# 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

#### Explanation:

Since the function has a zero of *also* be a zero.

Thus

Since

#(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

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