# The Unapologetic Mathematician

## 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 […]

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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