| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Integration or area using factorised polynomial |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing standard applications of the Factor/Remainder Theorem, polynomial division, and basic integration. All parts follow routine procedures: direct substitution for remainder, verification of a given factor, algebraic expansion to find coefficients, discriminant check, and a standard definite integral with area calculation. No novel problem-solving or insight required—purely procedural execution of well-practiced techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p(3) = 27 - 3 + 6\) | M1 | \(p(3)\) attempted |
| (Remainder) \(= 30\) | A1 | |
| Or long division up to remainder; Quotient \(= x^2 + 3x + 8\) and remainder \(= 30\) clearly stated | (M1)(A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(p(-2) = -8 + 2 + 6\) | M1 | \(p(-2)\) attempted: NOT long division |
| \(p(-2) = 0 \Rightarrow x + 2\) is factor | A1 | Shown \(= 0\) plus statement; May make statement *first* such as "\(x+2\) is a factor if \(p(-2) = 0\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = -2\) | B1 | No working required for B1 + B1 |
| \(c = 3\) | B1 | Try to mark first using B marks |
| or long division/comparing coefficients | (M1) | Award M1 if B0 earned and a clear method is used |
| \(p(x) = (x+2)(x^2 - 2x + 3)\) | (A1) | Must write final answer in this form if long division has been used to get A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b^2 - 4ac = (-2)^2 - 4\times3\) | M1 | Discriminant correct from their quadratic; M0 if \(b = -1\), \(c = 6\) used |
| \(b^2 - 4ac = -8\ (\text{or} < 0) \Rightarrow\) no (other) real roots | A1 | CSO All values must be correct plus statement |
| Or \((x-1)^2 + 2\) | (M1) | Completion of square for their quadratic |
| \((x-1)^2 + 2 > 0\) therefore no real roots | (A1) | Shown to be positive plus statement regarding no real roots |
| Or \((x-1)^2 = -2\) has no real roots |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((y_B =)\ 6\) | B1 | Condone \((0, 6)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{x^4}{4} - \frac{x^2}{2} + 6x\) | M1 | One term correct |
| A1 | Another term correct | |
| A1 | All correct (ignore \(+ c\) or limits) | |
| \(\left[\quad\right]_{-2}^{0} = 0 - (4 - 2 - 12)\) | m1 | \(F(-2)\) attempted |
| \(= 10\) | A1 | CSO Clearly from \(F(0) - F(-2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area of \(\Delta = \frac{1}{2}\times2\times6\) | M1 | Condone \(-2\) and ft their \(y_B\) value; Or \(\int_{-2}^{0}(3x+6)\,dx\) and attempt to integrate |
| \(= 6\) | A1 | Must be positive, allow \(-6\) converted to \(+6\) |
| Shaded region area \(= 10 - 6 = 4\) | A1 | CSO 10 must come from correct working |
## Question 4:
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p(3) = 27 - 3 + 6$ | M1 | $p(3)$ attempted |
| (Remainder) $= 30$ | A1 | |
| **Or** long division up to remainder; Quotient $= x^2 + 3x + 8$ and remainder $= 30$ clearly stated | (M1)(A1) | |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $p(-2) = -8 + 2 + 6$ | M1 | $p(-2)$ attempted: **NOT** long division |
| $p(-2) = 0 \Rightarrow x + 2$ is factor | A1 | Shown $= 0$ plus statement; May make statement *first* such as "$x+2$ is a factor if $p(-2) = 0$" |
### Part (a)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = -2$ | B1 | No working required for B1 + B1 |
| $c = 3$ | B1 | Try to mark first using B marks |
| or long division/comparing coefficients | (M1) | Award M1 if B0 earned and a clear method is used |
| $p(x) = (x+2)(x^2 - 2x + 3)$ | (A1) | Must write final answer in this form if long division has been used to get A1 |
### Part (a)(iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b^2 - 4ac = (-2)^2 - 4\times3$ | M1 | Discriminant correct from their quadratic; M0 if $b = -1$, $c = 6$ used |
| $b^2 - 4ac = -8\ (\text{or} < 0) \Rightarrow$ no (other) real roots | A1 | CSO All values must be correct plus statement |
| **Or** $(x-1)^2 + 2$ | (M1) | Completion of square for their quadratic |
| $(x-1)^2 + 2 > 0$ therefore no real roots | (A1) | Shown to be positive plus statement regarding no real roots |
| Or $(x-1)^2 = -2$ has no real roots | | |
### Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(y_B =)\ 6$ | B1 | Condone $(0, 6)$ |
### Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{x^4}{4} - \frac{x^2}{2} + 6x$ | M1 | One term correct |
| | A1 | Another term correct |
| | A1 | All correct (ignore $+ c$ or limits) |
| $\left[\quad\right]_{-2}^{0} = 0 - (4 - 2 - 12)$ | m1 | $F(-2)$ attempted |
| $= 10$ | A1 | CSO Clearly from $F(0) - F(-2)$ |
### Part (b)(iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area of $\Delta = \frac{1}{2}\times2\times6$ | M1 | Condone $-2$ and ft their $y_B$ value; **Or** $\int_{-2}^{0}(3x+6)\,dx$ and attempt to integrate |
| $= 6$ | A1 | Must be positive, allow $-6$ converted to $+6$ |
| Shaded region area $= 10 - 6 = 4$ | A1 | CSO 10 must come from correct working |4
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - x + 6$.
\begin{enumerate}[label=(\roman*)]
\item Find the remainder when $\mathrm { p } ( x )$ is divided by $x - 3$.
\item Use the Factor Theorem to show that $x + 2$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x ) = x ^ { 3 } - x + 6$ in the form $( x + 2 ) \left( x ^ { 2 } + b x + c \right)$, where $b$ and $c$ are integers.
\item The equation $\mathrm { p } ( x ) = 0$ has one root equal to - 2 . Show that the equation has no other real roots.
\end{enumerate}\item The curve with equation $y = x ^ { 3 } - x + 6$ is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{5f1ff5fa-b6e8-4c4f-aef7-63eb947b299f-3_529_702_945_667}
The curve cuts the $x$-axis at the point $A ( - 2,0 )$ and the $y$-axis at the point $B$.
\begin{enumerate}[label=(\roman*)]
\item State the $y$-coordinate of the point $B$.
\item Find $\int _ { - 2 } ^ { 0 } \left( x ^ { 3 } - x + 6 \right) \mathrm { d } x$.
\item Hence find the area of the shaded region bounded by the curve $y = x ^ { 3 } - x + 6$ and the line $A B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2009 Q4 [17]}}