SOLUTION: a^3-2a^2-4a-8=0 How would I solve this?

Question 346182: a^3-2a^2-4a-8=0
How would I solve this?
Found 2 solutions by Fombitz, CharlesG2:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
You can graph it to find out how many roots are real.
.
.
.

.
.
.
There is one real root.
The other two are complex.
I iterated using Newton's method,

.
.
.
The approximate complex roots are,
from www.wolframalpha.com

Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
a^3-2a^2-4a-8=0
How would I solve this?

you sure that it is not a^3 + 2a^2 - 4a - 8 = 0?
if it was:
(a + 2)(a^2 - 4) = 0
using FOIL:
a^3 - 4a + 2a^2 - 8 = 0
rearranging
a^3 + 2a^2 - 4a - 8 = 0

then: (a + 2)(a^2 - 4) = 0
(a + 2)(a + 2)(a - 2) = 0

then a = -2 (twice) and a = 2

RELATED QUESTIONS
The problem is A= 2A/3=8+4A The answers are either... A= -2.4 A= 2.4 A= 1.3... (answered by checkley75)
hi,I would like help with this equation,... (answered by PRMath)
2A/3=8+4A, solve for... (answered by stanbon)
solve... (answered by mando271)
How would I solve 4a - 7b = 13 2a - 7b = 3 Using... (answered by jim_thompson5910)
3a-1/a^2+4a+4 - 3/a^2+2a = 3/a this is a problem from my homework that is asking me to... (answered by longjonsilver)
How would you solve this equation?... (answered by jim_thompson5910,stanbon,bucky)
how do i solve... (answered by ReadingBoosters)
I need help with this equation 2A/3 = 8 + 4A (answered by stanbon)