| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Prove root count with given polynomial |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing standard applications of the Factor/Remainder Theorems with routine algebraic manipulation. Part (a) requires simple substitution p(3), part (b) verifies p(-1)=0, part (c)(i) uses polynomial division or inspection to find the quadratic factor, and part (c)(ii) applies the discriminant—all textbook procedures with no problem-solving insight required. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(3) = 3^3 - 2 \times 3^2 + 3(= 27 - 18 + 3) = 12\) | M1 | \(p(3)\) attempted; not long division |
| A1 | \(p(3)\) attempted; not long division |
| Answer | Marks | Guidance |
|---|---|---|
| \(p(-1) = (-1)^3 - 2(-1)^2 + 3\) | M1 | \(p(-1)\) attempted; not long division |
| \(p(-1) = -1 - 2 + 3 = 0 \Rightarrow x + 1\) is a factor | A1 also | correctly shown \(= 0\) plus statement |
| Answer | Marks | Guidance |
|---|---|---|
| Quadratic factor \((x^2 - 3x + 3)\) | M1 | \(b = -3\) or \(c = 3\) by inspection or full long division attempt or comparing coefficients |
| \(\left(p(x)\right) = (x+1)(x^2 - 3x + 3)\) | A1 | must see correct product |
| Answer | Marks | Guidance |
|---|---|---|
| Discriminant of quadratic \(b^2 - 4ac = (-3)^2 - 4 \times 3\) | M1 | 'their' discriminant considered possibly within quadratic equation formula |
| \(b^2 - 4ac < 0 \Rightarrow\) no real roots from quadratic \(\Rightarrow\) only one real root | A1 also |
**5(a)**
$p(3) = 3^3 - 2 \times 3^2 + 3(= 27 - 18 + 3) = 12$ | M1 | $p(3)$ attempted; not long division
| A1 | $p(3)$ attempted; not long division
**5(b)**
$p(-1) = (-1)^3 - 2(-1)^2 + 3$ | M1 | $p(-1)$ attempted; not long division
$p(-1) = -1 - 2 + 3 = 0 \Rightarrow x + 1$ is a factor | A1 also | correctly shown $= 0$ plus statement
**5(c)(i)**
Quadratic factor $(x^2 - 3x + 3)$ | M1 | $b = -3$ or $c = 3$ by inspection or full long division attempt or comparing coefficients
$\left(p(x)\right) = (x+1)(x^2 - 3x + 3)$ | A1 | must see correct product
**5(c)(ii)**
Discriminant of quadratic $b^2 - 4ac = (-3)^2 - 4 \times 3$ | M1 | 'their' discriminant considered possibly within quadratic equation formula
$b^2 - 4ac < 0 \Rightarrow$ no real roots from quadratic $\Rightarrow$ only one real root | A1 also |
---5 The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - 2 x ^ { 2 } + 3$.
\begin{enumerate}[label=(\alph*)]
\item Use the Remainder Theorem to find the remainder when $\mathrm { p } ( x )$ is divided by $x - 3$.
\item Use the Factor Theorem to show that $x + 1$ is a factor of $\mathrm { p } ( x )$.
\item \begin{enumerate}[label=(\roman*)]
\item Express $\mathrm { p } ( x ) = x ^ { 3 } - 2 x ^ { 2 } + 3$ in the form $( x + 1 ) \left( x ^ { 2 } + b x + c \right)$, where $b$ and $c$ are integers.
\item Hence show that the equation $\mathrm { p } ( x ) = 0$ has exactly one real root.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q5 [8]}}