| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Sketch curve using polynomial roots |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing standard applications of Factor and Remainder Theorems with routine algebraic manipulation. All parts are textbook exercises: showing a given factor works, factorising by division, evaluating p(2), and sketching using known roots. No problem-solving insight required, just methodical application of learned techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(-3) = (-3)^3 + 7(-3)^2 + 7(-3) - 15 = -27 + 63 - 21 - 15 = 0\); since \(p(-3)=0\), \((x+3)\) is a factor | M1 A1 | M1 for attempting \(p(-3)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(x) = (x+3)(x^2+4x-5) = (x+3)(x+5)(x-1)\) | M1 A1 A1 | M1 for attempting to find quadratic factor; A1 for \((x^2+4x-5)\); A1 for fully factorised form |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(2) = 8 + 28 + 14 - 15 = 35\) | M1 A1 | M1 for attempting \(p(2)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(-1) = -1 + 7 - 7 - 15 = -16\); \(p(0) = -15\); \(-16 < -15\) so \(p(-1) < p(0)\) ✓ | B1 | Both values required |
| Answer | Marks | Guidance |
|---|---|---|
| Cubic curve, correct orientation (positive leading coefficient); crosses \(x\)-axis at \(x=-5, -3, 1\); crosses \(y\)-axis at \((0,-15)\) | B1 B1 B1 B1 | B1 shape; B1 three \(x\)-intercepts marked; B1 \(y\)-intercept marked |
# Question 3:
## Part (a)(i)
| $p(-3) = (-3)^3 + 7(-3)^2 + 7(-3) - 15 = -27 + 63 - 21 - 15 = 0$; since $p(-3)=0$, $(x+3)$ is a factor | M1 A1 | M1 for attempting $p(-3)$ |
## Part (a)(ii)
| $p(x) = (x+3)(x^2+4x-5) = (x+3)(x+5)(x-1)$ | M1 A1 A1 | M1 for attempting to find quadratic factor; A1 for $(x^2+4x-5)$; A1 for fully factorised form |
## Part (b)
| $p(2) = 8 + 28 + 14 - 15 = 35$ | M1 A1 | M1 for attempting $p(2)$ |
## Part (c)(i)
| $p(-1) = -1 + 7 - 7 - 15 = -16$; $p(0) = -15$; $-16 < -15$ so $p(-1) < p(0)$ ✓ | B1 | Both values required |
## Part (c)(ii)
| Cubic curve, correct orientation (positive leading coefficient); crosses $x$-axis at $x=-5, -3, 1$; crosses $y$-axis at $(0,-15)$ | B1 B1 B1 B1 | B1 shape; B1 three $x$-intercepts marked; B1 $y$-intercept marked |
---3 The polynomial $\mathrm { p } ( x )$ is given by
$$\mathrm { p } ( x ) = x ^ { 3 } + 7 x ^ { 2 } + 7 x - 15$$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\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}\item Use the Remainder Theorem to find the remainder when $\mathrm { p } ( x )$ is divided by $x - 2$.
\item \begin{enumerate}[label=(\roman*)]
\item Verify that $\mathrm { p } ( - 1 ) < \mathrm { p } ( 0 )$.
\item Sketch the curve with equation $y = x ^ { 3 } + 7 x ^ { 2 } + 7 x - 15$, indicating the values where the curve crosses the coordinate axes.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q3 [12]}}