Easy -1.8 This is a straightforward factorisation requiring only recognition of a common factor (x) and difference of two squares (x² - 9). It's a single-step C1 question testing basic algebraic manipulation with no problem-solving element, making it significantly easier than average A-level questions.
\(x(x^2-9)\) or \((x\pm0)(x^2-9)\) or \((x-3)(x^2+3x)\) or \((x+3)(x^2-3x)\)
B1
B1: first factor taken out correctly; \(x(x^2+9)\) is B0
\(x(x-3)(x+3)\)
M1A1 (3)
M1: attempting to factorise a relevant quadratic; "ends" correct so \((x^2-9)=(x\pm p)(x\pm q)\) where \(pq=9\) is OK
A1: fully correct expression with all 3 factors; note \(-x(3-x)(x+3)\) scores A1
Treat any working to solve \(x^3-9x\) as ISW
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x(x^2-9)$ or $(x\pm0)(x^2-9)$ or $(x-3)(x^2+3x)$ or $(x+3)(x^2-3x)$ | B1 | B1: first factor taken out correctly; $x(x^2+9)$ is B0 |
| $x(x-3)(x+3)$ | M1A1 (3) | M1: attempting to factorise a relevant quadratic; "ends" correct so $(x^2-9)=(x\pm p)(x\pm q)$ where $pq=9$ is OK |
| | | A1: fully correct expression with all 3 factors; note $-x(3-x)(x+3)$ scores A1 |
| | | Treat any working to solve $x^3-9x$ as ISW |
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