f(z)=2z3−z2+4z+p=0By the factor theorem,f(1+2i)=02(1+2i)3−(1+2i)2+4(1+2i)+p=02(1+6i−12−8i)−(1+4i−4)+4+8i+p=02(−2i−11)−(−3+4i)+4+8i+p=0−4i−22+3−4i+4+8i+p=0∴p=15By the Fundamental theorem of Algebrathe conjugate of1+2iis also a root off(z)∴(z−(1+2i))(z−(1−2i))=0z2−(1+2i+1−2i)z+(1+4)=0z2−2z+5=0By Long division, we obtain the last factor2z+3z2−2z+5∣2z3−z2+4z+15−(2z3−4z2+10z)3z2−6z+15−(3z2−6z+15)=0⟹(z2−2z+5)(2z+3)=02z+3=0⟹z=−23∴z=−23is another solution.(i)−23(ii)p=15
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