June 2025 Paper 2 Q1
1.
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Factorise completely\[2x^3 - 24x^2 + 40x\](3)
| Scheme | Marks | AO |
|---|---|---|
| \(2x\left(x^2 - 12x + 20\right)\) or \(x\left(2x^2 - 24x + 40\right)\) | B1 | 1.1b |
| \(x^2 - 12x + 20 = (x - 2)(x - 10)\) or e.g. \(2x^2 - 24x + 40 = (2x - 4)(x - 10)\) | M1 | 1.1b |
| \(2x(x - 2)(x - 10)\) | A1 | 1.1b |
| (3) | ||
| (3 marks) |
Notes
B1: Takes a factor of \(2x\) or \(x\) out of the given cubic correctly.
Dividing by 2 and then achieving \(x\left(x^2 - 12x + 20\right)\) does not score this mark unless recovered.
M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for \((x \pm b)(x \pm d)\) where \(|bd| = 20\) coming from \(x^2 \pm 12x \pm 20\)
or for \((ax \pm b)(cx \pm d)\) where \(|ac| = 2\) and \(|bd| = 40\) coming from \(2x^2 \pm 24x \pm 40\)
May be scored if they have divided by \(x\) but not from e.g. \(2x^3 - 24x^2 + 40x \rightarrow (2x - 4)(x - 10)\) without clear indication that they have divided by or taken out a factor of \(x\) or \(2x\).
There may be incorrect intermediate steps such as \((2x - 20)(2x - 4)\) which do not score the mark on their own but may be ignored if they return to e.g. \((2x - 20)(x - 2)\).
A1: \(2x(x - 2)(x - 10)\) or e.g. \(2(x - 10)(x - 2)x\). Do not accept e.g. \(x(2x - 4)(x - 10)\). Ignore = 0
ISW after a fully correct factorisation e.g. \(2x(x - 2)(x - 10)\) that becomes \(x(x - 2)(x - 10)\) and ignore any attempt to find roots (before or after factorisation). Allow \((2x)(x - 2)(x - 10)\)
Alternative:
B1: Takes a factor of \((x - 2)\) or \((x - 10)\) out correctly i.e. \((x - 2)\left(2x^2 - 20x\right)\) or \((x - 10)\left(2x^2 - 4x\right)\). Must be seen as a product of factors and not just in a division attempt but may be implied by later work.
M1: Attempts to factorise their quadratic. Invisible brackets may be implied by later work.
Score for \(ax(bx \pm c)\) where \(|ab| = \text{``}2\text{''}\) and \(|ac| =\) their \(\text{``}20\text{''}\) or \(\text{``}4\text{''}\) coming from \(Ax^2 + Bx\)
A1: As main scheme.
Note: Solutions that solve the cubic = 0 to achieve \(x = 0\), 2 and 10 and then arrive at e.g. \(x(x - 2)(x - 10)\) score no marks without a prior line of working such as \(x\left(x^2 - 12x + 20\right)\) (would score M1) or \(x\left(2x^2 - 24x + 40\right)\) (would score B1).
Some examples:
- \(2x^3 - 24x^2 + 40x \rightarrow x(x - 2)(x - 10)\) scores B0M0A0
- \(2x^3 - 24x^2 + 40x \rightarrow (2x - 4)(x - 10)\) scores B0M0A0
- \(x^2 - 12x + 20 \rightarrow (x + 2)(x - 10)\) scores B0M1A0
- \(x^3 - 12x^2 + 20x \rightarrow x(x - 2)(x - 10)\) scores B0M1A0
- \(x\left(2x^2 - 24x + 40\right) \rightarrow x(x - 2)(x - 10)\) scores B1M0A0
- \(x\left(2x^2 - 24x + 40\right) \rightarrow \{x(2x + 2)(2x + 40) \rightarrow\}\ x(2x + 2)(x + 20)\) scores B1M1A0
- \(2x\left(x^2 - 12x + 20\right) \rightarrow 2x(x - 1)(x + 20)\) scores B1M1A0
- \(2x(x - 2)(x - 10)\) on its own scores B1M1A1