Properties of Polynomials

Properties of Polynomials

We are about to look at an important concept known as an eigenvalue shortly, but before then, we must secure a foundation of knowledge on polynomials. We have looked at polynomials throughout the Linear Algebra section on the site, for example, when we looked at $\wp (\mathbb{R})$ as the set of all polynomials.

 Definition: A Polynomial is a function in the form $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ where $a_0, a_1, ..., a_n \in \mathbb{F}$. The values $a_0, a_1, ..., a_n$ are called the Coefficients of The Polynomial $p$, and the Degree of $p$ denoted $\mathrm{deg} (p)$ is the largest exponent attached to a variable with a nonzero coefficient.

For example, the function $p(x) = 2x^2 + 3x^4$ is a polynomial with real coefficients $2$ and $3$ and whose degree $\mathrm{deg} (p) = 4$. Another example is the function $f(x) = 4 + x^2 + 4x^7$ which is also a polynomial with real coefficients $3$, $1$, and $4$ and whose degree $\mathrm{deg} (f) = 7$.

By convention, a polynomial $p(x) = 0$, that is the constant function that is zero is defined to have degree $-\infty$, that is $\mathrm{deg} (p) = - \infty$.

The following table shows the graphs of some arbitrary functions of increasing degree:

Degree $0$ Degree $1$ Degree $2$   Degree $3$ Degree $4$ Degree $5$   One important property of polynomials is that polynomials of even degree $2, 4, 6, ...$ tend to look similar, while polynomials of odd degree greater than $1$, in other words, $3, 5, 7, ...$ also tend to look similar.

We will now look at another important definition that the reader has already likely encountered.

 Definition: Let $p(x) = a_0 + a_1x + a_2x^2 + ... a_nx^n$ be a polynomial. Then $\lambda \in \mathbb{F}$ is called a Root, Solution, or Zero of $p$ if $p(\lambda) = 0$.

We will now look at an important theorem regarding the factorization of a polynomial and its roots.

 Theorem 1: If $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is a polynomial and $\mathrm{deg} (p) = n ≥ 1$ then $\lambda \in \mathbb{F}$ is a root of $p$ if and only if there exists a polynomial $q(x)$ where $\mathrm{deg} (q) = n - 1$ such that $p(x) = (x - \lambda) q(x)$.
• Proof: $\Leftarrow$ Suppose that $\lambda$ is a root of $p(x)$. Since $p(x)$ has degree $n$, $p(x)$ can be written as $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ where $a_n \neq 0$.
• Now since $\lambda$ is a root of $p$ then $p(\lambda) = 0$ and so $p(\lambda) = 0 = a_0 + a_1\lambda + a_2\lambda^2 + ... + a_n\lambda^n$. We thus get from subtracting the last two equations that:
(1)
\begin{align} \quad p(x) - p(\lambda) = (a_0 + a_1x + a_2x^2 + ... + a_nx^n) - (a_0 + a_1\lambda + a_2\lambda^2 + ... + a_m\lambda^m) \\ \quad p(x) - 0 = a_1(x - \lambda) + a_2(x^2 - \lambda^2) + ... + a_n(x^n - \lambda^n) \\ \quad p(x) = a_1(x - \lambda) + a_2(x^2 - \lambda^2) + ... + a_n(x^n - \lambda^n) \end{align}
• Now for some polynomial $q_2 (x)$ such that $\mathrm{deg} (q_2) = 1$ let $x^2 - \lambda^2 = (x - \lambda)q_2(x)$, and for some polynomial $q_3(x)$ such that $\mathrm{deg} (q_3) = 2$ let $x^3 - \lambda^3 = (x - \lambda)q_3(x)$, …, and for some polynomial $q_n(x)$ such that $\mathrm{deg} (q_n) = n-1$ let $x^n - \lambda^n = (x - \lambda)q_n(x)$. Then we have that:
(2)
\begin{align} \quad p(x) = a_1(x - \lambda) + a_2(x - \lambda)q_2(x) + ... + a_n(x - \lambda)q_n(x) \\ \quad p(x) = (x - \lambda)[a_1 + a_2q_2(x) + ... + a_nq_n(x)] \end{align}
• Therefore let $q(x) = a_1 + a_2q_2(x) + ... + a_nq_n(x)$, and so $p(x) = (x - \lambda)q(x)$ where $\mathrm{deg} (q) = n - 1$.
• $\Rightarrow$ Suppose that $p(x) = (x - \lambda)q(x)$. Then $p(\lambda) = (\lambda - \lambda)q(x) = 0q(x) = 0$ and so $\lambda$ is a root of $p$.
 Theorem 2: If $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ is a polynomial such that $\mathrm{deg} (p) = n ≥ 0$ then $p$ has at most $n$ distinct roots.

We should note that Theorem 2 pertains to distinct roots. There are some cases for which a polynomial has multiple roots but some may not be distinct. For example, the polynomial $p(x) = (x - 2)^2 = (x - 2)(x - 2)$ has the roots $\lambda_1 = 2$ and $\lambda_2 = 2$, but $\lambda_1 = \lambda_2$, so these roots are not distinct. There is only one distinct root for this quadratic equation, but notice this does not violate theorem 2 as $\mathrm{deg} (p) = 2$. We will not prove theorem 2.

• Proof: We will carry out this proof by induction. For $n \in \mathbb{N} \cup \{ 0 \}$ let $S(n)$ be the statement that the polynomial $p(x)$ with $\mathrm{deg} (p) = n$ has at most $n$ distinct roots.
• $S(0)$ says the polynomial $p(x) = a_0$ where $a_0 \neq 0$ has $0$ roots, which is true since $p(x)$ represents a horizontal line that does not intersect the $x$-axis.
• $S(1)$ says that the polynomial $p(x) = a_0 + a_1x$ where $a_1 \neq 0$ has $1$ root, which is also true since $\lambda = -\frac{a_0}{a_1}$ is the only root of $p$.
• Suppose that for some $k \in \mathbb{N}$, $k ≥ 1$ that $S(k-1)$ is true, that is every polynomial $q(x)$ with $\mathrm{deg} (q) = k-1$ has at most $k-1$ distinct roots. We want to show that $S(k+1)$ is true, that is show that every polynomial $p(x)$ with $\mathrm{deg} (p) = k$ has at most $k$ distinct roots. If $p$ has no roots, then we are done as $k ≥ 0$. If $p$ has a root, call it $\lambda$, and so by Theorem 1, for some polynomial $q(x)$ where $\mathrm{deg} (q) = k - 1$ we have that:
(3)
\begin{align} p(x) = (x - \lambda)q(x) \end{align}
• By the induction hypothesis, $q(x)$ has at most $k - 1$ distinct roots (which are also roots of $p$), and $\lambda$ is a root of $p$, so $p$ has at most $k$ distinct roots, so $S(k)$ is true.
• Therefore by the Principle of Mathematical Induction, any polynomial $p(x)$ with $\mathrm{deg} (p) = n ≥ 0$ has at most $n$ distinct roots. $\blacksquare$
 Corollary 1: If $p(x) = a_0 + a_1x + ... + a_mx^m$ where $a_0, a_1, ..., a_m \in \mathbb{F}$ and that $p(x) = 0$ for all $x \in \mathbb{F}$ then $a_0 = a_1 = ... = a_m = 0$.
• Proof: Suppose that $p(x) = a_0 + a_1x + ... + a_mx^m$ and that $p(x) = 0$ for all $x \in \mathbb{F}$. Then $p(x)$ has infinitely many roots, and so there does not exist an integer $m$ for which $\mathrm{deg} (p) = m$. Therefore $p(x)$ is precisely the zero polynomial and so $a_0 = a_1 = ... = a_m = 0$. $\blacksquare$