How Do You Factor the Polynomial in the Derivative of f(x) = (x^3 - 2x)e^x?

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Homework Statement


I need to find the critical points of
f(x) = (x^3 - 2x)e^x

I found the derivative, and set it equal to zero

ended up with e^x (x^3 +3x^2 -2x -2) = 0

I am having trouble factoring the second factor, any suggestions?
 
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let f(x)=x^3 +3x^2 -2x -2
try f(1) and see what happens...
 
yeah I found f(1) just by looking at it, but how can I solve it without guessing?
+ there might be more solutions?
 
Well if f(1)=0 then it means that (x-1) is a factor of f(x) just divide the polynomial by that linear factor and you'll get the other quadratic factor and then you can solve...Remember the remainder and factor theorem?
 
lol, wow I guess that was a while ago

thanks
 
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