x4-3x³+6x²+2x-60 = 0
Use synthetic division:
1+3i | 1 -3 6 2 -60
| 1+3i -11-3i 4-18i 60
1 -2+3i -5-3i 6-18i 0
So the first factorization is:
[x-(1+3i)][x³+(-2+3i)x²+(-5-3i)x+(6-18i)] = 0
Since 1+3i is a solution, so is its conjugate 1-3i
1-3i | 1 -2+3i -5-3i 6-18i
| 1-3i -1+3i -6+18i
1 -1 -6
The second factorization is:
[x-(1+3i)][x-(1-3i)](x²-x-6) = 0
The third and final factorization is:
[x-(1+3i)][x-(1-3i)](x-3)(x+2) = 0
x-(1+3i)=0
x=1+3i
x-(1-3i)=0
x=1-3i
x-3=0
x=3
x+2=0
x=-2
The four solutions are 1+3i, 1-3i, 3, -2
Edwin
