Given:
[tex]f(x)=x^3+2x^2+4x+8[/tex]
Let this as equation
[tex]x^3+2x^2+4x+8=0[/tex]
Set x=-1, we get
[tex](-1)^3+2(-1)^2_{}+4(-1)+8=0[/tex]
[tex]-1+2-4+8=0[/tex]
x=-1 is not a root.
Set x=-2, we get
[tex](-2)^3+2(-2)^2_{}+4(-2)+8=0[/tex]
[tex]-8+8+-8+8=0[/tex]
x=-2 is a root that is a real root.
Using a synthetic method to find the remaining roots.
The equation can be written as follows.
[tex]x^2+4=0[/tex][tex]x^2=-4[/tex][tex]x=\pm2i[/tex]
x=2i and x=-2i are complex roots.
Hence we get one real root and two complex roots.
