| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| 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 Remainder and Factor Theorems with routine algebraic manipulation. Part (a) is basic curve sketching, parts (b)(i)-(ii) are direct substitutions requiring no insight, part (iii) is polynomial division or comparison of coefficients, and part (iv) uses the discriminant—all textbook exercises with clear signposting and no problem-solving required. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Correct shape: cubic with repeated root at \(x=0\) (touching x-axis) | B1 | Touch at origin |
| Root/crossing at \(x = 3\) | B1 | Crosses x-axis at \(x=3\) |
| Correct overall shape (positive cubic, starting bottom-left) | B1 | General shape correct |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(p(4) = 4^2(4-3) + 20\) | M1 | Substituting \(x=4\) |
| \(= 16(1) + 20 = 36\) | A1 | Correct remainder = 36 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(p(-2) = (-2)^2(-2-3) + 20\) | M1 | Substituting \(x = -2\) |
| \(= 4(-5) + 20 = -20 + 20 = 0\), therefore \((x+2)\) is a factor | A1 | Correct working shown to conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((x+2)(x^2 + bx + c)\) by division or comparison | M1 | Attempt at factorisation/division |
| \(p(x) = (x+2)(x^2 - 2x + 10)\) so \(b=-2\), \(c=10\) | A1 | Both correct values |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Only real root from \((x+2)=0\) gives \(x = -2\) | B1 | Identifies \(x=-2\) as the one real root |
| Consider discriminant of \(x^2 - 2x + 10\): \(b^2 - 4ac = 4 - 40 = -36 < 0\) | M1 | Correct discriminant calculation |
| Since discriminant \(< 0\), quadratic has no real roots, so exactly one real root | A1 | Correct conclusion |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Correct shape: cubic with repeated root at $x=0$ (touching x-axis) | B1 | Touch at origin |
| Root/crossing at $x = 3$ | B1 | Crosses x-axis at $x=3$ |
| Correct overall shape (positive cubic, starting bottom-left) | B1 | General shape correct |
## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $p(4) = 4^2(4-3) + 20$ | M1 | Substituting $x=4$ |
| $= 16(1) + 20 = 36$ | A1 | Correct remainder = 36 |
## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $p(-2) = (-2)^2(-2-3) + 20$ | M1 | Substituting $x = -2$ |
| $= 4(-5) + 20 = -20 + 20 = 0$, therefore $(x+2)$ is a factor | A1 | Correct working shown to conclusion |
## Part (b)(iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(x+2)(x^2 + bx + c)$ by division or comparison | M1 | Attempt at factorisation/division |
| $p(x) = (x+2)(x^2 - 2x + 10)$ so $b=-2$, $c=10$ | A1 | Both correct values |
## Part (b)(iv):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Only real root from $(x+2)=0$ gives $x = -2$ | B1 | Identifies $x=-2$ as the one real root |
| Consider discriminant of $x^2 - 2x + 10$: $b^2 - 4ac = 4 - 40 = -36 < 0$ | M1 | Correct discriminant calculation |
| Since discriminant $< 0$, quadratic has no real roots, so exactly one real root | A1 | Correct conclusion |
I can see these are answer space pages from what appears to be an AQA MPC1 June 2015 exam paper. The pages shown (17-20) contain only blank answer spaces for questions 7 and 8 — there is no mark scheme content visible in these images.
To get the mark scheme content for these questions, I can work through the solutions based on the question text visible on page 18:
---7
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve with equation $y = x ^ { 2 } ( x - 3 )$.
\item The polynomial $\mathrm { p } ( x )$ is given by $\mathrm { p } ( x ) = x ^ { 2 } ( x - 3 ) + 20$.
\begin{enumerate}[label=(\roman*)]
\item Find the remainder when $\mathrm { p } ( x )$ is divided by $x - 4$.
\item Use the Factor Theorem to show that $x + 2$ is a factor of $\mathrm { p } ( x )$.
\item Express $\mathrm { p } ( x )$ in the form $( x + 2 ) \left( x ^ { 2 } + b x + c \right)$, where $b$ and $c$ are integers.
\item Hence show that the equation $\mathrm { p } ( x ) = 0$ has exactly one real root and state its value.\\[0pt]
[3 marks]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q7 [12]}}