a)10 have next dividers: "\\pm 1;\\pm2; \\pm 5; \\pm10;"
"For \\space 1:\\newline\n(1)^3+1+10\\not=0\\newline\n\nFor \\space -1:\\newline\n(-1)^3+(-1)+10\\not=0\\newline\n\nFor \\space 2:\\newline\n(2)^3+2+10\\not=0\\newline\n\nFor \\space -2:\\newline\n(-2)^3+(-2)+10=0\\newline"
So z=-2 is a root of this equation
b)"\\dfrac{z^3-3z^2-8z+30}{z-3-i}=z^2+iz+(-9+3i)\\newline"
"z^2+iz+(-9+3i)=0\\newline\nD=i^2-4(-9+3i)=35-12i\\newline\n\\sqrt{D}=\\sqrt{35-12i}\\newline\nz_1=\\dfrac{-i+\\sqrt{35-12i}}{2}\\newline\nz_2=\\dfrac{-i-\\sqrt{35-12i}}{2}\\newline"
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