (even a tick or Q.E.D.), or preamble e.g. "If $f(-4)=0$, $(x+4)$ is a factor". Long division scores no marks in (a); factor theorem required |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2x^3 - 3x^2 - 39x + 20 = (x+4)(2x^2 - 11x + 5)$ | M1 A1 | First M: use of $(x+4)$ to obtain $(2x^2 + ax + b)$, $a \neq 0$, $b \neq 0$, even with remainder |
| $(2x-1)(x-5)$ | M1 A1cso | Second M: attempt to factorise three-term quadratic. Usual rule: $(kx^2+ax+b)=(px+c)(qx+d)$ where $|cd|=|b|$ and $|pq|=|k|$. The 3 brackets need not be written together |
| or $\left(x - \frac{1}{2}\right)(2x-10)$ or equivalent | | Losing a factor of 2: $(x+4)\left(x-\frac{1}{2}\right)(x-5)$ scores M1 A1 M1 A0. Answer only one sign wrong e.g. $(x+4)(2x-1)(x+5)$ scores M1 A1 M1 A0 |
---1.
$$f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 39 x + 20$$
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to show that $( x + 4 )$ is a factor of $\mathrm { f } ( x )$.
\item Factorise f ( $x$ ) completely.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2008 Q1 [6]}}