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I haven't taken a math class in a few years and I have come to realize, as I am taking one again, that I do not know/remember how to factor equations.

One example from my textbook is:


Looking for a very detailed description of how to factor this, and just factoring in general.

EDIT: Isaac gave a very good description of how to factor this equation but I am looking for a general description of how to factor equations not just this equation.

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Oops, the first one is cubed. Typo. I fixed it. –  The.Anti.9 Jan 23 '11 at 6:39
I don't think this is an example where a basic factoring trick will help. There are formulas for finding roots of cubic polynomials (mathworld.wolfram.com/CubicFormula.html), but they are not typically covered in elementary algebra. –  Jonas Meyer Jan 23 '11 at 6:45
Just a quick tip: If you have a polynomial with integer coefficients, then if your polynomial have integer solutions, then this solution must divide the degree zero coefficient, in this case 1. So if this equation have integer solutions they must be $\pm 1$. (this does not have integer solutions) –  Fredrik Meyer Jan 23 '11 at 21:28

4 Answers 4

up vote 2 down vote accepted
  • There's general factoring of polynomials which is really not doable by humans. Given a random sequence of integer coefficients, it's just intractable (because the roots could be arbitrary complex numbers involving square roots and fractions).

  • Fortunately, what's usually given as exercises are polynomials that look pretty random but for teaching purposes are 'easy' (have only nice integer or fractional solutions, no roots or imaginaries). So the way to do these is to (intelligently) guess a solution $r$, then try to divide out $x - r$ from the original polynomial.

    • If that works then you try the same strategy again on the (necessarily) smaller degree polynomial. So really, what you're being asked to do is practice polynomial division.
    • If that guess didn't work, try another guess. Your initial guesses come from the rational root theorem (the first and last coefficients).

So for the above (fixed) example $3x^3-x^2-3x+1$, you'd guess maybe $1\over 3$ (from Peter Taylor's list; I picked it because I know that it happens to be a factor). Divide through by $(3x- 1)$ (because its root is $1\over3$) and get $x^2 + 0x - 1= x^2-1$ which you already know how to factor by the quadratic formula.

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Notice that you factored a different polynomial than the original polynomial given by the OP. His polynomial does not have any rational solutions. –  Fredrik Meyer Jan 23 '11 at 21:34
As in my comment on Peter Taylor's answer, beware that there are quartics (and higher-degree polynomials) with integer coefficients that have no real roots, yet still factor over the integers. –  Isaac Jan 24 '11 at 1:59
@Fredrik: Yes, I am aware of both these departures from the literal. I did the example on the fixed problem (the given example with a sign changed so that the resulting polynomial does have 'nice' roots. –  Mitch Jan 25 '11 at 0:49
@Isaac: Yes, I used your fixed polynomial, but pointing out how to do it in general. Also, as is often the case in the teaching of this procedure, the roots are implicitly guaranteed to have all integer (or rational) roots. –  Mitch Jan 25 '11 at 0:55

The polynomial given does not factor (with integer coefficients), but perhaps $3x^3-x^2-3x+1$ was intended. One technique for factoring polynomials with several terms (typically more than 3) is to group terms together and factor a common monomial factor out of each group. In this instance: $$\begin{align} 3x^3-x^2-3x+1 &= (3x^3-x^2)+(-3x+1) \\ &= x^2(3x-1)-1(3x-1) \\ &=(x^2-1)(3x-1) \\ &=(x+1)(x-1)(3x-1) \end{align}$$ The final step is one of several common factoring patterns—this one is sometimes referred to as a difference of squares: $a^2-b^2=(a+b)(a-b)$.

edit As to how to factor in general, my inclination is to say that you don't. If you pick a polynomial with random integer coefficients, the probability that it factors over the integers is 0. Typically, in textbooks, the factoring problems are common monomial factors, specific factoring patterns (difference of squares, sum/difference of cubes, perfect-square, perfect-cube, etc.), factoring-by-grouping, quadratics, or some other special-case technique (as below).

On a side note, since two other answers have suggested factoring by looking for roots, there are polynomials with integer coefficients that factor over the integers but have no real roots or no rational roots (though I think they must be of degree 4 or greater). For example: $$\begin{align} x^4+2x^2+81 &=x^4+18x^2+81-16x^2 \\ &=(x^2+9)^2-(4x)^2 \\ &=(x^2-4x+9)(x^2+4x+9) \end{align}$$ (This is an example of a class of such problems [quartics that are quadratics in $x^2$] that I have seen in multiple textbooks where the technique is a sort of "completing the square," but applied to the middle term, followed by a difference-of-squares factoring.)

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Can you give a little more detail on how you moved from line 2 two line 3? I kind of see it, but not quite. –  The.Anti.9 Jan 23 '11 at 6:51
@The.Anti.9: Replace $(3x-1)$ by $A$ so that the second line becomes $x^2A-1A$. Both terms have a common factor of $A$ (which is really $(3x-1)$, so $x^2A-1A=(x^2-1)A=(x^2-1)(3x-1)$. Does that make sense? –  Isaac Jan 23 '11 at 7:25
Ah yea, I get it now. –  The.Anti.9 Jan 23 '11 at 16:45
@Isaac: I thought factor over the integers' meant that the roots are integers. Do you mean 'there are polynomials with integer coefficients that don't factor over the integers' (I really am not sure)? Whatever the words are, your second example is fascinating...I see that your 2nd line equals the third but I don't see the easy move going from the difference of squares to the product of quadratics. –  Mitch Jan 25 '11 at 1:03
@Mitch: "factor over the integers" (or over another set) means to factor into polynomials with integer coefficients (or coefficients in that other set). In my quartic example, the second line can be thought of as $A^2-B^2$, which can be factored as $(A-B)(A+B)$ (difference of squares factoring pattern). Done more literally, that yields $((x^2+9)-4x)((x^2+9)+4x)$, which I rearranged to the form usually used when writing polynomials at this level. –  Isaac Jan 25 '11 at 1:07

Factoring over the integers, $(bx + c)$ can only divide into $a_n x^n + ... + a_1 x + a_0$ if $b | a_n$ and $c | a_0$. For small $a_n$ and $a_0$ you can easily try the different cases. Here you have $a_n$ = 3 and $a_0$ = 1, so you can evaluate your polynomial at $x \in \{-1, \frac{-1}{3}, \frac{1}{3}, 1\}$ to get the roots. If you don't find enough roots, divide through by the terms you have found and re-test the roots you found to see whether any are repeated.

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Though it's not an issue with cubics, beware that there are quartic polynomials that factor into two quadratics, neither of which have rational roots. –  Isaac Jan 23 '11 at 14:03

One thing you may always want to try is the rational root theorem (http://en.wikipedia.org/wiki/Rational_root_theorem), if your polynomial has integer coefficients. This is already touched on by the above answers. While this isn't a general trick, it may be able to help you find factors, and potentially then reduce this to something that you can easily recognize and factor.

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This was well-covered in two other answers. Additionally, as in my comments there and in the note in my answer, there are polynomials with integer coefficients that factor over the integers, but do not have rational roots. –  Isaac Jan 24 '11 at 13:05

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