| Exam Board | AQA |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Factor & Remainder Theorem |
| Type | Verify factor then solve related equation |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on the Factor Theorem requiring standard techniques: substituting x = -1/5 to verify the factor, polynomial division to find remaining factors, then a simple substitution n → 2n to show divisibility. While it has multiple parts (7+ marks total), each step follows directly from the previous with no novel insight required, making it slightly easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substitutes \(x = -\frac{1}{5}\) into \(125x^3 + 150x^2 + 55x + 6\): \(125\left(-\frac{1}{5}\right)^3 + 150\left(-\frac{1}{5}\right)^2 + 55\left(-\frac{1}{5}\right) + 6 = 0\) | M1 (AO 1.1a) | Must see \(-\frac{1}{5}\) bracketed correctly in cubed and squared terms. Or further step showing correct evaluation e.g. \(-1 + 6 - 11 + 6 = 0\) |
| Since \(P\left(-\frac{1}{5}\right) = 0\), \((5x+1)\) must be a factor of \(P(x)\) | R1 (AO 2.1) | Completes factor theorem argument. Statement can come first but must be in right direction AND accompanied by evaluation. Not: \((5x+1)\) is a factor \(\Rightarrow P\left(-\frac{1}{5}\right) = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Obtains quadratic factor of form \(25x^2 + bx + 6\) or states other roots | M1 (AO 1.1a) | PI by correct answer |
| Obtains second linear factor; condone \((x + 0.4)\) or \((x + 0.6)\) | M1 (AO 1.1a) | PI by correct answer |
| \((5x+1)(5x+2)(5x+3)\) | A1 (AO 1.1b) | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Deduces \(250n^3 + 300n^2 + 110n + 12 = 2(5n+1)(5n+2)(5n+3)\) | M1 (AO 2.2a) | FT three factors from part (b). Condone use of different letter to \(n\) |
| Explains \((5n+1)\), \((5n+2)\) and \((5n+3)\) are three consecutive positive whole numbers, so factors must contain a multiple of 3 and must also contain a multiple of 2 | R1 (AO 2.4) | Must have three factors in a form giving consecutive positive whole numbers |
| The extra 2 gives \(2 \times 2 \times 3 = 12\), therefore \(250n^3 + 300n^2 + 110n + 12\) is a multiple of 12 | R1 (AO 2.4) | Completes reasoned argument. Condone conclusion about \(2(5n+1)(5n+2)(5n+3)\) |
## Question 13(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substitutes $x = -\frac{1}{5}$ into $125x^3 + 150x^2 + 55x + 6$: $125\left(-\frac{1}{5}\right)^3 + 150\left(-\frac{1}{5}\right)^2 + 55\left(-\frac{1}{5}\right) + 6 = 0$ | M1 (AO 1.1a) | Must see $-\frac{1}{5}$ bracketed correctly in cubed and squared terms. Or further step showing correct evaluation e.g. $-1 + 6 - 11 + 6 = 0$ |
| Since $P\left(-\frac{1}{5}\right) = 0$, $(5x+1)$ must be a factor of $P(x)$ | R1 (AO 2.1) | Completes factor theorem argument. Statement can come first but must be in right direction AND accompanied by evaluation. Not: $(5x+1)$ is a factor $\Rightarrow P\left(-\frac{1}{5}\right) = 0$ |
**Subtotal: 2 marks**
---
## Question 13(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Obtains quadratic factor of form $25x^2 + bx + 6$ or states other roots | M1 (AO 1.1a) | PI by correct answer |
| Obtains second linear factor; condone $(x + 0.4)$ or $(x + 0.6)$ | M1 (AO 1.1a) | PI by correct answer |
| $(5x+1)(5x+2)(5x+3)$ | A1 (AO 1.1b) | OE |
**Subtotal: 3 marks**
---
## Question 13(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Deduces $250n^3 + 300n^2 + 110n + 12 = 2(5n+1)(5n+2)(5n+3)$ | M1 (AO 2.2a) | FT three factors from part (b). Condone use of different letter to $n$ |
| Explains $(5n+1)$, $(5n+2)$ and $(5n+3)$ are three consecutive positive whole numbers, so factors must contain a multiple of 3 and must also contain a multiple of 2 | R1 (AO 2.4) | Must have three factors in a form giving consecutive positive whole numbers |
| The extra 2 gives $2 \times 2 \times 3 = 12$, therefore $250n^3 + 300n^2 + 110n + 12$ is a multiple of 12 | R1 (AO 2.4) | Completes reasoned argument. Condone conclusion about $2(5n+1)(5n+2)(5n+3)$ |
**Subtotal: 3 marks**
**Question Total: 8 marks**13
\begin{enumerate}[label=(\alph*)]
\item Given that
$$P ( x ) = 125 x ^ { 3 } + 150 x ^ { 2 } + 55 x + 6$$
use the factor theorem to prove that ( $5 x + 1$ ) is a factor of $\mathrm { P } ( x )$.\\[0pt]
[2 marks]\\
13
\item Factorise $\mathrm { P } ( x )$ completely.\\
13
\item Hence, prove that $250 n ^ { 3 } + 300 n ^ { 2 } + 110 n + 12$ is a multiple of 12 when $n$ is a positive whole number.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 1 2021 Q13 [8]}}