all real and imaginary zeros. f(x)=x^3-4x^2-x+10. Show work. Thanks! Solution I thought ,,you are asking theoritically By graphing we find x=2 is a root f(x) f(x) = (x-2) ( x^2-2x-5) the zeros of x^2-2x-5 is x= 2±sqrt( 2^2 + 4*5) /2= 2±sqrt(24)/2 = 1±sqrt(6) this formula comes by let us consider quadratic equation ax^2+bx+c roots of the equation is x= [ -b± (b^2-4ac) ]/2a.

1. all real and imaginary zeros. f(x)=x^3-4x^2-x+10. Show work. Thanks!
Solution
I thought ,,you are asking theoritically
By graphing we find x=2 is a root f(x)
f(x) = (x-2) ( x^2-2x-5)
the zeros of x^2-2x-5 is
x= 2±sqrt( 2^2 + 4*5) /2= 2±sqrt(24)/2 = 1±sqrt(6)
this formula comes by
let us consider quadratic equation ax^2+bx+c
roots of the equation is
x= [ -b± (b^2-4ac) ]/2a