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

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Homework Help Overview

The discussion revolves around finding the critical points of the function f(x) = (x^3 - 2x)e^x by analyzing its derivative. The derivative has been set to zero, leading to the equation e^x (x^3 + 3x^2 - 2x - 2) = 0, where the focus is on factoring the polynomial part.

Discussion Character

  • Exploratory, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the difficulty of factoring the polynomial x^3 + 3x^2 - 2x - 2, with one suggesting testing specific values like f(1) to find roots. Others express concern about the possibility of multiple solutions and seek methods beyond guessing.

Discussion Status

The conversation is active, with participants exploring different approaches to factor the polynomial. Some guidance has been offered regarding the use of the remainder and factor theorem, but there is no explicit consensus on a complete method yet.

Contextual Notes

There is an implied need to find all critical points, and the discussion hints at the potential complexity of the polynomial, suggesting that more solutions may exist beyond the initial guess.

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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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