Real Polynomial Equations

Polynomial Equations

Definition: Real Polynomial Equation

A real polynomial equation is a polynomial equation $P=0$ where $P$ is a real polynomial.

Properties

Theorem: Maximum Number of Roots

The maximum number of distinct roots which the real polynomial equation

$$A(x)=0$$

can have is equal to the degree of $A$.

PROOF

TODO

Theorem: Roots and Divisibility

A number $p\in \mathbb{R}$ is a root of the real polynomial equation

$$A(x)=0$$

if and only if $A(x)$ is divisible by $(x-p)$.

PROOF

TODO

Theorem: Real Polynomial Equations with Integer Coefficients I

If the coefficients $a_0,\cdots ,a_n$ of the real polynomial equation

$$a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_{1}x+a_0=0$$

are integers and it has a rational root $x^{\ast }=\frac{p}{q}$ (where $p$ and $q$ are coprime), then $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$.

PROOF

Proof that $q$ is a divisor of $a_n$:

If $x^{\ast }=\frac{p}{q}$ is a root of the polynomial equation, then

$$a_n\frac{p^n}{q^n}+a_{n-1}\frac{p^{n-1}}{q^{n-1}}+\cdots +a_1\frac{p}{q}+a_0=0$$

Multiply by $q^{n-1}$.

$$a_n\frac{p^n}{q}+a_{n-1}{p}^{n-1}+\cdots +a_{1}p{q}^{n-2}+a_0{q}^{n-1}=0$$ $$a_n\frac{p^n}{q}=-a_{n-1}{p}^{n-1}-\cdots -a_{1}p{q}^{n-2}-a_0{q}^{n-1}$$

Since $p,q$ and $a_k$ are all integers, the right-hand side must be an integer as well. This means that $a_n\frac{p^n}{q}$ is an integer, but $p^n$ and $q$ have no common divisors, since they are coprime. This means that $q$ must divide $a_n$.

Proof that $p$ is a divisor of $a_0$:

If $x^{\ast }=\frac{p}{q}$ is a root of the polynomial equation, then

$$a_n\frac{p^n}{q^n}+a_{n-1}\frac{p^{n-1}}{q^{n-1}}+\cdots +a_1\frac{p}{q}+a_0=0$$

Multiply by $q^n$.

$$a_n{p}^n+a_{n-1}{p}^{n-1}q+\cdots +a_{1}p{q}^{n-1}+a_0{q}^n=0$$

Divide by $p$.

$$a_n{p}^n+a_{n-1}{p}^{n-1}q+\cdots +a_{1}p{q}^{n-1}+a_0{q}^n=0$$ $$a_n{p}^{n-1}+a_{n-1}{p}^{n-2}q+\cdots +a_1{q}^{n-1}+a_0\frac{q^n}{p}=0$$ $$a_0\frac{q^n}{p}=-a_n{p}^{n-1}-a_{n-1}{p}^{n-2}q-\cdots -a_1{q}^{n-1}$$

Once again, $p,q$ and $a_k$ are all integers and so the right-hand side must be an integer. This means that $a_0\frac{q^n}{p}$ is an integer. The numbers $q^n$ and $p$ have no common divisors, since $q$ and $p$ are coprime. This means that $p$ must be a divisor of $a_0$.

Theorem: Real Polynomial Equations with Integer Coefficients II

If the coefficients of the real polynomial equation

$$A(x)=0$$

are integers and it has a rational root $x^{\ast }=\frac{p}{q}$ (where $p$ and $q$ are coprime), then for every integer $m$, the number $A(m)$ is divisible by $p-mq$.

PROOF

TODO

Reciprocal Polynomial Equations

Definition: Reciprocal Polynomial Equations

A reciprocal polynomial equation is a polynomial equation which can be written either as

$$a_0{x}^{2n}+a_1{x}^{2n-1}+\cdots +a_{n-1}{x}^{n+1}+a_n{x}^n+a_{n-1}p{x}^{n-1}+\cdots +a_1{p}^{n-1}x+a_0{p}^n=0$$

or as

$$a_0{x}^{2n+1}+a_1{x}^{2n}+\cdots +a_n{x}^{n+1}+a_{n}p{x}^n+a_{n-1}{p}^3{x}^{n-1}+\cdots +a_1{p}^{2n-1}x+a_0{p}^{2n+1}=0.$$

Definition: Reciprocal Polynomial Equations

Properties

Algorithm: Degree Reduction for Reciprocal Polynomial Equations of Even Degree

We are given a reciprocal polynomial equation

$$a_0{x}^{2n}+a_1{x}^{2n-1}+\cdots +a_{n-1}{x}^{n+1}+a_n{x}^n+a_{n-1}p{x}^{n-1}+\cdots +a_1{p}^{n-1}x+a_0{p}^n=0$$

of degree $2n$. We can reduce it to a real polynomial equation of degree $n$ in the following way:

1. Divide by $x^n$.

2. Group the terms appropriately.

$$a_0\left(x^n+\frac{p^n}{x^n}\right)+a_1\left(x^{n-1}+\frac{p^{n-1}}{x^{n-1}}\right)+\cdots +a_{n-1}\left(x+\frac{p}{x}\right)+a_n=0$$

3. Substitute $y=x+\frac{p}{x}$.

$$a_0{y}^n+a_1{y}^{n-1}+\cdots +a_{n-1}y+a_n=0$$

Theorem: Roots of Reciprocal Polynomial Equations of Odd Degree

Every reciprocal polynomial equation

$$a_0{x}^{2n+1}+a_1{x}^{2n}+\cdots +a_n{x}^{n+1}+a_{n}p{x}^n+a_{n-1}{p}^3{x}^{n-1}+\cdots +a_1{p}^{2n-1}x+a_0{p}^{2n+1}=0.$$

of odd degree $2n+1$ has the root $x=-p$.

PROOF

TODO

Theorem: Reduction of Reciprocal Polynomial Equations of Odd Degree

Every reciprocal polynomial equation

$$a_0{x}^{2n+1}+a_1{x}^{2n}+\cdots +a_n{x}^{n+1}+a_{n}p{x}^n+a_{n-1}{p}^3{x}^{n-1}+\cdots +a_1{p}^{2n-1}x+a_0{p}^{2n+1}=0.$$

of odd degree $2n+1$ can be reduced to a reciprocal polynomial equation of even degree $2n$ by dividing it by $x+p$.

PROOF

TODO