| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Two unknowns with show-that step |
| Difficulty | Moderate -0.3 This is a straightforward application of the Remainder and Factor Theorems requiring substitution of x-values and solving simultaneous equations. While it has multiple parts and two unknowns, the techniques are standard C2 material with no novel insight required—slightly easier than average due to the 'show that' scaffolding in part (a). |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempting \(f(1)=45\) or \(f(-1)=45\) | M1 | 1 or \(-1\) substituted and expression put equal to \(\pm45\) |
| \(f(-1)=-6+3-A+B=45\) or \(-3-A+B=45\Rightarrow B-A=48\) | A1 *cso | Must have substituted \(-1\); correct equation with no errors [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((6x^3+3x^2+Ax+B)\div(x\pm1)=6x^2+px+q\) and sets remainder \(=45\) | M1 | Long division as far as remainder set equal to \(\pm45\) |
| Quotient is \(6x^2-3x+(A+3)\) and remainder is \(B-A-3=45\) so \(B-A=48\) | A1* | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Attempting \(f\!\left(-\frac{1}{2}\right)=0\) | M1 | Must see \(f(-\frac{1}{2})\) and "\(=0\)" unless subsequent work implies this |
| \(6\!\left(-\frac{1}{2}\right)^3+3\!\left(-\frac{1}{2}\right)^2+A\!\left(-\frac{1}{2}\right)+B=0\) or \(-\frac{1}{2}A+B=0\) or \(A=2B\) | A1 o.e. | Correct equation, even unsimplified |
| Solve to obtain \(B=-48\) and \(A=-96\) | M1 A1 | M1: attempts to solve; A1: both correct [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((6x^3+3x^2+Ax+B)\div(2x+1)=3x^2+px+q\) and sets remainder \(=0\) | M1 | |
| Quotient is \(3x^2+\frac{A}{2}\) and remainder is \(B-\frac{A}{2}=0\) | A1 | |
| Solve to obtain \(B=-48\) and \(A=-96\) | M1 A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Obtain \(\left(3x^2-48\right)\), \(\left(x^2-16\right)\), \(\left(6x^2-96\right)\), \(\left(3x^2+\frac{A}{2}\right)\), \(\left(3x^2+B\right)\), \(\left(x^2+\frac{A}{6}\right)\) or \(\left(x^2+\frac{B}{3}\right)\) as factor or quotient | B1ft | May be written straight down or from long division |
| Factorise \(\left(3x^2-48\right)\), \(\left(x^2-16\right)\), \(\left(48-3x^2\right)\), \(\left(16-x^2\right)\) or \(\left(6x^2-96\right)\) | M1 | Valid attempt to factorise a listed quadratic |
| \(=3(2x+1)(x+4)(x-4)\) | A1cso | Factor 3 must be shown; all terms together with brackets [3] |
| isw if they go on to solve: \(x=4,\ -4\) and \(-\frac{1}{2}\) | [9] |
# Question 3:
## $f(x)=6x^3+3x^2+Ax+B$
## Part (a) — Way 1
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempting $f(1)=45$ or $f(-1)=45$ | M1 | 1 or $-1$ substituted and expression put equal to $\pm45$ |
| $f(-1)=-6+3-A+B=45$ or $-3-A+B=45\Rightarrow B-A=48$ | A1 *cso | Must have substituted $-1$; correct equation with no errors **[2]** |
**Way 2 (a):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(6x^3+3x^2+Ax+B)\div(x\pm1)=6x^2+px+q$ and sets remainder $=45$ | M1 | Long division as far as remainder set equal to $\pm45$ |
| Quotient is $6x^2-3x+(A+3)$ and remainder is $B-A-3=45$ so $B-A=48$ | A1* | **[2]** |
## Part (b) — Way 1
| Working/Answer | Mark | Guidance |
|---|---|---|
| Attempting $f\!\left(-\frac{1}{2}\right)=0$ | M1 | Must see $f(-\frac{1}{2})$ and "$=0$" unless subsequent work implies this |
| $6\!\left(-\frac{1}{2}\right)^3+3\!\left(-\frac{1}{2}\right)^2+A\!\left(-\frac{1}{2}\right)+B=0$ or $-\frac{1}{2}A+B=0$ or $A=2B$ | A1 o.e. | Correct equation, even unsimplified |
| Solve to obtain $B=-48$ and $A=-96$ | M1 A1 | M1: attempts to solve; A1: **both** correct **[4]** |
**Way 2 (b):**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(6x^3+3x^2+Ax+B)\div(2x+1)=3x^2+px+q$ and sets remainder $=0$ | M1 | |
| Quotient is $3x^2+\frac{A}{2}$ and remainder is $B-\frac{A}{2}=0$ | A1 | |
| Solve to obtain $B=-48$ and $A=-96$ | M1 A1 | **[4]** |
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Obtain $\left(3x^2-48\right)$, $\left(x^2-16\right)$, $\left(6x^2-96\right)$, $\left(3x^2+\frac{A}{2}\right)$, $\left(3x^2+B\right)$, $\left(x^2+\frac{A}{6}\right)$ or $\left(x^2+\frac{B}{3}\right)$ as factor or quotient | B1ft | May be written straight down or from long division |
| Factorise $\left(3x^2-48\right)$, $\left(x^2-16\right)$, $\left(48-3x^2\right)$, $\left(16-x^2\right)$ or $\left(6x^2-96\right)$ | M1 | Valid attempt to factorise a listed quadratic |
| $=3(2x+1)(x+4)(x-4)$ | A1cso | Factor 3 must be shown; all terms together with brackets **[3]** |
| isw if they go on to solve: $x=4,\ -4$ and $-\frac{1}{2}$ | | **[9]** |
---3. $\mathrm { f } ( x ) = 6 x ^ { 3 } + 3 x ^ { 2 } + A x + B$, where $A$ and $B$ are constants.
Given that when $\mathrm { f } ( x )$ is divided by $( x + 1 )$ the remainder is 45 ,
\begin{enumerate}[label=(\alph*)]
\item show that $B - A = 48$
Given also that ( $2 x + 1$ ) is a factor of $\mathrm { f } ( x )$,
\item find the value of $A$ and the value of $B$.
\item Factorise f(x) fully.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2015 Q3 [9]}}