N2007 P1 Q7

The polynomial \(P(z)\) has real coefficients. The equation \(P(z)=0\) has a root \(r\mathrm e^{\mathrm i\theta}\), where \(r>0\) and \(0<\theta<\pi\).

1. Write down a second root in terms of \(r\) and \(\theta\), and hence show that a quadratic factor of \(P(z)\) is \(z^2-2rz\cos\theta+r^2\).[3]
2. Solve the equation \(z^6=-64\), expressing the solutions in the form \(r\mathrm e^{\mathrm i\theta}\), where \(r>0\) and \(-\pi<\theta\leq\pi\).[4]
3. Hence, or otherwise, express \(z^6+64\) as the product of three quadratic factors with real coefficients, giving each factor in non-trigonometrical form.[3]