# Polynomials and roots

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solution (if it exists): $u+$ null$(A)$
• Four fundamental subspaces of a matrix
• Column space, row space, null space, left null space
• Determinant of a square matrix
• Function with many interesting properties
• Change of basis for linear map
• Can result in simpler matrix representation

## Polynomials: degree 0,1,2

constant: $a$, $a\in F$

• Roots: all scalars if $a=0$; no roots for nonzero

• $a=0$: zero polynomial

linear: $ax+b$, $a,b\in F$, $a\ne 0$

• Roots: one root $=-b/a$

quadratic: $ax^2+bx+c$, $a,b,c\in F$, $a\ne 0$

• Roots: $(-b\pm\sqrt{b^2-4ac})/2a$

• Two roots (repeated or multiplicity 2 if $b^2=4ac$)

• $F=\mathbb{R}$: two complex roots or two real roots

## Polynomials: degree 3,4,5 and higher

cubic: $ax^3+bx^2+cx+d$, $a\ne0$

• Roots: there is a formula involving cube roots etc

• Three roots (multiplicities: 1,1,1 or 1,2 or 3)

• $F=\mathbb{R}$: at least one real root

quartic: $ax^4+bx^3+cx^2+dx+e$, $a\ne0$

• Roots: there is a formula involving fourth roots etc

• Four roots (multiplicities: 1,1,1,1 or 1,1,2 or 2,2 or 1,3 or 4)

quintic and higher: $p_nx^n+\cdots+p_1x+p_0$

• Roots: there is no formula involving radicals (Abel-Ruffini theorem)

modern numerical methods: compute roots for polynomials

• Try MATLAB, Mathematica, Python numpy/scipy

## Polynomials

Polynomials in one variable with coefficients from a field

$p(x)=p_0+p_1x+p_2x^2+\cdots+p_nx^n,\quad p_i\in F$

Degree of $p(x)$: deg $p(x)$ $=$ largest $n$ such that $p_i\ne 0$

degree(zero polynomial) $=-\infty$

Leading term of $p(x)$: LT $p(x)=p_nx^n$, where $n=$ deg $p(x)$

• deg $a(x)b(x)\ge$ deg $a(x)$, if $b(x)\ne 0$

Division algorithm: $p(x)$ by $a(x)$

• $i=0$, $q(x)=0$, $p_0(x)=p(x)$
• while deg $p_i(x)\ge$ deg $a(x)$
• $t_i(x)=\text{LT}(p_i(x))/\text{LT}(a(x))$
• $q(x)=q(x)+t_i(x)$
• $p_{i+1}(x)=p_i(x)-t_i(x)a(x)$
• $i=i+1$
• return $q(x)$, $r(x)=p_i(x)$

## Division, roots, factors

Given two polynomials $p(x)$ and $a(x)$, there exist unique polynomials $q(x)$ and $r(x)$ such that $p(x)=q(x)a(x)+r(x),$ with $r(x)=0$ or deg $r(x)<$ deg $a(x)$.

$q(x)$: quotient, $r(x)$: remainder

Proof: Use division algorithm with induction

Roots and factors

• $\lambda\in F$ is a root of $p(x)$ if $p(\lambda)=0$

• If $p(x)$ divided by $a(x)$ results in remainder zero,

• $p(x)=q(x)a(x)$

• $a(x)$ divides $p(x)$, denoted $a(x)\,|\,p(x)$

• $a(x)$ is a factor of $p(x)$

## Existence of roots and factorization

Fundamental theorem of algebra

Every non-constant polynomial with complex coefficients has a complex root.

Proof: Involves complex analysis.

$\lambda$ is a root of $p(x)$ iff $x-\lambda$ is a factor of $p(x)$.

degree-1 factor: called linear factor

Proof: Divide $p(x)$ by $x-\lambda$.

$p(x)=p_nx^n+\cdots+p_1x+p_0$, $p_i\in\mathbb{C}$, $p_n\ne 0$

There exists $n$ complex roots $\lambda_1,\ldots,\lambda_n$ and $p(x)=p_n(x-\lambda_1)\cdots(x-\lambda_n)$

Proof: Use fundamental theorem and degree-1 factor result repeatedly

## Polynomials with real coefficients

$p(x)=p_nx^n+\cdots+p_1x+p_0$, $p_i\in\mathbb{R}$, $p_n\ne 0$

• Need not have real roots (eg: $x^2+1$)

• If $\lambda$ is a root, $\bar{\lambda}$ is a root

Factors with real coefficients

• If $\lambda$ is a real root, $x-\lambda$ is a factor

• If $\lambda$ is a complex root, $x^2-(\lambda+\bar{\lambda})x+\lambda\bar{\lambda}$ is a factor

Factorization

$p(x)=$ (quadratic factors) (linear factors)