How do you find all the zeros of x4−4x3+14x2−4x+13 with the zero 2-3i?
1 Answer
Feb 27, 2016
Explanation:
Since the function has a zero of
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
=x2−4x+4+9
=x2−4x+13
The remaining factors of the polynomial can be found through
x4−4x3+14x2−4x+13x2−4x+13=x2+1
Thus the remaining zeros can be solved through
x2+1=0
x2=−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]}
