| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Factor condition (zero remainder) |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of the factor theorem requiring substitution of x=-2 and simple algebraic manipulation to show the result holds for all b. Part (b)(i) is a basic sketch requiring understanding that a repeated root gives a tangent to the x-axis. Part (b)(ii) requires recognizing that 'exactly two points' means a repeated root, so the discriminant of the quadratic factor must equal zero, leading to a simple quadratic equation. While multi-part, each component uses standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(x = -2\) into \(p(x)\) | M1 | Condone missing brackets |
| \(p(-2) = (-2)^3 + (b+2)(-2)^2 + 2(b+2)(-2) + 8 = -8+4b+8-4b-8+8 = 0\) | A1 | Must see numerical evaluation of powers of \(-2\) and either \(-8+4b+8-4b-8+8=0\) or \(-8+4(b+2)-4(b+2)+8=0\) or \(-8+(b+2)(4-4)+8=0\) |
| Concludes \((x+2)\) is a factor for all/any values of \(b\) | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Sketches cubic graph with correct orientation and two turning points | B1 | |
| Sketches any cubic that would only ever meet the \(x\)-axis at exactly two points | M1 | |
| Correctly orientated cubic with: single root labelled \(x=-2\); \(y\)-intercept labelled at \(8\); repeated root on positive \(x\)-axis | A1 | Ignore any value shown at the other root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Deduces a pair of possible factors of \((x^2+bx+4)\): either \((x+2)^2\) or \((x-2)^2\) or \((x+1)(x+4)\) or \((x-1)(x-4)\); or uses \(b^2-4ac\), PI \(b=4\) or \(b=-4\) or \(b^2-16\) seen | M1 | |
| Identifies quadratic factor as \((x-2)^2\), or obtains \(b^2-16=0\) OE, PI by \(b=\pm4\) or \(b=-4\) | R1 | \(b^2-4ac=0 \Rightarrow b^2-16=0 \Rightarrow b=\pm4\) |
| Obtains either \(b=\pm4\) or \(b=-4\) | M1 | |
| Rejects \(b=4\) giving a reason and concludes \(b=-4\); valid reasons: only one factor/root; only one point of intersection with \(x\)-axis; it would be \((x+2)^3\) which has only one point of intersection | R1 | \(b=4\) gives only one point of intersection, \(\therefore b=-4\) |
## Question 11(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $x = -2$ into $p(x)$ | M1 | Condone missing brackets |
| $p(-2) = (-2)^3 + (b+2)(-2)^2 + 2(b+2)(-2) + 8 = -8+4b+8-4b-8+8 = 0$ | A1 | Must see numerical evaluation of powers of $-2$ and either $-8+4b+8-4b-8+8=0$ or $-8+4(b+2)-4(b+2)+8=0$ or $-8+(b+2)(4-4)+8=0$ |
| Concludes $(x+2)$ is a factor for **all/any** values of $b$ | R1 | |
---
## Question 11(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Sketches cubic graph with correct orientation and two turning points | B1 | |
| Sketches any cubic that would only ever meet the $x$-axis at exactly two points | M1 | |
| Correctly orientated cubic with: single root labelled $x=-2$; $y$-intercept labelled at $8$; repeated root on positive $x$-axis | A1 | Ignore any value shown at the other root |
---
## Question 11(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Deduces a pair of possible factors of $(x^2+bx+4)$: either $(x+2)^2$ or $(x-2)^2$ or $(x+1)(x+4)$ or $(x-1)(x-4)$; or uses $b^2-4ac$, PI $b=4$ or $b=-4$ or $b^2-16$ seen | M1 | |
| Identifies quadratic factor as $(x-2)^2$, or obtains $b^2-16=0$ OE, PI by $b=\pm4$ or $b=-4$ | R1 | $b^2-4ac=0 \Rightarrow b^2-16=0 \Rightarrow b=\pm4$ |
| Obtains either $b=\pm4$ or $b=-4$ | M1 | |
| Rejects $b=4$ giving a reason and concludes $b=-4$; valid reasons: only one factor/root; only one point of intersection with $x$-axis; it would be $(x+2)^3$ which has only one point of intersection | R1 | $b=4$ gives only one point of intersection, $\therefore b=-4$ |
---11 The polynomial $\mathrm { p } ( x )$ is given by
$$\mathrm { p } ( x ) = x ^ { 3 } + ( b + 2 ) x ^ { 2 } + 2 ( b + 2 ) x + 8$$
where $b$ is a constant.\\
11
\begin{enumerate}[label=(\alph*)]
\item Use the factor theorem to prove that $( x + 2 )$ is a factor of $\mathrm { p } ( x )$ for all values of $b$.\\
11
\item The graph of $y = \mathrm { p } ( x )$ meets the $x$-axis at exactly two points.\\
11 (b) (i) Sketch a possible graph of $y = \mathrm { p } ( x )$\\
\includegraphics[max width=\textwidth, alt={}, center]{22ff390e-1360-43bd-8c7f-3d2b58627e91-20_1084_965_1619_532}
11 (b) (ii) Given $\mathrm { p } ( x )$ can be written as
$$\mathrm { p } ( x ) = ( x + 2 ) \left( x ^ { 2 } + b x + 4 \right)$$
find the value of $b$.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2022 Q11 [10]}}