Question

find an equation having -2, i, 3i as solutions

Answer 1

a polynomial equation?

Answer 2

(x+2)(x-i)(x-3i)=0

A polynomial equation in x

Answer 3

would look like
\[(x+2)(x-i)(x-3i)\] but if you want real coefficients then you need the conjugates as well
\[(x+2)(x-i)(x+i)(x-3i)(x+3i)\] will do it

Answer 4

\((x-i)(x+i)=x^2+1\) and \((x-3i)(x+3i)=x^2+9\) so if you want real coefficients multiply out \[(x+2)(x^2+1)(x^2+9)\]

Answer 5

thanks guys for your help

Answer 6

yw

Answer 7

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