| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then simplify rational expression |
| Difficulty | Moderate -0.3 This is a straightforward application of the Factor Theorem and polynomial division with standard techniques. Part (a) involves routine verification of a given factor, factorization by inspection/division, and algebraic simplification. Part (b) is a direct application of the Remainder Theorem. All steps are algorithmic with no novel problem-solving required, making it slightly easier than average but still requiring competent execution of multiple techniques. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division |
| Answer | Marks | Guidance |
|---|---|---|
| \(f\left(-\frac{1}{3}\right) = 9\left(-\frac{1}{3}\right)^3 + 18\left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right) - 2\) | M1 | Substituting \(x = -\frac{1}{3}\) |
| \(= -\frac{1}{3} + 2 + \frac{1}{3} - 2 = 0\), therefore \((3x+1)\) is a factor | A1 | Correct evaluation showing \(= 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x) = (3x+1)(3x^2 + 5x - 2)\) | M1 A1 | |
| \(= (3x+1)(3x-1)(x+2)\) | A1 | All three factors correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(9x^3 + 21x^2 + 6x = 3x(3x^2 + 7x + 2) = 3x(3x+1)(x+2)\) | M1 A1 | Factorising numerator |
| \(\frac{3x(3x+1)(x+2)}{(3x+1)(3x-1)(x+2)} = \frac{3x}{3x-1}\) | A1 | Correct simplification |
| Answer | Marks | Guidance |
|---|---|---|
| \(9\left(\frac{2}{3}\right)^3 + p\left(\frac{2}{3}\right)^2 - \frac{2}{3} - 2 = -4\) | M1 | Using remainder theorem with \(x = \frac{2}{3}\) |
| \(\frac{8}{3} + \frac{4p}{9} - \frac{2}{3} - 2 = -4\) | ||
| \(p = 18\) | A1 |
# Question 2:
## Part (a)(i)
| $f\left(-\frac{1}{3}\right) = 9\left(-\frac{1}{3}\right)^3 + 18\left(-\frac{1}{3}\right)^2 - \left(-\frac{1}{3}\right) - 2$ | M1 | Substituting $x = -\frac{1}{3}$ |
|---|---|---|
| $= -\frac{1}{3} + 2 + \frac{1}{3} - 2 = 0$, therefore $(3x+1)$ is a factor | A1 | Correct evaluation showing $= 0$ |
## Part (a)(ii)
| $f(x) = (3x+1)(3x^2 + 5x - 2)$ | M1 A1 | |
|---|---|---|
| $= (3x+1)(3x-1)(x+2)$ | A1 | All three factors correct |
## Part (a)(iii)
| $9x^3 + 21x^2 + 6x = 3x(3x^2 + 7x + 2) = 3x(3x+1)(x+2)$ | M1 A1 | Factorising numerator |
|---|---|---|
| $\frac{3x(3x+1)(x+2)}{(3x+1)(3x-1)(x+2)} = \frac{3x}{3x-1}$ | A1 | Correct simplification |
## Part (b)
| $9\left(\frac{2}{3}\right)^3 + p\left(\frac{2}{3}\right)^2 - \frac{2}{3} - 2 = -4$ | M1 | Using remainder theorem with $x = \frac{2}{3}$ |
|---|---|---|
| $\frac{8}{3} + \frac{4p}{9} - \frac{2}{3} - 2 = -4$ | | |
| $p = 18$ | A1 | |
---2
\begin{enumerate}[label=(\alph*)]
\item The polynomial $\mathrm { f } ( x )$ is defined by $\mathrm { f } ( x ) = 9 x ^ { 3 } + 18 x ^ { 2 } - x - 2$.
\begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $3 x + 1$ is a factor of $\mathrm { f } ( x )$.
\item Express $\mathrm { f } ( x )$ as a product of three linear factors.
\item Simplify $\frac { 9 x ^ { 3 } + 21 x ^ { 2 } + 6 x } { \mathrm { f } ( x ) }$.
\end{enumerate}\item When the polynomial $9 x ^ { 3 } + p x ^ { 2 } - x - 2$ is divided by $3 x - 2$, the remainder is - 4 . Find the value of the constant $p$.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2011 Q2 [10]}}