How do you factor #x^3-3x+2#?

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Answer 1 · 2015-05-31T18:52:01

First notice that substituting #x=1# results in #x^3-3x+2 = 0#

So #(x-1)# is a factor.

Use synthetic division to find the remaining factor...

#x^3-3x+2 = (x-1)(x^2+x-2)#

Notice that #x=1# is also a root of #x^2+x-2#

So we have another #(x-1)# factor...

#x^2+x-2 = (x-1)(x+2)#

So the complete factorization is:

#x^3-3x+2 = (x-1)(x-1)(x+2)#

Answer 2 · 2015-05-31T18:56:17

#f(x) = x^3 - 3x + 2 = 0#

Since #(a + b + c + d = 0)#, then one factor is #(x - 1)#

Factored form: #f(x) = (x - 1)(x^2 + x - 2) = (x - 1)(x - 1)(x + 2) =#

#= (x - 1)^2 * (x + 2)#