Answer:
Step-by-step explanation:
A graph shows the only real zero to be at x = 3.
Factoring that out gives the quadratic whose vertex form is ...
y = (x -10)² +1
The roots of this quadratic are the complex numbers x = 10 ± i.
_____
For y = (x -10)² +1, the zeros are ...
(x -10)² +1 = 0
(x -10)² = -1 . . . . . . . . . . subtract 1
x -10 = ±√(-1) = ±i . . . . .take the square root
x = 10 ± i . . . . . . . . . . . . add 10
Answer:
3, 10±i
Step-by-step explanation:
Given is a function [tex]f(x) = x^3 - 23x^2 + 161x -303.[/tex]
By rational roots theorem, this can have zeroes as ±1, ±3,±101
By trial and error checking we find f(3) =0
Hence x-3 is a factor
f(x) = [tex](x-3)(x^2-20x+101)[/tex]
II being a quadratic equation we find zeroes using formula
[tex]x=\frac{20±\sqrt{400-404} }{2} =10+i, 10-i[/tex]
zeroes are 3, 10±i
