A bus route's extra braking distance, in metres, beyond a safety threshold is modelled by \(3x^3 - 12x\), where \(x\) is the speed in tens of kilometres per...
Question 1 Report
A bus route's extra braking distance, in metres, beyond a safety threshold is modelled by \(3x^3 - 12x\), where \(x\) is the speed in tens of kilometres per hour above the threshold.
- Factorise \(3x^3 - 12x\) fully. (2)
- Hence write down the three values of \(x\) for which the extra braking distance is zero. (2)
Answer Details
Factorising a cubic expression usually begins by taking out any common factor from every term, which often leaves a simpler expression, such as a difference of two squares, that can be factorised further.
(a) Every term in \( 3x^3 - 12x \) shares a common factor of \( 3x \): taking this out gives \( 3x(x^2-4) \) [1 mark]. The remaining bracket, \( x^2 - 4 \), is a difference of two squares, \( x^2 - 2^2 \), which factorises as \( (x-2)(x+2) \), giving the full factorisation \( 3x^3-12x = 3x(x-2)(x+2) \) [1 mark].
(b) The product \( 3x(x-2)(x+2) \) is zero exactly when any one of its three factors is zero: \( 3x = 0 \) gives \( x = 0 \); \( x - 2 = 0 \) gives \( x = 2 \); \( x + 2 = 0 \) gives \( x = -2 \). So the extra braking distance is zero at \( x = 0 \), \( x = 2 \) and \( x = -2 \) [2 marks].
A product of several factors is zero only when at least one of the factors itself is zero; this "zero product rule" is what turns a fully factorised cubic directly into a list of solutions without any further algebra.
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