# Polynomials Review

- Recall that a
**Polynomial**if a function of the form $p(x) = a_0 + a_1x + a_2x^2 + …. + a_nx^n$ where $a_0, a_1, …, a_n$ are coefficients from the field $\mathbb{F}$. If $a_n \neq 0$, then the**Degree**of the polynomial $p$ is $\mathrm{deg} (p) = n$, that is, the largest exponent attached to the variable $x$.

- A
**Root**(also known as a solution or zero) of the polynomial $p$ is the number $\lambda \in \mathbb{F}$ such that $p(\lambda) = 0$. Recall from the Properties of Polynomials page that if $p$ is a polynomial such that $\mathrm{deg} (p) = n ≥ 1$ then $\lambda \in \mathbb{F}$ is a root of $p$ if and only if we can factor $p$ with a polynomial $q$ with $\mathrm{deg} (q) = n – 1$ as a factor such that:

(1)

\begin{align} \quad p(x) = (x – \lambda) q(x) \end{align}

- Furthermore, we saw that a polynomial $p$ with $\mathrm{deg} (p) = n$ has at most $n$
*distinct*roots. A quick way to see this is that if we have $m > n$ roots, then for each root $\lambda_i$ we have that $(x – \lambda_i)$ is a factor (for $i = 1, 2, …, m > n$, and the product of these factors would be a polynomial $q(x) = (x – \lambda_1)(x – \lambda_2)…(x – \lambda_m)$ whose degree is $m$ which implies that $p \neq q$ – a contradiction.

- We also note that if $p(x) = a_0 + a_1x + a_2x^2 + … + a_nx^n$ is a polynomial such that $p(x) = 0$ for each $x \in \mathbb{F}$ then $a_0 = a_1 = … = a_n = 0$.

- We then looked at one of the most notable theorems in mathematics known as The Fundamental Theorem of Algebra which says that if $p(x) = a_0 + a_1x + a_2x^2 + … + a_nx^n$ is a polynomial with $\mathrm{deg} (p) = n$ and the coefficients $a_0, a_1, …, a_n \in \mathbb{C}$ (that is the coefficients of $p$ are
*complex*numbers) then $p$ has exactly $n$ roots.

- Furthermore, we saw that if $p$ is a polynomial with
*complex*coefficients then there exists a unique factorization of $p$ of the following form (for $\lambda_1$, $\lambda_2$, …, $\lambda_m$ as the roots of $p$ and $c \in \mathbb{C}$):

(2)

\begin{align} \quad c(x – \lambda_1)(x – \lambda_2)…(x – \lambda_m) \end{align}

- On the Pairs of Complex Roots for Polynomials with Real Coefficients page we noted that if $p$ is a polynomial with
*real*coefficients and if $\lambda$ is a complex root of $p$ then $\bar{\lambda}$ is also a root of $p$. A prime example of this occurs in the polynomial $p(x) = 1 + x^2$. Clearly $p$ has real coefficients, and the roots of $p(x)$ are $\lambda_1 = i$ and $\lambda_2 = -i$. Note that $\lambda_1 = \bar{\lambda_2}$.

- As an important corollary, we saw that since complex roots come in pairs, then polynomials with odd degree must contain at least one real root since if all of the roots were complex, then there would be one complex root that was not paired with another contradicting what we just mentioned.

- Additionally, if $p$ is a nonconstant polynomial with
*real*coefficients then there exists a unique factorization of $p$ (where $\lambda_1, \lambda_2, …, \lambda_m \in \mathbb{R}$ are roots of $p$, and $c, \alpha_1, \alpha_2, …, \alpha_n, \beta_1, \beta_2, …, \beta_n \in \mathbb{R}$):

(3)

\begin{align} \quad p(x) = c(x – \lambda_1)(x – \lambda_2)…(x – \lambda_m)(x^2 + \alpha_1x + \beta_1)(x^2 + \alpha_2x + \beta_2)…(x^2 + \alpha_nx + \beta_n) \end{align}

- Each of the factors $(x^2 + \alpha_ix + \beta_i)$ for $i = 1, 2, …, n$ are the irreducible quadratic factors for $p$ and $\alpha_i^2 for each $i$.

- On the Root-Finding Techniques for Polynomials page we saw the Rational Root Theorem which says that if $r = \frac{p}{q}$ is a rational root of the polynomial $f(x) = a_0 + a_1x + a_2x^2 + … + a_nx^n$ whose coefficients $a_0, a_1, …, a_n \in \mathbb{Z}$ (having integer coefficients) then $p$ divides $a_0$ and $q$ divides $a_n$.

- We also mentioned Descarte’s Rule of Signs which says that if $p$ is a polynomial with real coefficients written as $p(x) = a_nxn + a_{n-1}x^{n-1} + … + a_1x + a_0$ then the number of positive real roots of $p$ is equal to the number of sign changes of $p(x)$ or the number of sign changes of $p(x)$ minus an even positive integer. Furthermore, the number of negative roots of $p(x)$ is equal to the number of sign changes in $p(-x)$ or the number of sign spaces of $p(-x)$ minus an even positive integer.