Find all other zeros of P(x)=x^3-2x-4, given that -1-i is a zero.

MATH HELP PLEASE!!!

If −1−i is a zero, then so is its complex conjugate −1+i
Therefore (x + 1 + i) and (x + 1 − i) are factors of P(x)

(x + 1 + i) (x + 1 − i)
= (x + 1)² − i²
= x² + 2x + 1 + 1
= x² + 2x + 2

Now divide P(x) by (x² + 2x + 2) to find third factor of P(x)
(x³ − 2x − 4) / (x² + 2x + 2) = x − 2 -----> third zero = 2

Zeros:
x = −1−i
x = −1+i
x = 2

Complex numbers come in conjugate pairs . If -1-i is a zero then -1+i will be a zero . Then factor is (x+1+i)((x+1-i) =(x+1)^-i^2=x^2+2x+2 . Now we divide x^3-2x-4 by the previous expression we get x-2 .the zeros will be -1-i or -1+i or 2

factorising ( x-2)(x² + 2x + 2 ) = 0

so x = 2

(x² + 2x + 2 ) = 0 will not factorise

solving gives ( x + 1 )² +2 - 1 = 0

(x + 1 )² = -1
x = -1 +/- √-1
x =( -1 - i) or (-1 +i)

x = -1 - i || x = -1 + i || x = 2