Question 362448
Since {{{x=1+2i}}} is a zero so is {{{x=1-2i}}} since complex solutions only  come in complex conjugate pairs.
However since the degree is 4, without additional information on the other real root, you can't determine the polynomial.
Just for illustration assume the fourth root is {{{x=a}}}.
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{{{f(x)=(x+1)(x-a)(x-(1+2i))(x-(1-2i))}}}
{{{f(x)=(x+1)(x-a)(x^2-2x+5)}}}
{{{f(x)=(x^2+(1-a)x-a)(x^2-2x+5)}}}
{{{highlight(f(x)=x^4-(a+1)x^3+(a+3)x^2-(3a-5)x-5a)}}}