The Unapologetic Mathematician

Mathematics for the interested outsider

Roots of Polynomials I

When we consider a polynomial as a function, we’re particularly interested in those field elements x so that p(x)=0. We call such an x a “zero” or a “root” of the polynomial p.

One easy way to get this to happen is for p to have a factor of X-x. Indeed, in that case if we write p=(X-x)q for some other polynomial q then we evaluate to find

p(x)=(X-x)q(x)=0

The interesting thing is that this is the only way for a root to occur, other than to have the zero polynomial. Let’s say we have the polynomial

p=c_0+c_1X+c_2X^2+...+c_nX^n

and let’s also say we’ve got a root x so that p(x)=0. But that means

0=c_0+c_1x+c_2x^2+...+c_nx^n

This is not just a field element — it’s the zero polynomial! So we can subtract it from p to find

\begin{aligned}p=\left(c_0+c_1X+c_2X^2+...+c_nX^n\right)-\left(c_0+c_1x+c_2x^2+...+c_nx^n\right)\\=c_1(X-x)+c_2(X^2-x^2)+...+c_n(X^n-x^n)\end{aligned}

Now for any k we can use the identity

X^k-x^k=(X-x)(X^{k-1}+X^{k-2}x+...+Xx^{k-2}+x^{k-1})

to factor out (X-x) from each term above. This gives the factorization we were looking for.

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July 30, 2008 - Posted by | Algebra, Polynomials, Ring theory

2 Comments »

  1. [...] can actually tease out more information from the factorization we constructed yesterday. Bur first we need a little [...]

    Pingback by Roots of Polynomials II « The Unapologetic Mathematician | July 31, 2008 | Reply

  2. [...] with Too Few Roots Okay, we saw that roots of polynomials exactly correspond to linear factors, and that a polynomial can have at most as many roots as its degree. In fact, there’s an [...]

    Pingback by Polynomials with Too Few Roots « The Unapologetic Mathematician | August 6, 2008 | Reply


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