| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 5 |
| 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 C1 question testing basic Factor Theorem application and polynomial factorisation. Part (a) requires only substituting x=-3 to verify the factor (routine recall), and part (b) involves standard polynomial division followed by factorising a quadratic—both are textbook exercises with no problem-solving insight required. Easier than average A-level questions. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\frac{3}{2}\) | B1 | Seeing \(-\frac{3}{2}\) OE |
| \(p(-1.5) = 2(-1.5)^4 + 3(-1.5)^3 - 8(-1.5)^2 - 14(-1.5) - 3\) | M1 | Attempting to evaluate \(p(-1.5)\) or \(p(1.5)\) |
| \(p(-1.5) = 10.125 - 10.125 - 18 + 21 - 3 = 0\), so \((2x+3)\) is a factor of \(p(x)\) | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^3 - 4x - 1 = 0 \Rightarrow x(x^2 - 4) - 1 = 0 \Rightarrow x^2 - 4 = \frac{1}{x}\) | M1 | Dividing throughout by \(x\) OE |
| \(x^2 = \frac{1}{x} + 4 \Rightarrow x = \sqrt{\frac{1}{x} + 4}\) (since \(x > 0\)) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_2 = 2.1213\) | B1 | AWRT 2.121 |
| \(x_3 = 2.1146\) | B1 | AWRT 2.1146 |
| \(x_4 = 2.1149\) | B1 | 3 |
# Question 1(a) [XMCA2]:
$x = -\frac{3}{2}$ | B1 | Seeing $-\frac{3}{2}$ OE
$p(-1.5) = 2(-1.5)^4 + 3(-1.5)^3 - 8(-1.5)^2 - 14(-1.5) - 3$ | M1 | Attempting to evaluate $p(-1.5)$ or $p(1.5)$
$p(-1.5) = 10.125 - 10.125 - 18 + 21 - 3 = 0$, so $(2x+3)$ is a factor of $p(x)$ | A1 | **3** | CSO — Need both the arithmetic to show $= 0$ and the conclusion
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# Question 1(b)(i) [XMCA2]:
$x^3 - 4x - 1 = 0 \Rightarrow x(x^2 - 4) - 1 = 0 \Rightarrow x^2 - 4 = \frac{1}{x}$ | M1 | Dividing throughout by $x$ OE
$x^2 = \frac{1}{x} + 4 \Rightarrow x = \sqrt{\frac{1}{x} + 4}$ (since $x > 0$) | A1 | **2** | CSO
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# Question 1(b)(ii) [XMCA2]:
$x_2 = 2.1213$ | B1 | AWRT 2.121
$x_3 = 2.1146$ | B1 | AWRT 2.1146
$x_4 = 2.1149$ | B1 | **3** | CAO
---1 The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - 13 x - 12$.
\begin{enumerate}[label=(\alph*)]
\item Use the Factor Theorem to show that $x + 3$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x )$ as the product of three linear factors.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q1 [5]}}