Do not use a calculator in this question. Find the coordinates of the points of intersection of the curve y = (2x+3)²(x-1) and the line y = 3 (2x+3).

Mark scheme

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

1. [M1] Either: (2x+3)² (x−1)=3(2x+3) (2x+3)(2x²+x−6)(= 0) For attempt to equate line and curve and attempt to simplify to 2x+3 × a quadratic factor or cancelling 2x + 3 and obtaining a quadratic factor
2. [M1] (2x+3)(2x²+x-6) = 0 (2x+3)(2x-3)(x+2) = 0 Dep for attempt at 3 linear factors from a linear term and a quadratic term
3. [B1] (-3/2, 0)
4. [A1] (3/2, 18) Dep on first M mark only
5. [A1] (-2, -3) Dep on first M mark only
6. [M1] Or: (2x+3)² (x−1)=3(2x+3) 4x³ +8x² -9x-18(=0) For attempt to equate line and curve and attempt to simplify to a cubic equation, by collecting like terms
7. [M1] (x+2)(4x²-9) (2x-3)(2x²+7x+6) (2x+3)(2x²+x−6) (2x+3)(2x-3)(x+2)(=0) Dep For attempt to find a factor from a 4 term cubic equation (usually x+2), do long division oe to obtain a quadratic factor and factorise this quadratic factor
8. [A1] (-3/2, 0)
9. [A1] (3/2, 18)
10. [A1] (-2, -3)