I was handed $x^3-x^2+x-2=0$ to factor, but I'm not sure how. I tried all the methods I know of--which, at the time of writing, are limited by my precalc math background (I'm working on that...). Is there some method to factor this?

  • 5
    $\begingroup$ Using the rational roots theorem we check $x = \pm 1, 2$, none of which give us equality. So this won't factor over the rationals. $\endgroup$ Sep 27, 2011 at 4:20
  • $\begingroup$ Great. But considering how complex this should be, how am I supposed to factor it? $\endgroup$
    – skeggse
    Sep 27, 2011 at 4:24
  • $\begingroup$ @CMC,wolframalpha.com/input/?i=factor+x^3-x^2%2bx-2 $\endgroup$
    – Pedja
    Sep 27, 2011 at 4:24
  • $\begingroup$ Sorry... just how complex should this be? What if you were handed $x + 5$ to factor?? $\endgroup$ Sep 27, 2011 at 4:27
  • 2
    $\begingroup$ Perhaps it's a misprint for $\rm\ x^3-x^2+2\ x-2\ =\ (x-1)\ (x^2 + 2)\:.$ $\endgroup$ Sep 27, 2011 at 5:12

1 Answer 1


The following answer omits numerical details. You can fill in these details, if you care to. The full answer is rather messy. It is possible that whoever handed you the problem had a simpler one in mind, and wrote down the wrong coefficients.

There are two factorizations that one might have in mind: (i) Factorization over the complex numbers and (ii) Factorization over the reals.

(i) Over the complex numbers: Let $a$, $b$, and $c$ be the roots of the equation $x^3-x^2+x-2=0$. Then we have $$x^3-x^2+x-2=(x-a)(x-b)(x-c).$$ Now we have the complete factorization over the complex numbers. There is, of course, a little missing detail: finding $a$, $b$, and $c$. We will get to that later.

(ii) Over the reals: It turns out that the equation $x^3-x^2+x-2=0$ has exactly one real root. Call that real root $a$. The other two roots are non-real. It turns out (not by accident) that these non-real roots are complex conjugates. One of them has shape $p+iq$ and the other has shape $p-iq$, where $p$ and $q$ are real and $q\ne 0$. Then the factorization over the reals is $$x^3-x^2+x-2=(x-a)(x^2-2px+p^2+q^2).$$ Again, we are finished, once we have found the roots. Actually, we only need to find $a$. For once we have found it, we can use division of polynomials to divide the polynomial $x^3-x^2+x-2$ by the polynomial $x-a$ to find the quadratic factor.

For either problem, finding the roots is more or less inescapable. But what does it mean to find the roots?

There are two reasonable interpretations of what it means to find the roots. (a) Find a good numerical approximation to the roots and (b) Find an exact expression for the roots.

(a) Numerical approximation: There are many ways to find a good numerical approximation to the real root of $x^3-x^2+x-2=0$. Many calculators (except the cheapest, like mine) will find a good numerical approximation. So will Wolfram Alpha. There are nice calculus-based methods that I will not describe since this is a pre-calculus question. You can also decide where the root might be, and get better approximations by intelligent trial-and-error, or by using a graphing calculator to zoom in on the relevant portion of the curve $y=x^3-x^2+x-2$. Once you have a good numerical approximation to the real root $a$, you can divide the original polynomial by $x-a$ and get an approximate factorization of your polynomial.

(b) Exact expression: The equation $x^3-x^2+x-2=0$ does not have any "nice" solutions, in particular it has no rational solutions. Luckily, a formula for the roots of a cubic was found by Cardano in the middle of the sixteenth century (I am lying a little, the history is more complicated than that.)

Wolfram Alpha knows that formula. For the details about the formula, you might want to look at this Wikipedia article (As usual, there are better sources, it depends on how much detail you want.)

Now you can use the formula in the Wikipedia article to find an expression for the real root $a$, and continue as suggested above. Or else you can use Wolfram Alpha. The expression for the real root is a mess, and would horrify more than it would enlighten. But actually the reasoning behind the formula is interesting, and really not all that hard. I can supply more detail if your searching does not satisfy your curiosity.


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