Find a cubic polynomial in standard form with real coefficients, having the zeros 4 and 5i. let the leading coefficient be 1.

If one of our zeros is 4, then the factor is x-4.  If the second zero is 5i, then the conjugate root theorem says there HAS to be a root that is -5i.  So our 3 factors are (x-4)(x+5i)(x-5i).  We will FOIL out these factors to get the polynomial.  Let's start with the ones that contain the imaginary numbers.  Doing that mutliplication we get x^2-25i^2.  i^2 is equal to -1, so what that expression simplifies down to is .  Now we will multiply in that last factor of (x-4):  .  FOILing out we have .  There you go!

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