Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.2 - 3i is a zero of f(x) = x4 - 4x3 + 14x2 - 4x + 13.*Can someone show the work I have the answers

Answer:One other zero is 2+3iStep-by-step explanation:If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.The conjugate of (a+b) is (a-b).The conjugate of (a-b) is (a+b).The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.Ok so we have two zeros 2-3i and 2+3i.This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).(x-(2-3i))(x-(2+3i))Foil!First: x(x)=x^2Outer: x*-(2+3i)=-x(2+3i)Inner:  -(2-3i)(x)=-x(2-3i)Last:  (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates) ---------------------------------Add together:x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2) Simplifying:x^2-2x-3ix-2x+3ix+4+9  (since i^2=-1)x^2-4x+13                     (since -3ix+3ix=0)So x^2-4x+13 is a factor of the given polynomial.I'm going to do long division to find another factor.Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.                 x^2+1               ---------------------------------------x^2-4x+13|  x^4-4x^3+14x^2-4x+13                                   -( x^4-4x^3+ 13x^2)             ------------------------------------------                                  x^2-4x+13                                -(x^2-4x+13)                                -----------------                                     0So the other factor is x^2+1.To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.[tex]x^2+1=0[/tex][tex]x^2=-1[/tex][tex]x=\pm \sqrt{-1}[/tex][tex]x=\pm i[/tex]So the zeros are i, -i , 2-3i , 2+3i