| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial with equal remainders |
| Difficulty | Standard +0.3 This is a structured multi-part question testing standard Remainder Theorem applications and basic polynomial analysis. Parts (a-b) involve routine substitution to set up simultaneous equations, (c) is direct verification, and (d) requires factorization and recognizing a quadratic has no real roots—all standard C2 techniques with clear scaffolding and no novel insight required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(f(-1) = r \therefore -1 + k + 7 - 15 = r\) | M1 | |
| \(k = r + 9\) | A1 | |
| \(f(3) = 3r \therefore 27 + 9k - 21 - 15 = 3r\) | M1 | |
| \(3k = r + 3\) | ||
| subtracting, \(2k = -6\) | M1 | |
| \(k = -3\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(r = -3 - 9 = -12\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(f(x) = x^3 - 3x^2 - 7x - 15\) | ||
| \(f(5) = 125 - 75 - 35 - 15 = 0 \therefore (x-5)\) is a factor | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Long division giving \(x^2 + 2x + 3\) | M1 A1 | |
| \(\therefore (x-5)(x^2 + 2x + 3) = 0\) | ||
| \(x = 5\) or \(x^2 + 2x + 3 = 0\) | ||
| \(b^2 - 4ac = 2^2 - (4 \times 1 \times 3) = -8\) | M1 | |
| \(b^2 - 4ac < 0 \therefore\) no real solutions to quadratic | ||
| \(\therefore\) only one real solution | A1 | (12) |
## Question 9:
**(a)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $f(-1) = r \therefore -1 + k + 7 - 15 = r$ | M1 | |
| $k = r + 9$ | A1 | |
| $f(3) = 3r \therefore 27 + 9k - 21 - 15 = 3r$ | M1 | |
| $3k = r + 3$ | | |
| subtracting, $2k = -6$ | M1 | |
| $k = -3$ | A1 | |
**(b)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $r = -3 - 9 = -12$ | B1 | |
**(c)**
| Answer/Working | Marks | Notes |
|---|---|---|
| $f(x) = x^3 - 3x^2 - 7x - 15$ | | |
| $f(5) = 125 - 75 - 35 - 15 = 0 \therefore (x-5)$ is a factor | M1 A1 | |
**(d)**
| Answer/Working | Marks | Notes |
|---|---|---|
| Long division giving $x^2 + 2x + 3$ | M1 A1 | |
| $\therefore (x-5)(x^2 + 2x + 3) = 0$ | | |
| $x = 5$ or $x^2 + 2x + 3 = 0$ | | |
| $b^2 - 4ac = 2^2 - (4 \times 1 \times 3) = -8$ | M1 | |
| $b^2 - 4ac < 0 \therefore$ no real solutions to quadratic | | |
| $\therefore$ only one real solution | A1 | **(12)** |
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**Total: 75 marks**9. The polynomial $\mathrm { f } ( x )$ is given by
$$f ( x ) = x ^ { 3 } + k x ^ { 2 } - 7 x - 15$$
where $k$ is a constant.\\
When $\mathrm { f } ( x )$ is divided by ( $x + 1$ ) the remainder is $r$.\\
When $\mathrm { f } ( x )$ is divided by $( x - 3 )$ the remainder is $3 r$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the value of $r$.
\item Show that $( x - 5 )$ is a factor of $\mathrm { f } ( x )$.
\item Show that there is only one real solution to the equation $\mathrm { f } ( x ) = 0$.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q9 [12]}}