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How can I factorize $x^3-3x+2$ ?

The answer that I got on the internet is $x^3-2x^2+x+2x^2-4x+2$=$(x-1)^2(x+2)$ It would be nice if anyone could also tell what these type of equations are called and where can I learn more?

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    @cardano: since Jim didn't answer, I will: often you can notice small roots just from examination. For example, any polynomial with no constant term will have 0 as a root. If the sum of the coefficients sum to zero (as they do in your question) then 1 is a root. It's usually worth trying -1 as well. Most "real" polynomials won't have nice roots like this, but contrived homework problems and textbook problems will.2011-05-07

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They are called cubic functions / cubic equations. A closed formula for the solutions exists but it is quite ugly so the common method to factorize the term is to guess one root $x_0$ and then do long division by $(x-x_0)$.

So the method one would be to use the formula on $x^3-3x+2=0$ and find that the roots are $1,1,-2$ and you are done. The trivial method is to guess $x_0=1$ and use the long division

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    An alternative to the long division method is the Ruffini's rule http://en.wikipedia.org/wiki/Ruffini's_rule.2011-05-05
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Just in case you're wondering what is meant by a rational "root": a root for a given polynomial can be thought of as a value $r$ such that when you set $x = r$, the value of the polynomial evaluates to zero. Some people refer to the roots as the "zeros" of a polynomial.
So, I'll use $P(x)$ to denote the equation $x^3-3x+2$; that is, we have
$P(x) = x^3 - 3x + 2 = (x-1)^2(x+2)$ Now, when is $P(x) = 0$?

When $(x-1)=0$ and/or when $(x+2)=0$, because when that happens, then $P(x) = (0)(x+1) =0\ \text{or}\ P(x) =(x-1)^2(0) = 0$ Now we simply solve for $x$ to find the roots:

$(x-1) = 0\implies x=1$, so $1$ is a root of $P(x)$, and
$(x+2) = 0 \implies x = -2$, so $-2$ is a root of $P(x)$,
and, voila, you have the roots, as given in the first answer.

In terms of learning how to factor a polynomial, the best advice I have to offer is practice!

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    I agree, @Babak! And the tools available NOW are so much more plentiful and powerful than those we learned or had access to "back in the day"!2013-08-01
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So I claim the polynomial $x^3 + 555x^2 + 52752x - 7197508$ has a double root. How do you find it?

Differentiate to get $3x^2 + 1110x + 52752$ then take the greatest common divisor of this with the other. If you don't know what greatest common divisor is or how to compute it you should study this because that is a very important theory. What matters here is that computing the greatest common divisor is a quick thing to do.

The gcd is found to be $x+314$ (if you want to see a calculation finding that I can post it) so we conclude that $x + 314$ is a factor of the polynomial twice! Now we can divide $(x + 314)^2$ out using long division to get $x - 73$ and so we have factored completely:

$x^3 + 555x^2 + 52752x - 7197508 = (x+314)^2(x - 73).$

If you like to solve these problems yourself here is one $x^3 - 1443x^2 + 431235x + 67013919$ you can do using the same techniques, GCD and long division.

By the way, as for the proof that this method works. Here it is: If polynomial $p$ has a double root then $p = q^2 r$ for some polynomials $q$ and $r$, differentiate to get p' = 2 q q' + q^2 r' = q (2 q' + q r') and then \text{gcd}(p,p')=\text{gcd}(q^2 r,q (2 q' + q r')) = q. Notice this proof does not use any properties of cubics, it applies to polynomials of any degree.


Now what if the cubic polynomial does not have a double root? For example $x^3 - 103x^2 - 21209x + 530127.$ One thing we do know is, if it's of the form $(x - r_1)(x - r_2)(x - r_3)$ then it equals $x^3 - (r_1 + r_2 + r_3)x^2 + (r_1 r_2 + r_1 r_3 + r_2 r_3)x - r_1 r_2 r_3$ and we want to find $r_1$, $r_2$, $r_3$.

We can factor $r_1 r_2 r_3 = 530127 = 3^2 \cdot 13 \cdot 23 \cdot 197$.. although that's really difficult to do by hand, it's even hard with a computer when the numbers get very large. It tells us that those $\pm$ primes need to be distributed into $r_1$, $r_2$, $r_3$ in such a way that their sum is $r_1 + r_2 + r_3 = 103$. Here's what I tried in the calculator

197 - 3*3 - 13*23 = -111 197 - 3*23 - 13 = 115 197 - 3*23 - 13*3 = 89 197 - 23 - 13*3*3 = 57 197 - 13 - 23*3*3 = -23 197 - 13*3 - 23*3 = 89 197 - 13*3*3 - 23 = 57 197 - 13*3*3 + 23 = 103 

as you can see it's tedious and boring, but it did work: we have factored the polynomial as $(x - 197)(x - 23)(x + 3^2 13)$!

Since it's difficult it will be nice to at least know, before doing a lot of work, whether we will succeed or not. For this we can use the discriminant.


The discriminant of a cubic polynomial $x^3 + ax^2 + bx + c = (x - \alpha)(x - \beta)(x - \gamma)$ (note that $\alpha,\beta,\gamma$ can be cubic irrational numbers or quadratic irrationals) is defined as $(\alpha - \beta)^2(\alpha - \gamma)^2(\beta - \gamma)^2$. Now note that if the three roots are integers then this is a square, if not it wont! That is the criteria to use.

To actually apply it we need to compute it from the coefficients, but this is possible because of the symmetry it has. The formula is $a^2b^2-4b^3-4a^3c-27c^2+18abc$ and we can use it to see what the discriminant of the polynomial above was: $58507812921600 = 7649040^2$.

But check for example, the discriminant of $x^3 - x - 1$ is $-23$ so we cannot factor it into integers. That can save a lot of work.

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    That's quite a generous answer.2011-11-15
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If the roots are rational you can get them by finding all positive and negative factors of the last digit and divide them by the same of the first digit. Then test them $\pm2$ and $\pm1$. We get 1 and $-2=x$. Then divide by their respective polynomials to see which one has order 2. Solved.