| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Solve p(exponential) = 0 |
| Difficulty | Standard +0.3 This is a structured multi-part question that guides students through standard Factor/Remainder Theorem techniques (parts a-c are routine C2 content), culminating in an exponential substitution in part (d). While part (d) requires recognizing that y = 3^t transforms the equation into f(y) = 0, this is a fairly standard 'hence' application once the factorization is complete. The question is slightly above average difficulty due to the exponential twist, but remains accessible to competent C2 students. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(3) = 2(3)^3 - 5(3)^2 - 23(3) - 10\) or long division by \((x-3)\) | M1 | Attempts to calculate \(f(\pm 3)\) or divides by \((x-3)\); for long division need minimum as shown with constant remainder |
| Remainder \(= -70\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f(-2) = 2(-2)^3 - 5(-2)^2 - 23(-2) - 10\) or long division by \((x+2)\) | M1 | Attempts \(f(\pm 2)\) or divides by \((x+2)\); for long division need minimum as shown with constant remainder |
| Remainder \(= 0\), hence \(x+2\) is a factor | A1* | Must obtain zero remainder and make a conclusion (not just a tick or QED); remainder need not be referenced in conclusion but zero remainder must have been obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{2x^3-5x^2-23x-10}{(x+2)} = ax^2+bx+c\) | M1 | Divides \(f(x)\) by \((x+2)\) or compares coefficients or uses inspection to obtain quadratic with \(2x^2\) as first term |
| \(2x^2 - 9x - 5\) | A1 | Correct quadratic seen |
| \(f(x) = (x+2)(2x+1)(x-5)\) | dM1A1 | dM1: attempt to factorise \(3TQ\) \((2x^2...)\); A1: all three factors together on one line, must appear here and not in (d) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3^t = 5 \Rightarrow t\log 3 = \log 5\) | M1 | Solves \(3^t = k\) where \(k > 0\) following from (c); accept sight of \(t = \log_3 k\) where \(k>0\) |
| \(t = \text{awrt } 1.465\) only | A1 | \(t = \text{awrt } 1.465\) and no other solutions |
## Question 8:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(3) = 2(3)^3 - 5(3)^2 - 23(3) - 10$ or long division by $(x-3)$ | M1 | Attempts to calculate $f(\pm 3)$ or divides by $(x-3)$; for long division need minimum as shown with constant remainder |
| Remainder $= -70$ | A1 | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(-2) = 2(-2)^3 - 5(-2)^2 - 23(-2) - 10$ or long division by $(x+2)$ | M1 | Attempts $f(\pm 2)$ or divides by $(x+2)$; for long division need minimum as shown with constant remainder |
| Remainder $= 0$, hence $x+2$ is a factor | A1* | Must obtain zero remainder and make a conclusion (not just a tick or QED); remainder need not be referenced in conclusion but zero remainder must have been obtained |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{2x^3-5x^2-23x-10}{(x+2)} = ax^2+bx+c$ | M1 | Divides $f(x)$ by $(x+2)$ or compares coefficients or uses inspection to obtain quadratic with $2x^2$ as first term |
| $2x^2 - 9x - 5$ | A1 | Correct quadratic seen |
| $f(x) = (x+2)(2x+1)(x-5)$ | dM1A1 | dM1: attempt to factorise $3TQ$ $(2x^2...)$; A1: all three factors together on one line, must appear here and not in (d) |
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3^t = 5 \Rightarrow t\log 3 = \log 5$ | M1 | Solves $3^t = k$ where $k > 0$ following from (c); accept sight of $t = \log_3 k$ where $k>0$ |
| $t = \text{awrt } 1.465$ only | A1 | $t = \text{awrt } 1.465$ and no other solutions |
---8.
$$f ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } - 23 x - 10$$
\begin{enumerate}[label=(\alph*)]
\item Find the remainder when $\mathrm { f } ( x )$ is divided by ( $x - 3$ ).
\item Show that ( $x + 2$ ) is a factor of $\mathrm { f } ( x )$.
\item Hence fully factorise $\mathrm { f } ( x )$.
\item Hence solve
$$2 \left( 3 ^ { 3 t } \right) - 5 \left( 3 ^ { 2 t } \right) - 23 \left( 3 ^ { t } \right) = 10$$
giving your answer to 3 decimal places.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q8 [10]}}