Question

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

Answer 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 $x^{2} +25$.  Now we will multiply in that last factor of (x-4):  $(x^2+25)(x-4)$.  FOILing out we have $x^3-4x^2+25x-100$.  There you go!