Polynomials

The canonical form of a polynomial with complex (respectively real) coefficients is P=a_nXⁿ + a_n-1X^n-1 + … + a₀, with a₀, a₁,…,${{a}_{n}}\in \mathbb{C}$ (respectively a₀, a₁, …, a_n$\in \mathbb{R}$) and a_n ≠0. In short we write P = $\displaystyle \sum\limits_{k=0}^{n}{{{a}_{k}}{{X}^{k}}}$.

If f = a_nXⁿ + … + a₀, with a_n≠0, we say the polynomial f has degree n.

1) Let f$\in$ $\mathbb{R}[X ]$ and z$\in$ $\mathbb{C}$. Then f($\bar{z}$) = $\overline{f(z)}$

2) Let f$\in$ $\mathbb{Q}[X ]$, a,b $\in \mathbb{Q}$ such that $\sqrt{b}$ $\in \mathbb{R}\backslash \mathbb{Q}$, b>0

Then f(a±$\sqrt{b}$) has the form A±B$\sqrt{b}$, A, B $\in \mathbb{Q}$

Operations with polynomials

Let f,g $\in \mathbb{C}[X ]$, f = $\displaystyle \sum\limits_{i=0}^{m}{{{a}_{i}}{{X}^{i}}}$ and g = $\displaystyle \sum\limits_{j=0}^{n}{{{b}_{j}}{{X}^{j}}}$, m<n.

The sum of the polynomials f and g is the polynomial written f +g, defined by:

f+g = g+f = $\displaystyle \sum\limits_{k=0}^{n}{{{c}_{k}}{{X}^{k}}}$, where c_k = $\left\{ \begin{aligned} & {{a}_{k}}+{{b}_{k}},k\le m \\ & {{b}_{k}},m<k\le n \\ \end{aligned} \right.$.

The product of the polynomials f and g is the polynomial $f\cdot g=g\cdot f={{c}_{m+n}}{{X}^{m+n}}+...+{{c}_{0}}$, where $\displaystyle {{c}_{k}}=\sum\limits_{k=0}^{{}}{{{a}_{i}}{{b}_{j}}}$, k=$\overline{0,n+m}$.

Let f and g be polynomials. Then deg(fg)=degf + degg.

For any polynomials f,g $\in \mathbb{C}\left[ X \right]$, g≠0, there exist unique polynomials

c, r $\in \mathbb{C}\left[ X \right]$ with the properties:

(1) $f=g\cdot c+r$;

(2) deg r < deg g.

*Let f,g $\in \mathbb{C}\left[ X \right]$. The polynomial f is divisible by the polynomial g if there is a polynomial h$\in \mathbb{C}\left[ X \right]$ such that f=gh. We write f$\vdots$g or g$\left| f \right.$.

The remainder theorem. The remainder on dividing a polynomial f by the binomial X-a is equal to the value f(a) of the polynomial at a.

Roots of polynomials

Bézout's theorem. Let f$\in \mathbb{C}\left[ X \right]$ be a non-zero polynomial and a$\in$ $\mathbb{C}$. Then a is a root of the polynomial f if and only if X-a divides f.

Let f$\in \mathbb{R}\left[ X \right]$, deg f=2, f=aX²+bX+c. Then:

1. f is reducible over $\mathbb{C}$

2. f is reducible over $\mathbb{R}$ if and only if $\Delta$=b²-4ac≥0.

The fundamental theorem of algebra (the d'Alembert–Gauss theorem). Every polynomial equation of degree greater than or equal to 1 has at least one complex root.

Let f $\in \mathbb{C}\left[ X \right]$. If degf=n, then f has exactly n complex roots (not necessarily distinct).

Let f = ${{a}_{0}}+{{a}_{1}}X+...+{{a}_{n}}{{X}^{n}}$ $\in \mathbb{C}\left[ X \right]$, n≥1, with roots x₁, x₂,...,x_n. We have

$f={{a}_{n}}(X-{{x}_{1}})(X-{{x}_{2}})...(X-{{x}_{n}})$ and the factorisation of f into linear factors is unique.

Viète's relations. Let $f={{a}_{0}}+{{a}_{1}}X+...+{{a}_{n}}{{X}^{n}}$ $\in \mathbb{C}\left[ X \right]$ with roots x₁, x₂ ,...,x_n; we have:

${{x}_{1}} + {{x}_{2}}+,...,+{{x}_{n}} =-\frac{{{a}_{n-1}}}{{{a}_{n}}}$;

${{x}_{1}}{{x}_{2}} + {{x}_{1}}{{x}_{3}} + ... + {{x}_{n-1}}{{x}_{n}}\text{ = }\frac{{{a}_{n-2}}}{{{a}_{n}}}$;

${{x}_{1}}{{x}_{2}}...{{x}_{n}}={{\left( -1 \right)}^{n}}\frac{{{a}_{0}}}{{{a}_{n}}}$^.

Let ${{\alpha }_{1}},{{\alpha }_{2}},...,{{\alpha }_{n}}$ $\in$ $\mathbb{C}$, n$\in$ $\mathbb{N}$, n≥2 and

${{S}_{1}}={{\alpha }_{1}}+{{\alpha }_{2}}+...+{{\alpha }_{n}};$

${{S}_{2}}={{\alpha }_{1}}{{\alpha }_{2}}+...+ {{\alpha }_{1}}{{\alpha }_{n}}+...+ {{\alpha }_{n-1}}{{\alpha }_{n}};...$

${{S}_{n}}={{\alpha }_{1}}\cdot {{\alpha }_{2}}\cdot ...\cdot {{\alpha }_{n}};$

Then ${{\alpha }_{1}},{{\alpha }_{2}},...,{{\alpha }_{n}}$ are the solutions of the equation

${{x}^{n}}-{{S}_{1}}{{x}^{n-1}}+{{S}_{2}}{{x}^{n-2}}+...+{{(-1)}^{n}}{{S}_{n}}=0.$

We say that $\alpha \in$ $\mathbb{C}$ is a root of multiplicity p of the polynomial f$\in \mathbb{C}\left[ X \right]$ if ${{\left( X-\alpha \right)}^{p}}$ divides f and ${{\left( X-\alpha \right)}^{p+1}}$ does not divide f.

Let f $\in \mathbb{R}\left[ X \right]$ be a non-zero polynomial with real coefficients and α a root of f.

1) $\bar{\alpha }$ is also a root of f.

2) α and $\bar{\alpha }$ have the same multiplicity.

Every polynomial f = ${{a}_{0}}+{{a}_{1}}X+...+{{a}_{n}}{{X}^{n}}$ of degree n≥1 with real coefficients can be decomposed into a product of first- or second-degree polynomials with real coefficients.

Let f be a non-zero polynomial with rational coefficients and a+b$\sqrt{d}$

(a,b,d $\in \mathbb{Q}$, d>0, $\sqrt{d}$ $\in \mathbb{Q}$) a root of f. Then:

1) a-b$\sqrt{d}$ is also a root of f.

2) a+b$\sqrt{d}$ and a-b$\sqrt{d}$ have the same multiplicity.

Let f = ${{a}_{0}}+{{a}_{1}}X+...+{{a}_{n}}{{X}^{n}}$ be a polynomial of degree n≥1 with integer coefficients, and α=$\frac{p}{q}$ a rational root of f, with p and q$\in \mathbb{Z}$, q≠0, (p,q)=1. Then:

1. p divides the constant term a₀

2. q divides the leading coefficient a_n.