How do you find all the zeros of #x^4 − 4x^3 + 14x^2 − 4x + 13# with the zero 2-3i?

1 Answer
Feb 27, 2016

Answer:

#x=i,-i,2+3i,2-3i#

Explanation:

Since the function has a zero of #2-3i#, and all the coefficients of the polynomial are real, you know that the complex conjugate of the zero will also be a zero.

Thus #2-3i# and #2+3i# are zeros of the polynomial.

Since #2-3i# and #2+3i# are roots, we know that two of the polynomial's factors are #(x-(2-3i))# and #(x-(2+3i))#.

#(x-(2-3i))(x-(2+3i))#

#=(x-2+3i)(x-2-3i)#

#=((x-2)+3i)((x-2)-3i)#

#=(x-2)^2-(3i)^2#

#=x^2-4x+4+9#

#=x^2-4x+13#

The remaining factors of the polynomial can be found through

#(x^4-4x^3+14x^2-4x+13)/(x^2-4x+13)=x^2+1#

Thus the remaining zeros can be solved through

#x^2+1=0#

#x^2=-1#

#x=+-i#

The function has four imaginary zeros and never crosses the #x#-axis.

graph{x^4-4x^3+14x^2-4x+13 [-5, 5, -21.47, 120]}