Is there any faster way to factor $x^3-3x+2$?

Question

$$x^3-3x+2$$ $$x^3-3x+x^2+2-x^2$$ $$x^2-3x+2+x^3-x^2$$ $$(x-2)(x-1)+x^2(x-1)$$ $$(x-1)[x^2+x-2]$$ $$(x-1)(x+2)(x-1)$$

Is there a better, faster way to factor this cubic trinomial?

Answer 1

Rewriting differently can help to find linear and quadratic factors: \begin{matrix}{x^3-3x+2=\\x^3-x-2x+2=\\ x(x^2-1)-2(x-1)=\\ (x-1)(x^2+x-2)} \end{matrix}

Answer 2

$$ \begin{align} & x^3-3x+2\\ =& x^3 +3x -3x^2 -1 +3x^2 -6x +3\\ =& (x-1)^3 +3(x-1)^2\\ =& (x-1)^2(x+2) \end{align} $$

Answer 3

We usually substitute the general value to $x$ when factorizing the rational coefficient polynomials.

As cited in the comment above, we can substitute $x=1$ having the polynomial $0$.

Therefore, the polynomial has $(x-1)$ for its factor.

$x^3-3x+2=(x-1)(x^2+x-2).$

Again, we can replace $x$ to $1$ in $x^2+x-2$ to have it $0$.

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

, so we are done.