Question 827428
<pre>

x<sup>4</sup>-3x³+6x²+2x-60 = 0

Use synthetic division:

1+3i | 1  -3      6     2      -60
     |<u>     1+3i -11-3i  4-18i   60</u>
       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
     |<u>     1-3i  -1+3i -6+18i</u>
       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</pre>