| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Single polynomial, two remainder/factor conditions |
| Difficulty | Moderate -0.3 This is a standard C2 Factor/Remainder Theorem question requiring systematic application of the theorem twice to form simultaneous equations, then factorisation. While it involves multiple steps (parts a-c), each step follows routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(p(-2) = 20 \quad \therefore -16 + 4 - 2a + b = 20\) | M1 | |
| \(b = 2a + 32\) | A1 | |
| (b) \(p(-3) = 0 \quad \therefore -54 + 9 - 3a + b = 0\) | M1 | |
| sub. \(-45 - 3a + (2a + 32) = 0\) | M1 | |
| \(a = -13, b = 6\) | A2 | |
| (c) Long division of \(2x^2 - 5x + 2\) by \(x + 3\) in \(2x^3 + x^2 - 13x + 6\) | M1 A1 | |
| \(p(x) = (x + 3)(2x^2 - 5x + 2)\) | ||
| \(p(x) = (x + 3)(2x - 1)(x - 2)\) | M1 A1 | (10) |
**(a)** $p(-2) = 20 \quad \therefore -16 + 4 - 2a + b = 20$ | M1 |
$b = 2a + 32$ | A1 |
**(b)** $p(-3) = 0 \quad \therefore -54 + 9 - 3a + b = 0$ | M1 |
sub. $-45 - 3a + (2a + 32) = 0$ | M1 |
$a = -13, b = 6$ | A2 |
**(c)** Long division of $2x^2 - 5x + 2$ by $x + 3$ in $2x^3 + x^2 - 13x + 6$ | M1 A1 |
$p(x) = (x + 3)(2x^2 - 5x + 2)$ | |
$p(x) = (x + 3)(2x - 1)(x - 2)$ | M1 A1 | **(10)**6. The polynomial $\mathrm { p } ( x )$ is defined by
$$\mathrm { p } ( x ) = 2 x ^ { 3 } + x ^ { 2 } + a x + b ,$$
where $a$ and $b$ are constants.\\
Given that when $\mathrm { p } ( x )$ is divided by $( x + 2 )$ there is a remainder of 20 ,
\begin{enumerate}[label=(\alph*)]
\item find an expression for $b$ in terms of $a$.
Given also that $( x + 3 )$ is a factor of $\mathrm { p } ( x )$,
\item find the values of $a$ and $b$,
\item fully factorise $\mathrm { p } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}