Question 370375
{{{G(x)=x^3-(1-i)x^2-(8-i)x+(12-6i)}}}
If 2-i is a zero, then (x-(2-i)) or (x - 2 + i) is a factor of G(x). We need to find the other factor and for this we can use Synthetic Division:
<pre>
2-i |   1  -1+i  -8+i  12-6i
-----       2-i   2-i -12+6i
       ----------------------
        1   1    -6     0
</pre>
Te remainder is zero so 2-i is indeed a factor. And the rest of that row of numbers, 1 1 -6, tells us the other factor: {{{1x^2 + 1x - 6}}}. So now
{{{G(x) = (x-2+i)(x^2+x-6)}}}
The second factor is a trinomial that is easily factored giving:
{{{G(x) = (x-2+i)(x+3)(x-2))}}}
G(x) is now fully factored.