| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Solve p(algebraic transform) = 0 |
| Difficulty | Standard +0.3 Part (a) requires factoring a cubic expression and solving, which is straightforward. Part (b) applies a substitution (y^(2/3) = x) to transform the equation into the form from part (a), then requires working backwards with fractional powers and simplifying surds. This is a standard 'hence' question testing algebraic manipulation and understanding of substitutions, slightly above average due to the fractional powers but still routine for P1. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2(2x+1) - 15x \Rightarrow 2x^3 + x^2 - 15x = x(2x^2 + x - 15)\) | M1 | Multiplies out bracket to get cubic, takes out linear factor or cancels \(x\) leading to quadratic. Award for \(...x(...x^2 \pm ...x \pm ...)\) or \(...x(...x \pm ...)(...x \pm ...)\). |
| \(x(2x-5)(x+3) = 0 \Rightarrow x = ...\) | dM1 | Attempts to solve quadratic by factorising, quadratic formula, or completing the square. Values of \(a\), \(b\), \(c\) must match their quadratic. Dependent on previous M. |
| Two of \(x = 0,\ \frac{5}{2},\ -3\) | B1 | |
| \(x = 0,\ \frac{5}{2},\ -3\) | A1 | All three correct, provided all previous marks scored. Check for \(x=0\) in earlier work. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y^{\frac{2}{3}} = \frac{5}{2} \Rightarrow y = \left(\frac{5}{2}\right)^{\frac{3}{2}}\) | M1 | Uses one of their positive solutions from (a). No marks for implied method with decimal answer or exact answer without method for fractional power. Allow notation \(y = \sqrt[3]{\frac{5}{2}}^2\). No marks without a positive solution from (a). |
| \(\frac{5}{4}\sqrt{10}\) | A1cso | Only this answer and no other solutions. Also accept \(1.25\sqrt{10}\) or \(1\frac{1}{4}\sqrt{10}\). Note: \(y^2 = \frac{125}{8} \Rightarrow y = \frac{5}{4}\sqrt{10}\) scores M1A1. |
# Question 4:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2(2x+1) - 15x \Rightarrow 2x^3 + x^2 - 15x = x(2x^2 + x - 15)$ | M1 | Multiplies out bracket to get cubic, takes out linear factor or cancels $x$ leading to quadratic. Award for $...x(...x^2 \pm ...x \pm ...)$ or $...x(...x \pm ...)(...x \pm ...)$. |
| $x(2x-5)(x+3) = 0 \Rightarrow x = ...$ | dM1 | Attempts to solve quadratic by factorising, quadratic formula, or completing the square. Values of $a$, $b$, $c$ must match their quadratic. Dependent on previous M. |
| Two of $x = 0,\ \frac{5}{2},\ -3$ | B1 | |
| $x = 0,\ \frac{5}{2},\ -3$ | A1 | All three correct, provided all previous marks scored. Check for $x=0$ in earlier work. |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y^{\frac{2}{3}} = \frac{5}{2} \Rightarrow y = \left(\frac{5}{2}\right)^{\frac{3}{2}}$ | M1 | Uses one of their positive solutions from (a). No marks for implied method with decimal answer or exact answer without method for fractional power. Allow notation $y = \sqrt[3]{\frac{5}{2}}^2$. No marks without a positive solution from (a). |
| $\frac{5}{4}\sqrt{10}$ | A1cso | Only this answer and no other solutions. Also accept $1.25\sqrt{10}$ or $1\frac{1}{4}\sqrt{10}$. Note: $y^2 = \frac{125}{8} \Rightarrow y = \frac{5}{4}\sqrt{10}$ scores M1A1. |\begin{enumerate}
\item In this question you must show detailed reasoning. Solutions relying on calculator technology are not acceptable.
\end{enumerate}
$$f ( x ) = x ^ { 2 } ( 2 x + 1 ) - 15 x$$
(a) Solve
$$\mathrm { f } ( x ) = 0$$
(b) Hence solve
$$y ^ { \frac { 4 } { 3 } } \left( 2 y ^ { \frac { 2 } { 3 } } + 1 \right) - 15 y ^ { \frac { 2 } { 3 } } = 0 \quad y > 0$$
giving your answer in simplified surd form.
\hfill \mbox{\textit{Edexcel P1 2022 Q4 [6]}}