Question 365323
find a degree 3 polynomial whose coefficients are real numbers and with given zeros: {{{-3}}}, {{{1-2i}}}

If {{{1-2i}}} is a zero, then so is {{{1+2i}}} so then {{{-3(1-2i)(1+2i)= 0}}}
Comment: That is not true; it is not what "zeroes" mean.
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Then using another example, I came up with this equation. But I'll be honest I don't understand why/how I did this??
Comment: The rule is "If a is a zero, (x-a) is a factor"
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f(x) = (x-(1+2i))(x-(1-2i))(x--3) 
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Rearrange to get:
f(x) = ((x-1)-2i)((x-1)+2i)(x+3)
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f(x) = [(x-1)^2 + 4][x+3]
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f(x) = [x^2-2x+5][x+3]
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f(x) = (x^2-2x+5)x + (x^2-2x+5)3
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f(x) = x^3-2x^2+5x+3x^2-6x+15
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f(x) = x^3+x^2-x+15
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You were correct.
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Cheers,
Stan H.