Question

How do you find all the real and complex roots of `x^3 + 8 = 0`?

Answer 1

Roots: `x in {-2, color(white)("X")1+sqrt(3)i, color(white)("X")1-sqrt(3)i)}`

Explanation:

Since `(-2)^3=-8`
`color(white)("XXX")x=(-2)` is an obvious root of `(x^3+8=0)`

Therefore `(x+2)` is a factor of `(x^3+8)`

`(x^3+8) div (x+2) = x^2-2x+4`
`color(white)("XXX")`use synthetic or polynomial long division to get
`color(white)("XXX")`this result if it is not apparent.

Use the quadratic formula:
`color(white)("XXX")x=(-b+-sqrt(b^2+4ac))/(2a)`
with `a=1, b=(-2), c=4`
to get the remaining (complex) roots
`color(white)("XXX")x=1+-sqrt(3)i`