How do you factor $x^{3} - 3 x + 2$ $x^3-3x+2$ ?

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Answer 1 · 2015-05-31T18:52:01

First notice that substituting $x = 1$ $x=1$ results in $x^{3} - 3 x + 2 = 0$ $x^3-3x+2 = 0$

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

Use synthetic division to find the remaining factor...

$x^{3} - 3 x + 2 = (x - 1) (x^{2} + x - 2)$ $x^3-3x+2 = (x-1)(x^2+x-2)$

Notice that $x = 1$ $x=1$ is also a root of $x^{2} + x - 2$ $x^2+x-2$

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

$x^{2} + x - 2 = (x - 1) (x + 2)$ $x^2+x-2 = (x-1)(x+2)$

So the complete factorization is:

$x^{3} - 3 x + 2 = (x - 1) (x - 1) (x + 2)$ $x^3-3x+2 = (x-1)(x-1)(x+2)$

Answer 2 · 2015-05-31T18:56:17

$f (x) = x^{3} - 3 x + 2 = 0$ $f(x) = x^3 - 3x + 2 = 0$

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

Factored form: $f (x) = (x - 1) (x^{2} + x - 2) = (x - 1) (x - 1) (x + 2) =$ $f(x) = (x - 1)(x^2 + x - 2) = (x - 1)(x - 1)(x + 2) =$

$= {(x - 1)}^{2} \cdot (x + 2)$ $= (x - 1)^2 * (x + 2)$

How do you factor x^3-3x+2x3−3x+2x^3-3x+2?