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Polynomials Extension.tex
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\documentclass[10pt]{article} \usepackage{amssymb,amsmath} \usepackage[hmargin=1cm,vmargin=0.5cm]{geometry} \begin{document} {\large Polynomials} \begin{align*} \text{\bf General Form:}\quad& P(x)\equiv\sum_{r=0}^n a_k x^k=a_n x^n+a_{n-1}x^{n-1}+\ldots+a_1 x+a_0\:,\quad \text{where $a_0, a_1, \ldots, a_n$ are constants and $a_n\neq 0$.}\\ \text{Notations:}\quad &\deg P(x)\equiv n\\ &q|r\:\text{where $q\neq 0$}\quad\text{if and only if}\quad r=kq\:\:\text{for some $k\in\mathbb{Z}$.}\quad\text{(It follows that $q|0$ for any integer $q\neq 0$.)}\\ &\gcd(p,q)\equiv g\quad\text{where $g\in\mathbb{Z}$ is the largest number that satisfy $g|p$ and $g|q$\: (the $HCF$ of $p$ and $q$).}\\ &\gcd(p,q)=1\quad\text{means that $p$ and $q$ are relative prime (and fraction $\tfrac{\:p\:}{q}$ cannot be simplified further).}\\ \\ \text{Division Form:}\quad&P(x)\equiv A(x)Q(x)+R(x)\:,\quad\text{where $A(x)$ is the devisor, $Q(x)$ is the quotient, and $R(x)$ is the remainder.}\\ &\deg P(x)=\deg A(x)+\deg Q(x)\:,\quad\deg P(x)\geq\deg Q(x)\:,\quad\deg R(x)<\deg Q(x)\\ \\ \text{Remainder }&\text{Theorem:}\quad\text{If $P(x)$ is divided by $(x-\alpha)$, then the remainder is $P(\alpha)$.}\\ \text{Proof:}\quad&\because\quad P(x)=A(x)(x-\alpha)+R\:,\quad(\deg R(x)=0)\:,\qquad \therefore\quad P(\alpha)=A(\alpha)(\alpha-\alpha)+R=R\:.\\ % \text{Corollary:}\quad&\text{If $P(x)$ is divided by $(qx-p)$, then the remainder is $P\left(\tfrac{\:p\:}{q}\right)$.}\\ \text{Proof:}\quad&\because\quad P(x)=A(x)(qx-p)+R\:,\qquad \therefore\quad P\left(\tfrac{\:p\:}{q}\right)=A\left(\tfrac{\:p\:}{q}\right)\left[q\left(\tfrac{\:p\:}{q}\right)-p\right]+R=R\:.\\ \\ \text{Factor Theorem:}\quad&\text{$(x-\alpha)$ is a factor of $P(x)$\quad\it if and only if\quad\rm $P(\alpha)=0$.}\\ \\ \text{\bf The Fundamental }&\text{\bf Theorem of Algebra: }\boxed{\text{Every polynomial of complex coefficient has at least one zero in the complex field.}}\\ % \text{Theorem 1:}\quad&\text{If $\tfrac{\:p\:}{q}$ is a zero of $P(x)\:,\quad\text{where }p,q\in\mathbb{Z}\:,\quad q\neq 0\:,\quad\gcd(p,q)=1$ , and $P(x)$ has integer coefficients,}\\ &\text{then}\quad p|a_0\:,\quad q|a_n\:,\quad\text{and}\quad qx-p\text{ is a factor of $P(x)$.}\\ &\text{i.e. If a polynomial has a zero in simple fractional form, the coefficient of the lowest degree term of the}\\ &\text{polynomial is divisible by the numerator, and the highest by the denominator.}\\ &\text{e.g. $6x^2-7x-3=(2x-3)(3x+1)$ , has two zeros: $\tfrac{\:3\:}{2}$ and $\tfrac{-1\:}{3}$.}\\ &\text{For the numerators, we have $3|-3$ , $-1|-3$ and for the denominators $2|6$ and $3|6$ .}\\ % \text{Corollary:}\quad&\text{If $a_n=1$\:,\quad$q=\pm 1$ and $p$ is an integer and $p|a_0$.}\\ &\text{i.e. If a monic polynomial with integer coefficients has a rational zero, it is an integer and a divisor of $a_0$.}\\ \\ \text{Theorem 2:}\quad&\text{If $p+\sqrt{q}$ is a zero of $P(x)$ ,\quad where $p,q\in\mathbb{Z}\:,\quad q\neq 0\:,$ and $P(x)$ has rational coefficients,}\\ &\text{then $p-\sqrt{q}$ is also a zero. (Likewise, if $p-\sqrt{q}$ is a zero, so is $p+\sqrt{q}$ .)}\\ &\text{i.e. If a polynomial with rational coefficients has irrational zeros, they are in conjugate pairs.}\\ \\ \text{Theorem 3:}\quad&\text{If $a+ib$ is a zero of $P(x)$ , where $a,b\in\mathbb{R}\:, b\neq 0\:,$ and $P(x)$ has real coefficients, then $a-ib$ is also a zero.}\\ &\text{i.e. If a polynomial with real coefficients has imaginery zeros, they are in conjugate pairs.}\\ \\ \text{Theorem 4:}\quad&\text{Every polynomial has exactly\quad$\deg P(x)$\quad linear factors over the complex field.}\\ \text{Proof:}\quad&\text{From the Fundamental Theorem of Algebra, $P(x)$ has at least one zero in the complex field, and therefore}\\ &\text{at least one linear factor. If it is $(x-\alpha)$, so that $P(x)=A(x)(x-\alpha)$, $A(x)$ of degree $n-1$ \big($n=\deg P(x)$\big)}\\ &\text{has at least one linear factor as well. By repeating the process, all $n$ linear factors will be found.}\\ &\text{i.e. Every polynomial equation of $n$ degrees has exactly $n$ roots in the complex field, including multiplicities.}\\ \\ \text{Theorem 5:}\quad&\text{If $P(x)=0$ has a root of multiplicity $m$, then $P'(x)$ has the same root of multiplicity of $(m-1)$.}\\ \text{Proof:}\quad&\text{Consider }P(x)=(x-\alpha)^m Q(x)\:,\quad P'(x)=(x-\alpha)^{m-1}\big(mQ(x)+(x-\alpha)Q'(x)\big)\:.\\ \end{align*} \end{document}