How many real and complex roots exist for the polynomial F(x) = x2 + 2x2 + 4x+8 ?

Question

asked Aug 18, 2023 100k views

1 Answer

Andriy Ivaneyko · Answer 1 · 2023-08-25T15:41:24+0000

Given:

Let this as equation

Set x=-1, we get

$(-1)^3+2(-1)^2_{}+4(-1)+8=0$

x=-1 is not a root.

Set x=-2, we get

$(-2)^3+2(-2)^2_{}+4(-2)+8=0$

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.

$x=\pm2i$

x=2i and x=-2i are complex roots.

Hence we get one real root and two complex roots.