| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Solve p(algebraic transform) = 0 |
| Difficulty | Moderate -0.3 Part (a) is straightforward factorization (factor out x, then solve quadratic), requiring only routine algebraic manipulation. Part (b) applies the same solution via substitution u=(y-5)², which is a standard 'hence' extension requiring recognition of the structural similarity but no additional conceptual difficulty beyond careful algebraic handling of the transform. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(2x^3 + 3x^2 - 35x = 0 \Rightarrow x(2x^2 + 3x - 35) = 0\) | M1 | Takes out common factor of \(x\); score if each term divided by \(x\) |
| \((2x-7)(x+5) = 0 \Rightarrow x = \ldots\) | dM1 | Attempts to solve resulting quadratic via algebra; allow factorisation, formula or completion of square |
| \(x = -5, 0, \frac{7}{2}\) | A1 | All three values correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| States \(y = 5\) is a solution | B1 | |
| \((y-5)^2 = ``\frac{7}{2}`` \Rightarrow y = \ldots\) | M1 | Realises \(x = (y-5)^2\); follow through on any positive value from (a) |
| \(y = 5 + \sqrt{\frac{7}{2}}\) or \(y = 5 - \sqrt{\frac{7}{2}}\) | A1ft | One solution; allow decimals (awrt 6.87 or awrt 3.13) |
| Both \(y = 5 + \sqrt{\frac{7}{2}}\) and \(y = 5 - \sqrt{\frac{7}{2}}\) | A1 | Both exact values; no other solutions; no decimal equivalents |
## Question 5:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $2x^3 + 3x^2 - 35x = 0 \Rightarrow x(2x^2 + 3x - 35) = 0$ | M1 | Takes out common factor of $x$; score if each term divided by $x$ |
| $(2x-7)(x+5) = 0 \Rightarrow x = \ldots$ | dM1 | Attempts to solve resulting quadratic via algebra; allow factorisation, formula or completion of square |
| $x = -5, 0, \frac{7}{2}$ | A1 | All three values correct |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| States $y = 5$ is a solution | B1 | |
| $(y-5)^2 = ``\frac{7}{2}`` \Rightarrow y = \ldots$ | M1 | Realises $x = (y-5)^2$; follow through on any positive value from (a) |
| $y = 5 + \sqrt{\frac{7}{2}}$ or $y = 5 - \sqrt{\frac{7}{2}}$ | A1ft | One solution; allow decimals (awrt 6.87 or awrt 3.13) |
| Both $y = 5 + \sqrt{\frac{7}{2}}$ and $y = 5 - \sqrt{\frac{7}{2}}$ | A1 | Both exact values; no other solutions; no decimal equivalents |
---\begin{enumerate}
\item (a) Find, using algebra, all real solutions of
\end{enumerate}
$$2 x ^ { 3 } + 3 x ^ { 2 } - 35 x = 0$$
(b) Hence find all real solutions of
$$2 ( y - 5 ) ^ { 6 } + 3 ( y - 5 ) ^ { 4 } - 35 ( y - 5 ) ^ { 2 } = 0$$
\hfill \mbox{\textit{Edexcel P1 2019 Q5 [7]}}