The polynomial 6x³ + ax² + bx + 9 is denoted by p(x), where a and b are constants. It is given that (x-3) is a factor of p(x), and when the first derivative p'(x) is divided by (x – 3) the remainder is 72.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2025). The mark or mark type is shown in brackets.

1. [M1*] Substitute x = 3 or –3 into p(x) and equate to 0 or into p'(x) and equate to 72
2. [A1 OE] Obtain 162 + 9a + 3b + 9 = 0
3. [A1 OE] Obtain 162 + 6a + b = 72
4. [DM1] Solve simultaneous equations to obtain either a or b after using p(±3) = 0 and p'(±3) = 72
5. [A1] Obtain a = −11 and b = -24
6. [M1] Equate (x-3)(6x² + Ax + B) to 6x³ – 11x² – 24x + 9 and obtain equations to solve for A and B Using their a and their b. A - 18 = a = − 11, B – 3A = - b = - 24, - 3B = 9. A = 7 and B = -3. Or divide 6x³ – 11x² – 24x + 9 by x – 3 and reach 6x² ± 7x.
7. [A1 SOI] (x − 3)(6x² + 7x – 3)
8. [A1] Obtain (x - 3)(2x + 3)(3x – 1) Special Case: If only (x-3)(x+3/2)(x-1/3) or (x - 3)(2x + 3)(3x – 1) seen, SC B1 only (but can gain two marks in (c)).
9. [B1 FT] Obtain one correct region x < -3/2 or -1/3 < x < 3 Must be final answer not in working. FT is on the last two brackets (not (x-3)).
10. [B1 FT] Obtain both regions x < -3/2 and -1/3 < x < 3 Allow x < -3/2 and -1/3 < x < 3. SC B1 for x≤-3/2, -1/3≤x≤3. FT is on the last two brackets (not (x-3)). If incorrect factor or factors in (b) but correct regions here, allow SC B1 only.