Question 1010315

find all other zeros give 2i is one zero.
f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12
<pre>One zero is 2i, so the other, its conjugate is: - 2i
This gives us: (x - 2i)(x + 2i), or {{{x^2 + 4}}}
Dividing {{{x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12}}} by {{{x^2 + 4}}} results in:
{{{x^3 - 3x^2 + x - 3}}}, which can be factored into: 
{{{x^2(x - 3) + 1(x - 3)}}}
{{{(x^2 + 1)(x - 3)}}}
{{{(x + i)(x - i)(x - 3)}}}.
This leads to zeroes of: - i, i, and 3.
Thus the zeroes of {{{f(x) = x^5 - 3x^4 + 5x^3 - 15x^2 + 4x - 12}}} are: {{{highlight_green(system(x =" "+- 2i, x =" " +-i, x = 3))}}}