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

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Question 346182: a^3-2a^2-4a-8=0
How would I solve this?
Found 2 solutions by Fombitz, CharlesG2:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
You can graph it to find out how many roots are real.
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graph%28300%2C300%2C-5%2C5%2C-5%2C5%2Cx%5E3-2x%5E2-4x-8%29
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There is one real root.
The other two are complex.
I iterated using Newton's method,
x=3.6786
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The approximate complex roots are,
x=-0.8393+%2B-+1.2126i%29 from www.wolframalpha.com

Answer by CharlesG2(834) About Me  (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