Easy -1.8 This is a straightforward factorisation requiring only basic algebraic manipulation: extract common factor x, then recognise difference of two squares. No problem-solving or conceptual depth required—purely routine algebraic technique for C1 level.
Take out common factor, usually \(x\), to produce \(x(25-9x^2)\). Accept \((x\pm0)(25-9x^2)\) or \(-x(9x^2-25)\). Other options: \((5+3x)(5x-3x^2)\) or \((5-3x)(5x+3x^2)\). Must be correct.
\((25-9x^2) = (5+3x)(5-3x)\)
M1
For factorising their quadratic term, usually \((25-9x^2)=(5+3x)(5-3x)\). Accept sign errors. If \((5\pm3x)\) taken out first, this is for attempt to factorise \((5x\mp3x^2)\)
\(25x-9x^3 = x(5+3x)(5-3x)\)
A1
cao \(x(5+3x)(5-3x)\) or any equivalent with three factors e.g. \(x(5+3x)(-3x+5)\) or \(x(3x-5)(-3x-5)\) etc including \(-x(3x+5)(3x-5)\). isw if they go on to show \(x=0, \pm\frac{5}{3}\)
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $25x - 9x^3 = x(25 - 9x^2)$ | B1 | Take out common factor, usually $x$, to produce $x(25-9x^2)$. Accept $(x\pm0)(25-9x^2)$ or $-x(9x^2-25)$. Other options: $(5+3x)(5x-3x^2)$ or $(5-3x)(5x+3x^2)$. Must be correct. |
| $(25-9x^2) = (5+3x)(5-3x)$ | M1 | For factorising their quadratic term, usually $(25-9x^2)=(5+3x)(5-3x)$. Accept sign errors. If $(5\pm3x)$ taken out first, this is for attempt to factorise $(5x\mp3x^2)$ |
| $25x-9x^3 = x(5+3x)(5-3x)$ | A1 | cao $x(5+3x)(5-3x)$ or any equivalent with three factors e.g. $x(5+3x)(-3x+5)$ or $x(3x-5)(-3x-5)$ etc including $-x(3x+5)(3x-5)$. isw if they go on to show $x=0, \pm\frac{5}{3}$ |
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