Find the $x$-coordinates of all the points on the curve $y = 2{x^4} + 6{x^3} + \frac{7}{2}{x^2} - 5x + \frac{3}{2}$ at which

the tangent to the curve is parallel to the tangent at $( - 1,{\text{ }}6)$.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 8{x^3} + 18{x^2} + 7x - 5$    A1

when $x =  - 1,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 2$     A1

$8{x^3} + 18{x^2} + 7x - 5 =  - 2$    M1

$8{x^3} + 18{x^2} + 7x - 3 = 0$

$(x + 1)$ is a factor     A1

$8{x^3} + 18{x^2} + 7x - 3 = (x + 1)(8{x^2} + 10x - 3)$    (M1)

Note:     M1 is for attempting to find the quadratic factor.

$(x + 1)(4x - 1)(2x + 3) = 0$

$(x =  - 1),{\text{ }}x = 0.25,{\text{ }}x =  - 1.5$    (M1)A1

Note:     M1 is for an attempt to solve their quadratic factor.

[7 marks]

The first half of the question was accessible to all the candidates. Some though saw the word ‘tangent’ and lost time calculating the equation of this. It was a pity that so many failed to spot that $x + 1$ was a factor of the cubic and so did not make much progress with the final part of this question.