We have been given that 4 is a zero of the polynomial function [tex]f(x)=x^3-6x^2+13x-20[/tex]. We are asked to find the remaining zeros of function.
Since 4 is a zero of f(x), so [tex]x-4[/tex] will be a factor of f(x).
Let us divide our function f(x) by [tex]x-4[/tex].
[tex]\frac{x^3-6x^2+13x-20}{x-4}[/tex]
[tex]\frac{(x-4)(x^2-2x+5)}{x-4}[/tex]
Now we will cancel out [tex]x-4[/tex] from numerator and denominator.
[tex](x^2-2x+5)[/tex]
Now we will use quadratic formula to solve for x as:
[tex](x^2-2x+5)=0[/tex]
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(5)}}{2(1)}[/tex]
[tex]x=\frac{2\pm\sqrt{4-20}}{2}[/tex]
[tex]x=\frac{2\pm\sqrt{-16}}{2}[/tex]
[tex]x=\frac{2\pm\sqrt{-1\cdot 16}}{2}[/tex]
Now we will use [tex]i^2=-1[/tex].
[tex]x=\frac{2\pm\sqrt{i^2\cdot 16}}{2}[/tex]
[tex]x=\frac{2\pm 4i}{2}[/tex]
[tex]x=\frac{2(1\pm 2i)}{2}[/tex]
[tex]x=1\pm 2i[/tex]
[tex]x=1-2i\text{ or }x= 1+2i[/tex]
Therefore, other two zeros of function are [tex]1-2i\text{ or } 1+2i[/tex].