A2 June 2025 Paper 1 Q7

AQACurrent spec4 marksUsing Roots of Polynomials

7 The cubic equation

\[5z^3 + 4z^2 - z + 3 = 0\]

has roots \(\alpha\), \(\beta\) and \(\gamma\)

Find an equation, with integer coefficients, that has roots \(2\alpha - 1\), \(2\beta - 1\) and \(2\gamma - 1\) [4 marks]

Mark scheme

Scheme	Marks	AO
Obtains \(\dfrac{w + 1}{2}\) Accept any letter for \(w\) or Obtains \(2(\alpha + \beta + \gamma) - 3\) PI by \(-\dfrac{23}{5}\)	B1	1.1b
Substitutes \(\dfrac{w \pm 1}{2}\) or \(\dfrac{w}{2} \pm 1\) into the cubic expression. Accept any letter for \(w\) or Writes the new pairwise sum of roots as \(4(\alpha\beta + \beta\gamma + \gamma\alpha) - 4(\alpha + \beta + \gamma) + 3\) PI by \(\dfrac{27}{5}\)	M1	3.1a
Obtains a four-term cubic expression from the substitution method or Writes the new product of roots as \(8\alpha\beta\gamma - 4(\alpha\beta + \beta\gamma + \gamma\alpha)\) \(+ 2(\alpha + \beta + \gamma) - 1\) PI by \(-\dfrac{33}{5}\)	M1	1.1a
Obtains \(k(5w^3 + 23w^2 + 27w + 33) = 0\) Where \(k\) is an integer	A1	1.1b
	(4 marks)

Typical solution

\[w = 2z - 1 \Rightarrow z = \frac{w + 1}{2}\]\[5\left(\frac{w + 1}{2}\right)^3 + 4\left(\frac{w + 1}{2}\right)^2 - \left(\frac{w + 1}{2}\right) + 3 = 0\]\[\frac{5}{8}(w^3 + 3w^2 + 3w + 1) + (w^2 + 2w + 1) - \frac{1}{2}(w + 1) + 3 = 0\]\[\frac{1}{8}(5w^3 + 15w^2 + 15w + 5 + 8w^2 + 16w + 8 - 4w - 4 + 24) = 0\]\[5w^3 + 23w^2 + 27w + 33 = 0\]