| Exam Board | OCR |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two factors given |
| Difficulty | Standard +0.3 This is a straightforward application of the Factor Theorem requiring substitution of x=2 into both polynomials to form simultaneous equations, then factorising a cubic with a known factor. While multi-step, it follows standard C2 procedures without requiring novel insight or complex algebraic manipulation. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(2) = 8 + 2a - 6 + 2b = 0\); \(g(2) = 24 + 4 + 10a + 4b = 0\) | M1 | Attempt at least one of \(f(2)\), \(g(2)\). Allow substituting \(x=2\) into either equation – no need to simplify. Division – complete attempt to divide by \((x-2)\). Coeff matching – attempt all 3 coeffs of quadratic factor. |
| Equate at least one of \(f(2)\) and \(g(2)\) to 0 | M1 | Just need to equate their substitution attempt to 0 (but just writing e.g. \(f(2)=0\) is not enough). Could be implied by later working, even after attempt to solve equations. Division – equating remainder to 0. Coeff matching – equate constant terms. |
| \(2a + 2b = -2\), \(5a + 2b = -14\) | A1 | Obtain two correct equations in \(a\) and \(b\). Could be unsimplified equations. Could be \(8a + 2b = -26\) (from \(f(2) = g(2)\)). |
| hence \(3a = -12\) | M1 | Attempt to find \(a\) (or \(b\)) from two simultaneous equations. M1 for eliminating \(a\) or \(b\) from 2 sim eqns – allow sign slips only. |
| so \(a = -4\) AG | A1 | Obtain \(a = -4\), with necessary working shown. If finding \(b\) first, must show at least one line of working to find \(a\) (unless earlier shown explicitly e.g. \(a = -1 - b\)). |
| \(b = 3\) | A1 | Correct working only. |
| [6] | SR Assuming \(a = -4\): Either use this scheme, or the original, but don't mix elements from both. M1 Attempt either \(f(2)\) or \(g(2)\). M1 Equate \(f(2)\) or \(g(2)\) to 0 (also allow \(f(2) = g(2)\)). A1 Obtain \(b = 3\). A1 Use second equation to confirm \(a = -4\), \(b = 3\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x) = (x-2)(x^2 + 2x - 3) = (x-2)(x+3)(x-1)\) | M1 | Attempt full division of their \(f(x)\) by \((x-2)\). Could also be for full division attempt by \((x-1)\) or \((x+3)\) if identified as factors. Must be using \(f(x) = x^3 - 7x + k\). Must be complete method – ie all 3 terms attempted. Long division – must subtract lower line (allow one slip). Inspection – expansion must give at least three correct terms. Coefficient matching – must be valid attempt at all 3 quadratic coeffs. Factor theorem – must be finding 2 more factors/roots. |
| Obtain \(x^2\) and at least one other correct term, from correct \(f(x)\) | A1 | Could be middle or final term depending on method. Coeff matching – allow for stating values e.g. \(A=1\) etc. Factor theorem – state factors of \((x+3)\) and \((x-1)\). |
| Obtain \((x-2)(x+3)(x-1)\) | A1 | Must be seen as a product of three linear factors. Answer only gains all 3 marks. |
| \(g(x) = (x-2)(3x^2 + 7x - 6) = (x-2)(x+3)(3x-2)\) OR \(g(1) = -4\), \(g(-3) = 0\) | M1 | Attempt to verify two common factors. Possible methods: Factorise \(g(x)\) completely – \(f(x)\) must have been factorised. Find quadratic factor of \(g(x)\) and identify \(x = -3\) as root. Test roots of \(f(x)\) in \(g(x)\). Just stating e.g. \(g(-3) = 0\) is not enough – working required. If \(f(x)\) hasn't been factorised, allow M1 for using factor thm on both functions to find common factor, or for factorising \(g(x)\) and testing roots in \(f(x)\). |
| Hence common factor of \((x+3)\) | A1 | Identify \((x+3)\) as a common factor. Just need to identify \((x+3)\) – no need to see \((x-2)\) or explicitly state 'two common factors'. Need to see \((x+3)\) as factor of \(g(x)\) – just showing \(g(-3)=0\) and stating 'common factor' is not enough. CWO (inc A0 for \(g(x) = (x-2)(x+3)(x-\frac{2}{3})\)). If using factor thm, no need to find \(g(1)\) if \(g(-3)\) done first. Just stating \((x+3)\) with no supporting evidence is M0A0. A0 if referring to \(-3\) (and 2) as 'factors'. A0 if additional incorrect factor given. |
| [5] |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(2) = 8 + 2a - 6 + 2b = 0$; $g(2) = 24 + 4 + 10a + 4b = 0$ | M1 | Attempt at least one of $f(2)$, $g(2)$. Allow substituting $x=2$ into either equation – no need to simplify. Division – complete attempt to divide by $(x-2)$. Coeff matching – attempt all 3 coeffs of quadratic factor. |
| Equate at least one of $f(2)$ and $g(2)$ to 0 | M1 | Just need to equate their substitution attempt to 0 (but just writing e.g. $f(2)=0$ is not enough). Could be implied by later working, even after attempt to solve equations. Division – equating remainder to 0. Coeff matching – equate constant terms. |
| $2a + 2b = -2$, $5a + 2b = -14$ | A1 | Obtain two correct equations in $a$ and $b$. Could be unsimplified equations. Could be $8a + 2b = -26$ (from $f(2) = g(2)$). |
| hence $3a = -12$ | M1 | Attempt to find $a$ (or $b$) from two simultaneous equations. M1 for eliminating $a$ or $b$ from 2 sim eqns – allow sign slips only. |
| so $a = -4$ **AG** | A1 | Obtain $a = -4$, with necessary working shown. If finding $b$ first, must show at least one line of working to find $a$ (unless earlier shown explicitly e.g. $a = -1 - b$). |
| $b = 3$ | A1 | Correct working only. |
| **[6]** | | **SR** Assuming $a = -4$: Either use this scheme, or the original, but don't mix elements from both. M1 Attempt either $f(2)$ or $g(2)$. M1 Equate $f(2)$ or $g(2)$ to 0 (also allow $f(2) = g(2)$). A1 Obtain $b = 3$. A1 Use second equation to confirm $a = -4$, $b = 3$. |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x) = (x-2)(x^2 + 2x - 3) = (x-2)(x+3)(x-1)$ | M1 | Attempt full division of their $f(x)$ by $(x-2)$. Could also be for full division attempt by $(x-1)$ or $(x+3)$ if identified as factors. Must be using $f(x) = x^3 - 7x + k$. Must be complete method – ie all 3 terms attempted. Long division – must subtract lower line (allow one slip). Inspection – expansion must give at least three correct terms. Coefficient matching – must be valid attempt at all 3 quadratic coeffs. Factor theorem – must be finding 2 more factors/roots. |
| Obtain $x^2$ and at least one other correct term, from correct $f(x)$ | A1 | Could be middle or final term depending on method. Coeff matching – allow for stating values e.g. $A=1$ etc. Factor theorem – state factors of $(x+3)$ and $(x-1)$. |
| Obtain $(x-2)(x+3)(x-1)$ | A1 | Must be seen as a product of three linear factors. Answer only gains all 3 marks. |
| $g(x) = (x-2)(3x^2 + 7x - 6) = (x-2)(x+3)(3x-2)$ **OR** $g(1) = -4$, $g(-3) = 0$ | M1 | Attempt to verify two common factors. Possible methods: Factorise $g(x)$ completely – $f(x)$ must have been factorised. Find quadratic factor of $g(x)$ and identify $x = -3$ as root. Test roots of $f(x)$ in $g(x)$. Just stating e.g. $g(-3) = 0$ is not enough – working required. If $f(x)$ hasn't been factorised, allow M1 for using factor thm on both functions to find common factor, or for factorising $g(x)$ and testing roots in $f(x)$. |
| Hence common factor of $(x+3)$ | A1 | Identify $(x+3)$ as a common factor. Just need to identify $(x+3)$ – no need to see $(x-2)$ or explicitly state 'two common factors'. Need to see $(x+3)$ as factor of $g(x)$ – just showing $g(-3)=0$ and stating 'common factor' is not enough. CWO (inc A0 for $g(x) = (x-2)(x+3)(x-\frac{2}{3})$). If using factor thm, no need to find $g(1)$ if $g(-3)$ done first. Just stating $(x+3)$ with no supporting evidence is M0A0. A0 if referring to $-3$ (and 2) as 'factors'. A0 if additional incorrect factor given. |
| **[5]** | | |
---8 Two cubic polynomials are defined by
$$\mathrm { f } ( x ) = x ^ { 3 } + ( a - 3 ) x + 2 b , \quad \mathrm {~g} ( x ) = 3 x ^ { 3 } + x ^ { 2 } + 5 a x + 4 b$$
where $a$ and $b$ are constants.\\
(i) Given that $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ have a common factor of ( $x - 2$ ), show that $a = - 4$ and find the value of $b$.\\
(ii) Using these values of $a$ and $b$, factorise $\mathrm { f } ( x )$ fully. Hence show that $\mathrm { f } ( x )$ and $\mathrm { g } ( x )$ have two common factors.
\hfill \mbox{\textit{OCR C2 2012 Q8 [11]}}