| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Solve p(algebraic transform) = 0 |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring substitution to find a constant, factorizing a cubic (with common factor x), then applying algebraic substitutions. Part (a) uses basic algebra and factor theorem techniques. Parts (b)(i) and (b)(ii) involve standard substitutions (y^(1/3) = x and 9^z = x) that are routine for P1 level. All steps are predictable with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-a + 6a + 8 + a^2 = 32 \Rightarrow a^2 + 5a - 24 = 0\) | M1 | Substitutes \(x = \pm1\), \(y=32\), proceeds to 3TQ in \(a\) with all terms on one side. Condone lack of \(=0\) and slips in rearrangement |
| \((a+8)(a-3) = 0\) | dM1 | Attempts to solve quadratic by factorising, completing the square, or formula. Must show working (factorised form, or values in formula). Dependent on first M1 |
| \(a = 3\) or \(a = -8\), chooses \(a = 3\) with reason | A1* | Minimal reason required e.g. "\(a\) is a positive constant" or "\(a>0\)". Just crossing out \(-8\) without reason is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3x^3 + 26x^2 - 9x = 0 \Rightarrow x(3x^2 + 26x - 9) = 0\) then \(x(3x-1)(x+9)\) | M1 | Takes factor of \(x\) from cubic (with \(a=3\)), attempts to solve resulting quadratic. Must show at least one intermediate line of working |
| \((x =)\ 0,\ \frac{1}{3},\ -9\) | A1 | All three correct, provided M1 scored. Solutions with no working score 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y =)\ 0\) | B1 | |
| \(y^{\frac{1}{3}} = \frac{1}{3}\) or \(y^{\frac{1}{3}} = -9 \Rightarrow y = ...\) | M1 | Sets \(y^{\frac{1}{3}}\) equal to any non-zero solution from (a) and cubes to find \(y\). Must show at least one stage of working. \((-9)^3 = ...\) on its own scores M1 |
| \((y =)\ \frac{1}{27},\ -729\) | A1 | Both values and no others other than 0. Max B1M0A0 without working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(9^z = \frac{1}{3} \to z = ...\) | M1 | Sets \(9^z\) equal to any positive solution from (a), proceeds to find \(z\). May write \(9^z\) as \(3^{2z}\). Must show at least one stage of working |
| \((z =)\ -\frac{1}{2}\) only | A1 | Provided M1 scored. Answer without working or evidence of calculator use is A0 |
# Question 2:
## Part (a)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-a + 6a + 8 + a^2 = 32 \Rightarrow a^2 + 5a - 24 = 0$ | M1 | Substitutes $x = \pm1$, $y=32$, proceeds to 3TQ in $a$ with all terms on one side. Condone lack of $=0$ and slips in rearrangement |
| $(a+8)(a-3) = 0$ | dM1 | Attempts to solve quadratic by factorising, completing the square, or formula. Must show working (factorised form, or values in formula). Dependent on first M1 |
| $a = 3$ or $a = -8$, chooses $a = 3$ with reason | A1* | Minimal reason required e.g. "$a$ is a positive constant" or "$a>0$". Just crossing out $-8$ without reason is A0 |
## Part (a)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3x^3 + 26x^2 - 9x = 0 \Rightarrow x(3x^2 + 26x - 9) = 0$ then $x(3x-1)(x+9)$ | M1 | Takes factor of $x$ from cubic (with $a=3$), attempts to solve resulting quadratic. Must show at least one intermediate line of working |
| $(x =)\ 0,\ \frac{1}{3},\ -9$ | A1 | All three correct, provided M1 scored. Solutions with no working score 0 |
## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y =)\ 0$ | B1 | |
| $y^{\frac{1}{3}} = \frac{1}{3}$ or $y^{\frac{1}{3}} = -9 \Rightarrow y = ...$ | M1 | Sets $y^{\frac{1}{3}}$ equal to any non-zero solution from (a) and cubes to find $y$. Must show at least one stage of working. $(-9)^3 = ...$ on its own scores M1 |
| $(y =)\ \frac{1}{27},\ -729$ | A1 | Both values and no others other than 0. Max B1M0A0 without working |
## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $9^z = \frac{1}{3} \to z = ...$ | M1 | Sets $9^z$ equal to any positive solution from (a), proceeds to find $z$. May write $9^z$ as $3^{2z}$. Must show at least one stage of working |
| $(z =)\ -\frac{1}{2}$ only | A1 | Provided M1 scored. Answer without working or evidence of calculator use is A0 |
---2. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
$$f ( x ) = a x ^ { 3 } + ( 6 a + 8 ) x ^ { 2 } - a ^ { 2 } x$$
where $a$ is a positive constant.
Given $\mathrm { f } ( - 1 ) = 32$
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item show that the only possible value for $a$ is 3
\item Using $a = 3$ solve the equation
$$\mathrm { f } ( x ) = 0$$
\end{enumerate}\item Hence find all real solutions of
\begin{enumerate}[label=(\roman*)]
\item $3 y + 26 y ^ { \frac { 2 } { 3 } } - 9 y ^ { \frac { 1 } { 3 } } = 0$
\item $3 \left( 9 ^ { 3 z } \right) + 26 \left( 9 ^ { 2 z } \right) - 9 \left( 9 ^ { z } \right) = 0$
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q2 [10]}}