Andrew Thangaraj
Aug-Nov 2020
constant: \(a\), \(a\in F\)
Roots: all scalars if \(a=0\); no roots for nonzero
\(a=0\): zero polynomial
linear: \(ax+b\), \(a,b\in F\), \(a\ne 0\)
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
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
quintic and higher: \(p_nx^n+\cdots+p_1x+p_0\)
modern numerical methods: compute roots for 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 \(p(x)\)
Addition, Multiplication (well-known)
Division algorithm: \(p(x)\) by \(a(x)\)
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 \(p(x)\)
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
\(p(x)=p_nx^n+\cdots+p_1x+p_0\), \(p_i\in\mathbb{R}\), \(p_n\ne 0\)
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)