\chapter{Polynomials} \section{Introduction to polynomials} We're now familiar with the notation of algebra, basic manipulation techniques, and how it reflects properties of arithmetic. One of the main goals of elementary algebra is finding solutions for 'zeros of polynomials'. For example, consider $x^2-7x+10=0$; this cannot be solved by simply 'moving' every term to the other side of the equation to get $x$ by itself. It requires the use of the higher level techniques; working by means of algebraic identities. Indeed, this equation has more than one solution, again coming from the subtle fact that $x^2=y$ has the multiple solutions $\pm \sqrt{y}$. This part will synthesize the more advanced algebraic techniques covered to study the properties of polynomials and algebraic equalities involving them. A 'function' in mathematics is something that takes a.Functions are one of the most important mathematical objects. Functions take ; they're like machines that take one object as input, process this object, and return another object as output. Set theory is an area of mathematics that gives a precise mathematical meaning for functions, but this explanation will suffice. We'll often use the following notation for functions. \[ f(x)=y \] $f(x)=y$ means 'you get $y$ when you put $x$ into the function $f$'. \begin{definition} A \emph{zero} or \emph{root of a function $f$} is some number $r$ that satisfies the following. \[ f(r)=0 \] \end{definition} A simple form of function are polynomials, and algebra \begin{definition} A \emph{polynomial} is a function $P$ of the folloing form. \[ P(x) = \sum^{n}_{k=0} a_k x^k\] The \emph{degree of a polynomial} is the largest power in the polynomial \[ \mathrm{deg}(P) = n \] \begin{definition} \section{Linear polynomials} We first introduce linear polynomials (for completeness, and to help ease the reader into ideas of functions) \begin{definition} A \emph{linear polynomial} is a function of the following form where $a,b\in \mathbb{R}$ are real numbers. \[p(x)=ax+b\] \end{definition} One problem of historical mathematical interest was to find formulae to solve for the zeros of any polynomial. Given the problem $ax+b=0$, basic algebraic manipulation is sufficient to obtain a general solution of $x$. \begin{proposition}[Linear formula] \[a,b \in \mathbb{R} \land a \neq 0 \land ax+b=0 \implies x=-\frac{b}{a}\] \end{proposition} However consiering polynomials of larger degrees brings much more of a challenge; we'll now start considering \emph{quadratics}. \section{Quadratic polynomials} \begin{definition} A \emph{quadratic polynomial} is a function of the following form where $a,b\in \mathbb{R}$ are real numbers. \[p(x)=ax^2+bx+c\] \end{definition} - factorization method for quadratic equations - completing the square - Vieta's formulae for quadratic equations \section{Quadratic formula} - Quadratic formula \begin{theorem} \[a,b,c \in \mathbb{R} \land a \neq 0 \land ax^2+bx+c=0 \implies x=\frac{-b \pm \sqrt{b^2 -4ac}}{2a}\] \end{theorem} %\chapter{Advanced techniques} We've discussed the basic theory of polynomials, however a bit more insight can really light the way in understanding the nature of polynomials and even help with solving more advanced polynomial equations. Some of the techniques here are accessible with the background given by this text, however some material alludes to material yet to be presented in this text or material that is supplemented by knowlegde in calculus. \section{Discriminant} There exists a useful tool for studying the nature of a polynomial's zeros, even if the zeros themselves are unknown. It relies soley of the coefficients of the polynomial and is called the \emph{discriminant}. \begin{definition}[Discriminant] The \emph{discriminant of a polynomial} is a quantity that characterises the existance of real roots for a polynomial. \end{definition} \subsection{Discriminant of degree 2}

Quadratic discriminant

\(\Delta = b^2 - 4ac\)

\(\Delta = 0 \iff r_{0} \in \mathbb{R} \land r_1 \notin \mathbb{R}\)

\(\Delta > 0 \iff r_{0},r_{1} \in \mathbb{R} \)

\(\Delta < 0 \iff r_{0},r_{1} \notin \mathbb{R} \)

- discriminant of degree 2 \section{Vieta's formulae} \begin{theorem}[Vieta's formulae] \end{theorem} \section{Theorems relating to zeros of polynomials} \begin{theorem} Let $P$ be a polynomial with $\mathrm{deg}(P)=n$. Then $P$ cannot have more than $n$ real zeros. \end{theorem} \begin{lemma} Let $P$ be a polynomial such that $\mathrm{deg}(P)$ is odd, then there is at least one real number that is a zero of $P$. \end{lemma} \begin{theorem}[Rational root theorem] Let $P$ be a polynomial $P(x)=a_n x^{n} + \hdots + a_1$, $\frac{p}{q}$ be a rational number in lowest terms. If $\frac{p}{q}$ is a zero of $P$, then $p$ divides $a_1$ and $q$ divides $a_n$ \[ \frac{p}{q} \in \mathbb{Q} \land \gcd(p,q) = 1 \land P(\frac{p}{q}) = 0 \implies p|a_0 \land q|a_{\deg(P)} \] \end{theorem} \section{Polynomial division} \begin{proposition}[Euclid's lemma for polynomials] For any polynomials $P,Q$, there exists polynomials $D,R$ with $\mathrm{deg}(R) < \mathrm{deg}(Q)$ such that the following holds. $P(x)=Q(x)D(x)+R(x)$ \end{proposition} \begin{definition}[Divisibility for polynomials] \[D(x) | P(x) \iff \exists Q [ P(x)=D(x)Q(x) ] \] \end{definition} \begin{definition}[Remainder theorem] \[P(r)=0 \iff\] \end{definition} And as a natural extension to the remainder theorem, $(x-a)$ can only be a factor of $P(x)$ if and only if $P(a)=0$ \begin{theorem}[Factor theorem] \[ (x-r) | P(x) \iff P(r)=0 \] \end{theorem} \begin{theorem}[Remainder theorem] \[ P(x) = Q(x)(x-a)+P(a) \] \end{theorem} The remainder must be of degree 0 since its degree must be strictly less than the degree of $(x-a)$, but when we set $x=a$, we see that the remainder must be $P(a)$. \begin{definition}[Multiplicity of a zero] THe multiplicity of a zero is the amount of times a polynomial has that zero, or in other terms, how many times it can be factored our of the polynomial without remainders. \[$r$\iff [ (x-r)^n | P(x) ] \] \end{definition} The following proposition requires calculus to prove. \begin{proposition} If a zero has a multiplicity higher than one, it is a stationary point. \end{proposition} \section{Fundamental theorem of algebra} \begin{proposition} Let $P$ be a polynomial and $\mathrm{deg}(P)=n$, then there is a sequence $(r_k)_{k=1}^{m}$ of real zeros and a polynomial $Q$ of degree $n-m$ such that the following holds. \[ P(x)=Q(x)\prod_{k=1}^{n} (x - r_{k}) \] \end{proposition} The following theorem is quite advanced, but greatly improves the way we view polynomials. \begin{theorem}[Fundamental theorem of algebra] Let $P$ be a polynomial and $\mathrm{deg}(P)=n$, then there is a sequence $(z_k)_{k=1}^{n}$ of complex numbers such that the following holds. \[ P(x)=\prod_{k=1}^{n} (x - z_{k}) \] \end{theorem} Although this theorem is 'fundamental' in the theory of polynomials, the scope of algebra has widened significantly since this theorem's inception, hence many say that the name of this theorem is misleading; perhaps it is more suitable to be called the fundamental theorem of polynomials. \[ x^n - y^n = (x-y)(\sum_{k=0}^{n-1} x^{n-1-k} y^{k} ) \]

\(P(x)=\prod_{n=0}^{\deg (P)} (x - c_{n})\)

Zero

Domain element of a function that evaluates to zero, also called a root

\(c \text{ is a zero of }f \iff f(c)=0\)

Multiplicity

Property of a zero regarding how many times a polynomial has that same root

For instance, the polynomial \(P(x)=(x-6)^2 (x-2)\) has the root 6 with a multiplicity of 2

Polynomial conditions

All zeroes theorem

If a polynomial has the same amount of real zeroes as its degree, then the polynomial has some model \(P(x) = a \prod^{\deg (P)}_{k=1} (x-a_k)\), where \(a_{k}\) represents each zero. This is because if even one of these equations in the product series zero out due to x being equal to one of the \(a_{k}\), the whole polynomial zeroes out due to multplication of the rest of the function by zero

Vieta's formulae

For \(ax^2 + bx + c\), it can be shown with a bit of manipulation that with zeroes \(\alpha,\beta\):

For \(ax^3 + bx^2 + cx + d\), it can be shown with a bit of manipulation that with zeroes \(\alpha,\beta,\gamma\):

For \(ax^4 + bx^3 + cx^2 + dx + e\), it can be shown with a bit of manipulation that with zeroes \(\alpha,\beta,\gamma,\delta\):

Let \(d= \deg(P(x))\), \(r_{ij}\) is the sequence of the combinations of \(i\) zeroes in multiplication, \(a_n\) be a sequence of polynomial coefficients and \(k \in \mathbb{N} \cap [1,\deg(P(x))]\)

\( \sum_{i=1}^{\binom{d}{k}} (\prod^{k}_{j=1} r_{ij}) = (-1)^{k} \frac{a_{d-k}}{a_{d}} \)

Multiplicity of a zero theorem

A zero has a multiplicity of m if \(P(x) = (x-a)^{m} Q(x), Q(a) \neq 0\). It can be shown with calculus that zeroes with multiplicity above 1 are stationary points. Because with some deriving and substituting you end up at \( P'(x) = (x-a)^{m-1}R(x) \) where \(R(x) = mQ(x)+(x-a)Q'(x)\), and since we know that \(Q(a) \neq 0\) we can substitute to find that \(R(a) \neq 0\). This proves that the \((x-a)^{m-1}\) is doing the zeroing out, and so these roots are also stationary points. And intuitively, if the multiplicity is even then the stationary point is not a turning point

Polynomials as linear and irreducable quadratic factors

All polynomials with real coefficients can be written as the product of linear factors or irreducible quadratics (y'know, in the case that there are no zeroes). A basic form of this as well as proof is below:

When all zeroes are real, \( P_{n} (x) = a_{n} \prod_{k=1}^{n} (x-\alpha_{k}) \)

When all zeroes are complex, \( P_{n} (x) = a{n} \prod_{k=1}^{\frac{n}{2}} (x-\alpha_{k})(x-\overline{\alpha_{k}}) \implies a{n} \prod_{k=1}^{\frac{n}{2}} (x^2-2 \Re (\alpha_{k})x) + |\alpha_{k}|^2) \)

With real and complex zeroes, \( P_{n} (x) = a_{n} \times (\prod_{k=1}^{j} (x-\alpha_{k})(x-\overline{\alpha_{k}})) (\prod_{l=2j+1}^{n} (x-\alpha_{l})) \) (The first pi product deals with all complex roots with their conjugate pair in one iteration, this is why we start the real roots at \(2j+1\). We can see from the case where all zeroes are complex that this also reduces to linear factors and irreducible quadratics)