Factoring $x^3-6x^2+11x-6$ without using Rational Roots Theorem

Question

To factor $P(x)=x^3-6x^2+11x-6$ one way is using Rational Roots Theorem and recognizing that $x=1$ makes $P(x)$ zero. But I want to factor without using it. I tried,

$$x^3-6x^2+11x-6=x^2(x-6)+11(x-6)+60=(x^2+11)(x-6)+60$$ But $+60$ remains at the end.

Answer 1

To the best of my knowledge, what you want is not really feasible. That is, to the best of my knowledge, the only viable approaches to attacking the Cubic are the Rational Root Theorem, and Cardano's approach, which I will describe at the end of my answer.

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Since you are excluding the Rational Root Theorem, you are excluding the natural approach indicated by the comment of carraig, following the question, which would (also) be my first try.

Suppose that you try to artistically derive the factors of

$$(x-r_1)(x-r_2)(x-r_3) = x^3 - 6x^2 + 11x - 6.\tag1 $$

Using Vieta's formulas, you know that:

• $r_1 + r_2 + r_3 = (-1) \times (-6).$

• $r_1r_2 + r_1r_3 + r_2r_3 = 11.$

• $r_1r_2r_3 = (-1) \times (-6).$

If you then try to solve for one of $r_1, r_2, r_3$, you end up inevitably chasing your tail. How do I know this? Because, for centuries, this is exactly what Mathematicians did, before Cardano (or whoever) figured out the following alternative:

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The following is a brief description of Cardano's method, as more fully described in this article.

First, the cubic should be expressed in the format of

$$x^3 + bx^2 + cx + d = 0 \tag2 $$

which is consistent with the RHS of (1) above.

Then, you depress the cubic into the format of

$$y^3 + py + q = 0 \tag3 $$

via the change of variable

$$y = x + \frac{b}{3} \implies y^3 = x^3 + bx^2 + \frac{b^2x}{3} + \frac{b^3}{27}.$$

Then, you convert (3) above into a bizarre quadratic equation, through the further change of variable:

$$y = (s + t) ~~\text{and}~~ 3st = -p. \tag4 $$

The idea is that because you are using two variables $s$ and $t$ to express $y$, you are permitted to introduce the artificial constraint that $3st = -p$.

The reason that this is relevant is because then, you have that

$$y^3 = (s + t)^3 = s^3 + t^3 + 3st(s+t) = s^3 + t^3 + 3st(y). \tag5 $$

Thus, using (5) above, (3) above is transformed into

$$s^3 + \left(\frac{-p}{3s}\right)^3 + q = 0. \tag6 $$

You end up with a quadratic in $s^3$, which will succumb to the quadratic equation. However, subsequently turning the computed roots of $(s)^3$ into the roots of $x$ is potentially, extremely ugly.

This is why the Rational Root Theorem is generally the first port of call. Another way of saying the same thing is:

• Certainly, there is nothing inherent in the Cubic equation that will prevent roots from being irrational.
  
  So, logically, the general attack on the Cubic can't be limited to the Rational Root Theorem.

• Attempts to artistically factor the Cubic, beyond the Rational Root theorem, aren't generally taught. Your use of Vieta's formulas generally involve chasing your tail.

• The fallback to the Rational Root theorem therefore has to be the ugly Cardano's approach.

Answer 2

For a cubic with integer coefficients to find the factorization we can either use an integer factorization algorithm, or do something very easy beforehand. So we assume that \begin{align*} P & =x^3-6x^2+11x-6 \\ & =(x-a)(x-b)(x-c) \\ & =x^3 - x^2(a + b + c) + x(ab + ac + bc) - abc. \end{align*} So by comparison of the coefficients we obtain easy equations in $a,b,c$, namely \begin{align*} a+b+c -6 & =0,\\ ab + ac + bc - 11 & =0,\\ abc - 6 & =0. \end{align*} It is easy to see that $(a,b,c)$, up to permutation is $(1,2,3)$ and we are done.