Question 396308
If 2-i and 3i are the zeros how do you get to an equation from that
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Complex solutions occur in conjugate pairs, so 2+i and -3i are also solutions.
{{{(x - (2-i))*(x - (2+i))*(x - 3i)*(x + 3i) = 0}}}
{{{(x*2 -4x + 5)*(x^2 + 9) = 0}}}
You can multiply that.