4

$$x^3 + 1$$

factors as

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

It would have taken me a few minutes to identify this. What are the various approaches to determining rapidly that it is factorable, and factoring it?

Christopher King
  • 10,365
Eric Johnson
  • 606
  • 2
    Remembering the high school identity $a^3+b^3=(a+b)(a^2-ab+b^2)$, and more generally $a^{2n+1}+b^{2n+1}=(a+b)(…)$, is enough. – Bernard Feb 28 '16 at 01:18
  • @Bernard Yes this is the answer I was looking for !!! – Eric Johnson Feb 28 '16 at 01:22
  • Related:http://math.stackexchange.com/questions/1675083/determining-whether-there-are-solutions-to-the-cubic-polynomial-equation-x3 – NoChance Feb 28 '16 at 02:14

4 Answers4

7

In this case, the Rational Root Theorem tells us that if $r$ is a rational root of $x^3 + 1$, then $r$ is either $+1$ or $-1$. Substituting shows that $-1$ is indeed a root (and that $+1$ is not), so $(x - (-1))$ is a factor of $x^3 + 1$. Applying polynomial long division gives the desired factorization: $$\color{#bf0000}{\boxed{x^3 + 1 = (x + 1)(x^2 - x + 1)}}.$$

Note that on the other hand, the discriminant of the quadratic on the r.h.s. of the equation is $-3 < 0$, so it does not factor any further (over $\Bbb R$).

In general, any cubic polynomial $p(x)$ over $\Bbb R$ has a root, so in principle we can always factor it as a product $$p(x) = (a x + b) (A x^2 + B x + C),$$ and we can attempt to solve this equation by distributing and comparing like coefficients to produce a quadratic system in $a, b, A, B, C$. For general cubic polynomials, however, the solutions of this system are extremely unpleasant. Essentially, in the cases in which they are not (at least, when the coefficients of $p$ are rational), the above rational root-finding method generally works.

Note too that substituting $x = -y$ in the above formula and rearranging gives the similar and occasionally useful formula $$y^3 - 1 = (y - 1) (y^2 + y +1).$$

Travis Willse
  • 99,363
  • I'm sorry. I'm doing precalculus. I understood maybe 5% of what you said. – Eric Johnson Feb 28 '16 at 01:20
  • If you're doing precalculus, presumably you saw the Rational Root Theorem, polynomial long division, and the discriminant of a quadratic polynomial in a previous course. – Travis Willse Feb 28 '16 at 01:27
  • 1
    @EricJohnson Rational Root Theorem is critical for factorizing a polynomial at your level. You might have used a different name (or not given it a name at all) but it boils down to this: Any possible rational roots of a polynomial must be a factor of the constant term divided by a factor of the co-efficient of the highest power term. – Ian Miller Feb 28 '16 at 01:38
  • @IanMiller that's a mouthful! I will learn it. Can you recommend some good resources for learning the theorem? – Eric Johnson Feb 28 '16 at 01:45
  • Its much easier explained in an example. Consider the polynomial $f(x)=12x^3+28x^2-59x-35$. To find the rational roots (i.e. solution to $f(x)=0$) you need to look at the first and last number: 12 and 35. The roots (if they are rational) will be something which divides into 35 over something which divides into 12. So you need to try $x$ equal $\pm\frac{a}{b}$ where $a=1,5,7,35$ and $b=1,2,3,4,6,12$. – Ian Miller Feb 28 '16 at 01:59
5

Here is one approach.

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

Another approach would be noticing that $x=-1$ is a solution to $x^3+1=0$. Then you can divide $x^3+1$ by $x+1$ with polynomial long division.

The third approach would be to set $x^3+1=(x+a)(x^2+bx+c)$ and solving the system of equations.

A fourth approach would be using the formula for geometric series.

The sum of the geometric series $1$, $-x$, $x^2$ is $\frac{1-(-x)^3}{1-(-x)}=\frac{x^3+1}{x+1}$.

Multiplying $x+1$ on each side gives us that $(1-x+x^2)(1+x)=1+x^3$.

S.C.B.
  • 22,768
  • This seems almost too lucky that you could factor $x^3+1$ in the manner you did, a more general approach should be taken to answer this question. – Simply Beautiful Art Feb 28 '16 at 02:36
  • 1
    @SimpleArt I would like to point out the fact that this is the approach taken in many textbooks: and the OP did not ask for a general method, but asked for a variety of methods. Also, how is finding the solution to $x^3+1$ lucky? One can take the cube root of $-1$, and I don't think that it is related to luck. I think that only my first approach is related to luck: the second and third approach seems generalizable in my opinion, especially considering the fact that there is a general form for solving cubic equations. – S.C.B. Feb 28 '16 at 03:05
  • @SimpleArt Of course, perhaps the other answers are better. And I respect your opinion. But note the fact that this is the method that I prefer: and while I have considered this method difficult to find, I had thought from the fact that many factorizable equations cannot be solved by the rational root theorem but instead can just require luck, a lucky solution could also be important. (also, the second approach is how I learned to factor $a^n+b^n$ when $n$ is a odd number). – S.C.B. Feb 28 '16 at 03:11
  • Yes, I meant your first approach is a bit of luck, and I do respect your other approaches, as they are generalized. Personally, a similar approach to your second approach would be to notice that $x=\sqrt[3]{_k(-1)}$ and proceed to find the 3 cube root solutions to that. In $a^n+b^n$, for $n$ that is even, try using $x^2-(iy)^2=(x+iy)(x-iy)$. By the way, how do I write $k$th branch of a root in LaTeX? – Simply Beautiful Art Feb 28 '16 at 13:16
  • @SimpleArt I don't know. I still use http://functionspace.com/equationeditor and Detexify for my Latex. Why don't you use Tex Stackexchange? – S.C.B. Feb 28 '16 at 14:06
  • @SimpleArt: \sqrt[k]{x}. Thanks to Knuth for such a wonderfully intuitive LaTeX command name! – user21820 Mar 07 '16 at 04:34
  • @user21820 No, I meant to represent the $k$th branch of the $n$th root. The $n$th root of $x$ is \sqrt[n]{x}, but I don't know how to include the branches. – Simply Beautiful Art Mar 07 '16 at 18:19
  • @SimpleArt: Oh that. In the past, I made up my own notation for it; $\sqrt{{_α} x} = \exp(\frac{1}{2}\ln_α(x))$, and so $\sqrt{{_π} x}$ would be the principal square-root. Similarly for other roots. This is of course incompatible with your notation of branch. I guess we just have to define whichever we want if necessary. – user21820 Mar 08 '16 at 00:21
  • @user21820 Thanks. – Simply Beautiful Art Mar 08 '16 at 00:30
1

If you notice that $x=-1$ is a root (this is reasonably easy to see by inspection and is something most people would look out for), then you can write

$x^3+1 = (x+1)(Ax^2 + Bx + C)$.

The coefficients $A$ and $C$ are easily determined, since the the only way to get an $x^3$ term is by multiplying $Ax^2$ by $x$, so this gives $A = 1$. In the same manner, $C$ is easily determined, since the only term independent of $x$ is $C$, so that $C = 1$. This means that you only need to find $B$.

If you multiply the rhs and look at the coefficient of $x$ you have $B + C = 0$ so that $B = -C = -1$. The final result is

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

jim
  • 1,214
0

As Travis has noted, the rational roots theorem is probably the most powerful/easy method to trying to factor a polynomial of your type.

You have some polynomial:

$$ax^3+bx^2+cx+d$$

If there are any rational roots, they will be of the form

$$x=\pm\frac{Factor(a)}{Factor(d)}$$

For example:

$$x^3+1$$

$$x=\pm\frac{Factor(1)}{Factor(1)}=\pm\frac11=\pm1$$

This is the only possible rational root. You have to actually check if it works, sadly.

$$x\ne1,x=-1$$

If you had $6x^3-3x^2-x-2$, then:

$$x=\pm\frac{Factor(6)}{Factor(2)}=\pm\frac{1,2,3,6}{1,2}=\pm\frac{1,2,3,6}{1},\pm\frac{1,2,3,6}{2}$$

After you attempt to see if any of the solutions actually works be substituting them back in for $x$, you find that $x=1$ is the only one that works.

Proceed to factor the polynomial to find the other non-rational roots.

I note that if you allow complex roots that are with many cube roots and other radicals, then all cubic polynomials and quartic polynomials are factorable.

When it comes down to it, group theory and things like discriments are most likely the best way to go if you have really hard polynomials to factor.

Simply Beautiful Art
  • 74,685