Root-Coefficient Theorem

Section 7.10 Root-Coefficient Theorem

Another useful tool is the following theorem which shows how the coefficients of a polynomial relate to the roots.

Theorem 7.10.1. Root-Coefficient Theorem.

If \(p(x)=x^{n}+a_{n-1}x^{n-1}+\cdots + a_{1}x+a_{0}\text{,}\) has roots \(r_{1},\ldots,r_{n}\) (counting multiplicities), then

\begin{equation*} r_{1}+\cdots + r_{n} = -a_{n-1} \end{equation*}

\begin{equation*} r_{1}\cdots r_{n} = (-1)^{n}a_{0}\text{.} \end{equation*}

In general, if \(s_{j}\) denotes the sum of all products of \(j\)-tuples of the roots (e.g. \(s_{2} = r_{1}r_{2}+r_{1}r_{3}+r_{2}r_{3}+\cdots\)), then

\begin{equation*} s_{j} = (-1)^{j}a_{n-j}\text{.} \end{equation*}

A “tuple” is a finite ordered list of items, here the roots. The phrase “\(j\)-tuples” means a list of items of length \(j\text{.}\)

Proof.

First, factorize

\begin{equation*} p(x)=x^{n}+a_{n-1}x^{n-1}+\cdots + a_{1}x+a_{0}=a(x-r_{1})\cdots (x-r_{n})\text{.} \end{equation*}

Note that as the coefficient of \(x^{n}=1\text{,}\) we know \(a=1\) (since otherwise the right side, when multiplied out, wouldn't equal the left). We can establish the formulas in the theorem now by multiplying out the product on the right.

This theorem looks complicated, but in fact it just encodes a pattern.

\begin{equation*} (x-\alpha)=x-\alpha \end{equation*}

\begin{equation*} (x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x + \alpha\beta \end{equation*}

\begin{equation*} (x-\alpha)(x-\beta)(x-\gamma) =x^3-(\alpha+\beta+\gamma)x^2 +(\alpha\beta+\beta\gamma +\alpha\gamma)x^2 - \alpha\beta\gamma \end{equation*}

In the language of the theorem, a single \(2\)-tuple from the set \(\{\alpha,\beta,\gamma\}\) would be a two of these items. The set of all different \(2\)-tuples is

\begin{equation*} \alpha\beta,\beta\gamma,\alpha\gamma \end{equation*}

Example 7.10.2.

Suppose \(x^{3}+ax^{2}+bx+c\) has roots \(\alpha,\beta\text{,}\) and \(\gamma\text{.}\) Find a polynomial with roots \(\alpha\beta\text{,}\) \(\beta\gamma\text{,}\) and \(\gamma\alpha\) in terms of \(a,b,c\) (that is, the coefficients of your polynomial should only be described using \(a,b\text{,}\) and \(c\text{,}\) not \(\alpha,\beta\text{,}\) and \(\gamma\)).

By the Root Coefficient Theorem,

\begin{equation*} \alpha+\beta+\gamma = -a\text{,} \end{equation*}

\begin{equation*} \alpha\beta+\beta\gamma+\gamma\alpha = b \end{equation*}

and

\begin{equation*} \alpha\beta\gamma = -c\text{.} \end{equation*}

Let \(x^{3}+Ax^{2}+Bx+C\) be a polynomial with roots \(\alpha\beta\text{,}\) \(\beta\gamma\) and \(\gamma\alpha\text{.}\) Then we know

\begin{equation*} -A=\alpha\beta+\beta\gamma+\gamma\alpha = b \end{equation*}

\begin{equation*} B=\alpha\beta^{2}\gamma+\alpha\beta\gamma^{2}+\alpha^{2}\beta\gamma=-c(\alpha+\beta+\gamma)=ac \end{equation*}

and

\begin{equation*} -C=\alpha\beta \cdot \beta\gamma\cdot \gamma\alpha = (\alpha\beta\gamma)^{2}=c^{2} \end{equation*}

Hence, the polynomial is

\begin{equation*} x^{3}-bx^{2}+acx-c^{2}\text{.} \end{equation*}

Example 7.10.3.

Are all the roots of \(x^{3}+11x^2+7\) integers?

Suppose they were. Notice that by the Factorization Theorem, this polynomial has three roots \(a,b,c\) (where some of these roots could repeat). By the Root-Coefficient Theorem, \(abc=-7\text{,}\) so \(7=|abc|=|a|\cdot|b|\cdot |c|\text{.}\) Since \(7\) is prime and we are assuming the roots are integers, the only way this is possible is if one of these absolute values is \(7\) and the others are \(1\text{,}\) say \(|a|=7\) and \(|b|=|c|=1\text{.}\) The Root-Coefficient Theorem also says \(a+b+c=-11\text{,}\) so by the triangle inequality,

\begin{equation*} 11=|-11|=|a+b+c|\leq |a|+|b|+|c|=1+1+7=9\text{,} \end{equation*}

which is impossible. Thus, at least one of the roots is not an integer.