September 12, 2008 / wj32

Factorising polynomials from their roots

If you have a polynomial

$\displaystyle p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{2}x^{2}+a_{1}x+a_{0}$

, you can factorise it by finding it’s roots. The easiest method is Newton’s method, which can also be used on a scientific calculator:

$\displaystyle x_{n+1}=x_{n}-\frac{a_{n}x_{n}^{n}+a_{n-1}x_{n}^{n-1}+\cdots+a_{2}x_{n}^{2}+a_{1}x_{n}+a_{0}}{\frac{a_{n}}{n}x_{n}^{n-1}+\frac{a_{n-1}}{n-1}x_{n}^{n-2}+\cdots+\frac{a_{2}}{2}x_{n}+a_{1}}$

On a calculator, type in a number, press equals, press AC, type in the formula, replacing the $x_{n}$ with Ans (or whatever your calculator uses), and keep on hitting the equals button until you get a stable number. That’s one root. Now, press AC, type in another number, press equals, press AC, type in the formula, and so on. Hopefully, you’ll get another root. Keep on repeating with different starting numbers until you get the same number of roots as the degree of the polynomial (maybe you should check for multiple roots…).

The fundamental theorem of algebra states that every nth degree polynomial can be written as (where x_n are the polynomial’s roots):

$\displaystyle C(x-x_{1})(x-x_{2})\cdots(x-x_{n})$

So that means $p(x)$ equals that (replace the x_ns with the roots you found). We’re almost there, but we still don’t know what C is. Notice that when you expand $(x-x_{1})(x-x_{2})\cdots(x-x_{n})$ , the last term (the constant) in the result is equal to $(-1)^{n}x_{1}x_{2}\cdots x_{n}$ . So to find C we use: