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

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

Blue
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SirMrpirateroberts
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    $1$ can easily be guessed as a root , so $(x-1)$ must be a factor – Peter Jun 11 '23 at 16:09
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    In general, the fastest way is Wolfram... – MathFail Jun 11 '23 at 16:09
  • @MathFail I would like to see more methods for solving polynomials – SirMrpirateroberts Jun 11 '23 at 16:11
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    see rational root theorem – MathFail Jun 11 '23 at 16:11
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    It's hard to quantify the "speed" of these pen-and-paper methods, because they essentially rely on how fast the person is at making calculations and how many lines they can skip without making an error of sign. For polynomials of degree 3 in $\Bbb Z[x]$, using the rational root test and then some form of polynomial division (be it long division or Ruffini's rule https://en.wikipedia.org/wiki/Ruffini%27s_rule ) is essentially as good as it gets whenever you don't need to resort to Cardano. Then you are left with $(x-r)(ax^2+bx+c)$, which is easy. –  Jun 11 '23 at 16:21

3 Answers3

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

ananta
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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}

user376343
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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.

RDK
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