| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Polynomial identity or expansion |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing routine application of Factor and Remainder Theorems with standard polynomial manipulation. Part (a) involves direct substitution to verify a factor then simple algebraic division/factorisation, while part (b) requires basic remainder calculation and expressing in quotient form—all standard textbook exercises requiring minimal problem-solving beyond method recall. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(p(-3) = (-3)^3 - 5(-3)^2 - 8(-3) + 48\) | M1 | Clear attempt at \(p(-3)\), NOT long division; must see powers of \(-3\) simplified correctly |
| \(= -27 - 45 + 24 + 48 = 0\), therefore \(x+3\) is a factor | A1 | Working showing \(p(-3)=0\) and correct statement |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x^2 + bx + c\) with \(b=-8\) or \(c=16\) | M1 | By inspection; may see as quotient in long division |
| \(x^2 - 8x + 16\) | A1 | |
| \(p(x) = (x+3)(x-4)(x-4)\) | A1 | Must see product |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(p(2) = 2^3 - 5\times2^2 - 8\times2 + 48 = 8-20-16+48\) | M1 | Clear attempt at \(p(2)\), NOT long division |
| (Remainder \(=\)) 20 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Quadratic factor \(x^2+bx+c\), \(b=-3\) or \(c=-14\) | M1 | By inspection; may see as quotient in long division |
| \(x^2 - 3x - 14\) | A1 | |
| \(p(x) = (x-2)(x^2-3x-14)+20\) | A1 | Must see full correct expression |
# Question 4:
## Part (a)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $p(-3) = (-3)^3 - 5(-3)^2 - 8(-3) + 48$ | M1 | Clear attempt at $p(-3)$, NOT long division; must see powers of $-3$ simplified correctly |
| $= -27 - 45 + 24 + 48 = 0$, therefore $x+3$ is a factor | A1 | Working showing $p(-3)=0$ and correct statement |
## Part (a)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| $x^2 + bx + c$ with $b=-8$ or $c=16$ | M1 | By inspection; may see as quotient in long division |
| $x^2 - 8x + 16$ | A1 | |
| $p(x) = (x+3)(x-4)(x-4)$ | A1 | Must see product |
## Part (b)(i)
| Working | Mark | Guidance |
|---------|------|----------|
| $p(2) = 2^3 - 5\times2^2 - 8\times2 + 48 = 8-20-16+48$ | M1 | Clear attempt at $p(2)$, NOT long division |
| (Remainder $=$) 20 | A1 | |
## Part (b)(ii)
| Working | Mark | Guidance |
|---------|------|----------|
| Quadratic factor $x^2+bx+c$, $b=-3$ or $c=-14$ | M1 | By inspection; may see as quotient in long division |
| $x^2 - 3x - 14$ | A1 | |
| $p(x) = (x-2)(x^2-3x-14)+20$ | A1 | Must see full correct expression |
---4 The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 3 } - 5 x ^ { 2 } - 8 x + 48$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Use the Factor Theorem to show that $x + 3$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x )$ as a product of three linear factors.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the Remainder Theorem to find the remainder when $\mathrm { p } ( x )$ is divided by $x - 2$.
\item Express $\mathrm { p } ( x )$ in the form $( x - 2 ) \left( x ^ { 2 } + b x + c \right) + r$, where $b , c$ and $r$ are integers. [3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2016 Q4 [10]}}