a)10 have next dividers: ±1;±2;±5;±10;\pm 1;\pm2; \pm 5; \pm10;±1;±2;±5;±10;
For 1:(1)3+1+10≠0For −1:(−1)3+(−1)+10≠0For 2:(2)3+2+10≠0For −2:(−2)3+(−2)+10=0For \space 1:\newline (1)^3+1+10\not=0\newline For \space -1:\newline (-1)^3+(-1)+10\not=0\newline For \space 2:\newline (2)^3+2+10\not=0\newline For \space -2:\newline (-2)^3+(-2)+10=0\newlineFor 1:(1)3+1+10=0For −1:(−1)3+(−1)+10=0For 2:(2)3+2+10=0For −2:(−2)3+(−2)+10=0
So z=-2 is a root of this equation
b)z3−3z2−8z+30z−3−i=z2+iz+(−9+3i)\dfrac{z^3-3z^2-8z+30}{z-3-i}=z^2+iz+(-9+3i)\newlinez−3−iz3−3z2−8z+30=z2+iz+(−9+3i)
z2+iz+(−9+3i)=0D=i2−4(−9+3i)=35−12iD=35−12iz1=−i+35−12i2z2=−i−35−12i2z^2+iz+(-9+3i)=0\newline D=i^2-4(-9+3i)=35-12i\newline \sqrt{D}=\sqrt{35-12i}\newline z_1=\dfrac{-i+\sqrt{35-12i}}{2}\newline z_2=\dfrac{-i-\sqrt{35-12i}}{2}\newlinez2+iz+(−9+3i)=0D=i2−4(−9+3i)=35−12iD=35−12iz1=2−i+35−12iz2=2−i−35−12i