Question 62290
find a polynomial of lowest degree that has real coefficients, a leading coefficient of 1, and with 5-2i and 3 as two of its zeroes.
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If 5-2i is a zero and it has real coefficients then 5+2i is also
a zero.
So f(x)=(x-3)(x-(5-2i))(x-(5+2i))
f(x)= (x-3)((x-5)^2+4)
f(x)= (x-3)(x^2-10x+29)
f(x)=x^3-10x^2-3x^2+29x-30x-12
f(x)=x^3-13x^2-x-12
Cheers,
stan H.