Question

Find a third-degree polynomial function with real coefficients and with zeros at -4 and -3 5i

Answer 1

If -3+5i is a solution, then by the conjugate root theorem, -3-5i is also a solution. We find the polynomial by multiplying together the factors. If x = -4, then x + 4 is the factor. If x = -3+5i, then (x-(-3+5i)) is a factor, and so is (x-(-3-5i)). Simplifying those down gives us as the first factor as (x+3-5i) and the second as (x+3+5i). We can FOIL those 2 together to get their product, and then FOIL in x+4. FOILing the 2 complex factors together gives us $x^2+3x+15ix+3x+9+15i-15ix-15i-25i^2$. If we combine like terms and cross out things that cancel it's much easier than what it looks like there! It simplifies down to $x^2+6x-25i^2$. Since i^2 = -1, it simplifies further to $x^2+6x-25(-1)$ and finally, to $x^2+6x+25$. Now we will FOIL in x+4. $(x^2+6x+25)(x+4)=x^3+6x^2+25x+4x^2+24x+100$. Our final simplified third degree polynomial is $x^3+10x^2+49x+100$