# Factoring $x^3 - x^2 - x + 1$

I'm familiar with basic forms of polynomial factoring like completing the square or factoring (e.g. finding that $x^2+x-6=(x+3)(x-2)$, but I'm currently working on integration of rational functions by partial fractions and I find myself encountering the factorizations of polynomials unlike those I've worked with before.

Given: $x^3 - x^2 - x + 1$

When I first look at this, I'm unsure of where to start. It's very unlike the example above. It's degree $3$, for one thing, and it has $4$ terms.

When I look ahead to the solution, the book says $f(1) = 0$, so we know that $(x-1)$ is a factor. Okay, now that they mention that, I can see it fits, but what?! How did they come up with that? Is that a general rule? Does anyone have a link to where I can learn about it?

Further, once it's known that $(x-1)$ is a factor, how do I then factor an $(x-1)$ out of a longer polynomial like this?

In other words, how does $x^3 - x^2 - x + 1 = (x-1)(x^2-1)$?

• polynomial division. – Alexander Grothendieck Mar 3 '14 at 16:59
• $f(x) = x^{3} - x^{2} - x + 1$ so by substituting 1 in for all of the $x$ values you obtain $f(1) = 0$ so you have $x - 1$ is a factor. Then use this example purplemath.com/modules/polydiv2.htm for the other factor – Ahrz Mar 3 '14 at 17:00
• – Alraxite Mar 3 '14 at 17:02

Answer to your 1st question: if a polynomial is degree $n$, it can have up to $n+1$ terms, so having $4$ terms in a degree $3$ polynomial is nothing out of the ordinary.

Answer to your 2nd question: finding a factor of a polynomial of degree $3$ or more is usually just guesswork. In fact, it's been proven that it's IMPOSSIBLE to find a formula to find all solutions of a polynomial with degree of $5$ or above. And unless you're willing to memorize this gruesome formula for polynomials of degree 3 (and let's not even talk about degree 4...), guesswork is your best friend.

• Why memorize, if computer algebra packages will factor for you (if there are nice factors)? – vonbrand Mar 3 '14 at 17:56

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

• I have edited your post for readability. In the future, to make multiple lines in MathJax, put \\ after the end of one line and &= in place of = to align the equations like above. – Cameron Williams Mar 3 '14 at 17:45

Three answers and nobody has mentioned the Factor Theorem!

Factoring cubics can be tricky. But, if you know a few tricks, then it might not be so bad.

One trick for this cubic: there appears to be a pattern in the coefficients: $+1,-1,-1,+1$, and this sums to zero, so it's logical to look at $\pm 1$ as a root.

More generally, you can use the rational root theorem to impose conditions on any rational root of the polynomial. In particular, the rational root theorem says that if $\frac{p}{q}$ is a root of the polynomial, then $q$ must divide the leading coefficient ($1$), and $p$ must divide the unit coefficient (also $1$). Therefore, if the polynomial has a rational root, $1$ is a candidate ($-1$ might also work).

Once you impose conditions on nicely-behaved roots, simply try to factor it out by polynomial division!

• In fact, $1$ is a root of any polynomial whose coefficients sum to zero! – Emily Mar 3 '14 at 17:04

One thing that can be done for polynomials of degree 3 is that you can estimate the roots by differentiating the polynomial and equationg that expression to 0. That will hopefully help you to find a domain in which one root lies. For example here, $$\frac d{dx}(x^3-x^2-x+1)=3x^2-2x-1=0$$$$=>(3x+1)(x-1)=0 \\ =>x=1 \space and \space x=\frac {-1}3.$$ Therefore $\frac {-1}3 \le \alpha\le 1$. Then you may try integer values in the domain that satisfy the polynomial. In this case it is 1 and since 1 is also a root of the derivative, hence it is a double root. Now you can easily find the other root that is -1.

There's some theorem (rational root theorem/ factor theorem?) that says that for all rational factors of the form p/q, p divides into the constant in your polynomial and q divides into the leading coefficient. That leaves to with 1 and -1. Testing 1, first, we find it's a root. -1 is also a root. Using polynomial division it becomes easier to test these numbers. Since it's a cubic it has three roots - it turns out that 1 is a double root.