AQA AS Paper 2 2022 June — Question 5 5 marks

Exam BoardAQA
ModuleAS Paper 2 (AS Paper 2)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeFactorise polynomial completely
DifficultyModerate -0.8 This is a straightforward application of the factor theorem requiring verification of f(3)=0, identification of a sign error in Statement 2 (should be x-3, not x+3), and routine polynomial division to complete the factorisation. The question tests basic understanding rather than problem-solving skills, making it easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

5 Kaya is investigating the function $$f ( x ) = 2 x ^ { 3 } - 7 x ^ { 2 } - 12 x + 45$$ Kaya makes two statements.
Statement 1: \(\mathrm { f } ( 3 ) = 0\) Statement 2: this shows that ( \(x + 3\) ) must be a factor of \(\mathrm { f } ( x )\).
5
  1. State, with a reason, whether each of Kaya's statements is correct. Statement 1: \(\_\_\_\_\) Statement 2: \(\_\_\_\_\) 5
  2. Fully factorise f (x).

Question 5(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f(3) = 54 - 63 - 36 + 45 = 0\), Kaya is correct that \(f(3)=0\)E1 AO 2.4 — shows 3 substituted and \(f(3)=0\) stated
But Kaya's conclusion is wrong; \((x-3)\) is a factor of \(f(x)\)E1 AO 2.3 — states Kaya incorrect and gives correct conclusion. Or states \(f(-3) \neq 0\). Or states \((x+3)\) is not a factor
Question 5(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\((2x^3 - 7x^2 - 12x + 45) \div (x-3) = 2x^2 - x - 15\)M1 AO 1.1a — divides \(f(x)\) by \((x-3)\) or uses inspection; must obtain \(2x^2\) or \(15\)
\((x-3)^2(2x+5)\) — correct quadratic factor and both remaining roots of \(f(x)=0\)A1 AO 1.1b
Fully correct factorisation of \(f(x)\)A1 AO 1.1b — ISW
## Question 5(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(3) = 54 - 63 - 36 + 45 = 0$, Kaya is correct that $f(3)=0$ | E1 | AO 2.4 — shows 3 substituted and $f(3)=0$ stated |
| But Kaya's conclusion is wrong; $(x-3)$ is a factor of $f(x)$ | E1 | AO 2.3 — states Kaya incorrect and gives correct conclusion. Or states $f(-3) \neq 0$. Or states $(x+3)$ is not a factor |

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## Question 5(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(2x^3 - 7x^2 - 12x + 45) \div (x-3) = 2x^2 - x - 15$ | M1 | AO 1.1a — divides $f(x)$ by $(x-3)$ or uses inspection; must obtain $2x^2$ or $15$ |
| $(x-3)^2(2x+5)$ — correct quadratic factor and **both** remaining roots of $f(x)=0$ | A1 | AO 1.1b |
| Fully correct factorisation of $f(x)$ | A1 | AO 1.1b — ISW |

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5 Kaya is investigating the function

$$f ( x ) = 2 x ^ { 3 } - 7 x ^ { 2 } - 12 x + 45$$

Kaya makes two statements.\\
Statement 1: $\mathrm { f } ( 3 ) = 0$\\
Statement 2: this shows that ( $x + 3$ ) must be a factor of $\mathrm { f } ( x )$.\\
5
\begin{enumerate}[label=(\alph*)]
\item State, with a reason, whether each of Kaya's statements is correct.

Statement 1: $\_\_\_\_$\\

Statement 2: $\_\_\_\_$\\

5
\item Fully factorise f (x).
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2022 Q5 [5]}}