(v) z2−2z−1z^2 - 2z - 1z2−2z−1 Solution Find roots using quadratic formula:z=2±4+42=2±82=2±222=1±2Therefore:z2−2z−1=(z−(1+2))(z−(1−2))\begin{aligned} & \boxed{\text{Find roots using quadratic formula:}} \\ \\ & z = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2} \\ \\ & \boxed{\text{Therefore:}} \\ \\ & z^2 - 2z - 1 = (z - (1 + \sqrt{2}))(z - (1 - \sqrt{2})) \end{aligned}Find roots using quadratic formula:z=22±4+4=22±8=22±22=1±2Therefore:z2−2z−1=(z−(1+2))(z−(1−2))