The equation is given as:
z3−3z2−8z+30=0
Let's type the equation in the form of:
z3−6z2+3z2+10z−18z+30=0
After some substitutions:
z3−6z2+10z+3z2−18z+30=0
Realease z from the first three components, and 3 from the remaining ones:
z(z2−6z+10)+3(z2−6z+10)=0
Then, release z2−6z+10 from both multiplications, yields
(z+3)(z2−6z+10)=0
The root of the first bracket is:
z=−3 ;
As for the second bracket, we know that complex roots occur in conjugate pairs:
as 3+i is a root, so must the other be 3−i
Now, sum of the roots =3+i+3−i=6 and,
the product of the roots =(3+i)(3−i)=9−3i+3i−i2=9+1=10
Hence, the roots of the equation are: −3;3−i;3+i
Comments