Given that x-2 is a factor of ax³-12x²+5x+6, use the factor theorem to show that a = 4.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [M1] (i) Uses x = 2 as a root: a(2³) -12(2²) +5(2) + 6 = 0 or synthetic division setup
2. [A1] Solves 8a-48+10+6=0 to find a = 4 or solves -32+8a = 0 to find a = 4. If M0 then SC1 for 4(2³) -12(2²) +5(2) + 6 = 0 or showing that the synthetic division with a = 4 results in a remainder of 0
3. [B2] (ii) (x-2)(4x² - 4x - 3) (B1 for any two terms correct in quadratic factor)
4. [B1] Product of three correct linear factors: (x-2)(2x-3)(2x+1)
5. [B1] x = 2, x = 1.5, x = -0.5 oe (dep on all previous marks having been earned. If B2 B0 then award SC1 for correct factorisation of correct quadratic factor leading to 3 correct roots)