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Section 7.8 Factoring Polynomials

Using this, we can prove that any polynomial can be factored completely.

In the above theorem some roots may repeat. If a root appears \(m\) times in \(r_{1},\ldots,r_{n}\text{,}\) then we say it has multiplicity \(m\text{.}\)

Notice that if we multiply out the above polynomial, then we get a polynomial like \(p(z)=az^{n}+\cdots\text{,}\) so if \(p\) was as in (7.7.1), then we actually know that \(a=a_n\text{.}\)

The above theorem tells us that an \(n\)-degree polynomial always has \(n\) roots if we also take into account the multiplicity of the roots (that is, how often it appears in the factorization above). For example, \((z-1)^2\) is a \(2\)-degree polynomial with only one root, but that root has multiplicity \(2\text{.}\)

We prove this by induction on \(n\text{,}\) the degree of the polynomial \(p\text{.}\) Let \(P(n)\) be the statement

A degree \(n\) polynomial \(p\) has \(n\) roots \(r_{1},\ldots,r_{n}\in\mathbb{C}\) and there is a number \(a\in\mathbb{C}\) so that
\begin{equation*} p(z) = a(z-r_{1})(z-r_{2})\cdots (z-r_{n})\text{.} \end{equation*}

Base Case: Assume \(p\) is any degree \(1\) polynomial. If we write \(p\) as \(p(z)=az+b\text{,}\) we can re-write \(p(z) = a(z-\frac{-b}{a})\text{.}\) Then \(p\) has precisely one root \(r_1=\frac{-b}{a}\) and \(p\) can be written in the form \(p(z)=a(z-r_1)\text{.}\) This proves \(P(1)\text{.}\)

Induction Step: Suppose the theorem holds true for some integer \(n\geq 1\text{,}\) i.e. \(P(n)\) is true. Let \(p\) be a degree \(n+1\) polynomial. By the Fundamental Theorem of Algebra there is a root \(w\) of \(p\text{,}\) i.e. \(p(w)=0\text{.}\)

Consider the polynomial \(q(z):=p(z+w)\text{.}\) Then \(q\) is also degree \(n+1\) and has a root at \(0\text{.}\) Write \(q\) out as a polynomial then because it has a root at \(z=0\) the constant term is also zero and we can write

\begin{equation*} q(z) = q_{1}z+\cdots + q_{n+1}z^{n+1} = z\underbrace{(q_{1}+q_{2}z+\cdots + q_{n+1}z^{n})}_{=g(z)}\text{.} \end{equation*}

where \(g\) is a polynomial of degree \(n\text{.}\) By our induction hypothesis \(P(n)\text{,}\) there are \(a,z_{1},\ldots,z_{n}\in\mathbb{C}\) so that

\begin{equation*} g(z) = a(z-z_{1})(z-z_{2})\cdots (z-z_{n})\text{.} \end{equation*}

Note if \(q(z)=p(z+w)\) then

\begin{equation*} p(z)=q(z-w)=(z-w)g(z-w) \end{equation*}

and so if \(r_{i}=z_{i}+w\) for \(i=1,\cdots, n\) and \(r_{n+1}=w\) then

\begin{equation*} p(z) = a(z-w-z_{1})(z-w-z_{2})\cdots (z-w-z_{n})(z-w) =a(z-r_{1})(z-r_{2})\cdots (z-r_{n+1})\text{.} \end{equation*}

This proves \(p\) has \(n+1\) roots and there is a number \(a\in\mathbb{C}\) so that we can write

\begin{equation*} p(z) = a(z-r_{1})(z-r_{2})\cdots (z-r_{n+1})\text{.} \end{equation*}

Since \(P(1)\) is true, and \(P(n)\Rightarrow P(n+1)\) it follows that \(P(n)\) is true for all \(n\in\NN\) by the principle of mathematical induction.