(viii) 2z2−22z+652z^2 - 22z + 65 2z2−22z+65 Solution Find roots using quadratic formula:z=22±484−5204=22±−364=22±6i4=11±3i2Therefore:2z2−22z+65=2(z−11+3i2)(z−11−3i2)=2(2z−11−3i2)(2z−11+3i2)=24(2z−11−3i)(2z−11+3i)=12(2z−11−3i)(2z−11+3i)\begin{aligned} & \boxed{\text{Find roots using quadratic formula:}} \\ \\ & z = \frac{22 \pm \sqrt{484 - 520}}{4} = \frac{22 \pm \sqrt{-36}}{4} = \frac{22 \pm 6i}{4} = \frac{11 \pm 3i}{2} \\ \\ & \boxed{\text{Therefore:}} \\ \\ & 2z^2 - 22z + 65 = 2\left(z - \frac{11 + 3i}{2}\right)\left(z - \frac{11 - 3i}{2}\right) \\ \\ & = 2\left(\frac{2z - 11 - 3i}{2}\right)\left(\frac{2z - 11 + 3i}{2}\right) \\ \\ & = \frac{2}{4}(2z - 11 - 3i)(2z - 11 + 3i) \\ \\ & = \frac{1}{2}(2z - 11 - 3i)(2z - 11 + 3i) \end{aligned}Find roots using quadratic formula:z=422±484−520=422±−36=422±6i=211±3iTherefore:2z2−22z+65=2(z−211+3i)(z−211−3i)=2(22z−11−3i)(22z−11+3i)=42(2z−11−3i)(2z−11+3i)=21(2z−11−3i)(2z−11+3i)