Polynomial Functions

A function of the form $f(x) = a_nx^n, a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$ is callled a polynomial function where $a_i\in\mathbb{C}$, where $i = 0, 1, 2, \ldots, n$ i.e. $i \geq 0$ and $i\in\mathbb{I}$. Since $a_i\in\mathbb{C}$, it is evident that $a_i\in\mathbb{R}$ because $\mathbb{R}\subset\mathbb{C}$. This equation will be called an equation of degree $n$ if and only if $a_n\neq 0$. $a_n$ is called leading coefficient of the polynomial. If the leading coefficient is $1$ then the polynomial is also callled monic polynomial. A polynomial with one term is called monomial, with two terms, a binomial and with three terms it is called a trinomial. The most useful trinomials are quadratic equations, which we will study further in this chapter. If $f(x) = a_0$, then it is called a constant polynomial. If $n = 0$ implies $f(x) = a_0$, which will be a polynomial of degree $0$. If $f(x) = 0$, then it is callled zero polynomial, in this case the degree is defined as $-\infty$ to satisfy the first two properties given below. We take domain and range of these polynimoials or functions as set of complex numbers, $\mathbb{C}$. A real number $r$ or a complex number $z$, for which $f(r) = 0$ or $f(z) = 0$, then $r$ and $z$ are called zeros, roots or solutions of the polynomial.

If $f(x)$ is a polynomial of degree $p$, and $g(x)$ is a polynomial of degree $q$, then

1. $f(x)\pm g(x)$ is a polynomial of degree $\max(p, q)$,
2. $f(x).g(x)$ is a polynomial of degree $p + q$, and
3. $f(g(x))$ is a polynomial of degree $p.q$, where $g(x)$ is not a constant polynomial.

The $f(x)$ shown at the beginning is a polynomial in one variable, and similarly, we can have polynomials in $2, 3, \ldots, m$ variables. The domain of such a polynomial of $m$ variables is set of ordered $m$ tuple of complex numbers and range is $\mathbb{C}$.

Division of Polynomials

If $P(x)$ and $D(x)$ are any two polynomials such that $D(x)\not\equiv 0$, then two unique polynomilas $Q(x)$ and $R(x)$ can be found such that $P(x) = D(x).Q(x) + R(x)$. Here, the degree of $R(x)$ would be less than the degree of $D(x)$ or $R(x)\equiv 0$. Like numbers $Q(x)$ denotes the quotient, and is called so, while $R(x)$ is called the remainder.

Particulalrly, if $P(x)$ is a polynomial with complex coefficients and $z$ is a complex number, then a polynomial $Q(x)$ of degree $1$ less than $P(x)$ will exist such that $P(x) = (x - z)Q(x) + R$, where $R$ is a complex number.

Remainder Theorem

If $f(x)$, a polynomial, is divided by $(x - \alpha)$, then the remainder is $f(\alpha)$.

Proof: $f(x) = (x - \alpha)Q(x) + R\Rightarrow f(\alpha) = (\alpha - \alpha)Q(x) + R \Rightarrow R = f(\alpha)$.$\square$

Factor Theorem

$f(x)$ has a factor $(x - \alpha)$, if and only if, $f(\alpha) = 0$

Proof: Following from remainder theorem, described above, if $R = f(\alpha) = 0$, then $f(\alpha) = (x - \alpha)Q(x)$, and thus, $f(x)$ has a factor $(x - \alpha)$.$\square$

Fundamental Theorem of ALgebra

Every polynomial of degree greater or equal than one has at least one root/solution/zero in the complex numbers. We can also say that for $f(x)$ introduced in the beginning with $n\geq 1$, then there exists a $z\in\mathbb{C}$, such that $$f(z) = a_nz^n + a_{n - 1}z^{n - 1} + \cdots + a_1z + a_0 = 0.$$ Now it is trivial to deduce that an $n$th degree polynomial will have exactly $n$ roots i.e. $f(x) = a(x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_{n - 1})(x - \alpha_n)$.

Notes:

1. Some of the roots of the polynomial may have repetition.
2. If a root $\alpha$ repeats $m$ times, then $m$ is called {\it multiplicity} of the root $\alpha$ or $\alpha$ is called $m$ fold root.
3. Quadratic surds of the form $\sqrt{a} + \sqrt{b}$, where $\sqrt{a}$ and $\sqrt{b}$ are irrational numbers, then it will have its conjugate as a root. Similarly, if a complex root occurs, then it always occurs in pair with its complex conjugate as another root of the polynomial. However, if the coefficients are complex numbers then it is not mandatory for complex roots to appear in conjugate pairs.

Identity Theorem

If $f(x)$, a polynomial of degree $n$, vanishes for at least $n + 1$ distinct values of $x$, then it is identically $0$.

Proof: We have $f(x) = a(x - \alpha_1)(x - \alpha_2)\cdots(x - \alpha_{n - 1})(x - \alpha_n)$, and we let that it vanishes for $\alpha_{n + 1}$, then

$$ f(x) = a(\alpha_{n + 1} - \alpha_1)(\alpha_{n + 1} - \alpha_2)\cdots(\alpha_{n + 1} - \alpha_{n - 1})(\alpha_{n + 1} - \alpha_n) = 0$$

Because $\alpha_{n + 1}$ is different from $\alpha_1, \alpha_2, \ldots, \alpha_{n - 1}, \alpha_n$ none of the terms will vanish, which implies that $a = 0 \Rightarrow f(x) = 0$.$\square$

Corollary: Consider two poynomials $f(x)$ and $g(x)$ having degrees $p$ and $q$ respectively, such that $p\leq q$. If both of them have equal value for $q + 1$ distinct values of $x$, then they must be equal.

Proof: Let $h(x) = f(x) - g(x)$. This implies that the degree of $h(x)$ is at most $q$ and it vanishes for $q + 1$ distinct values of $x$. $\Rightarrow h(x) = f(x) - g(x) = 0 \Rightarrow f(x) = g(x)$.$\square$

Corollary: If $f(x)$ is a periodic polynomial with some constant period $T$ i.e. $f(x) = f(x + T)\;\forall\;x\in\mathbb{R}$, then $f(x) = c$.

Proof: Let $f(0) = x$, then $f(0) = f(T) = f(2T) = \cdots = c$. Thus, polynomials $f(x)$ and constant polynomial $g(x) = c$ take same values for infinite number of points. Hence, they must be identical.

Rational Root Theorem

If $p, q\in\mathbb{Z}, q\neq = 0$ such that they are relatively prime i.e. $gcd(p, q) = 1$, then if $\frac{p}{q}$ is a root of the equation $a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 = 0$, where $a_0, a_1, \ldots, a_{n - 1}, a_n \in\mathbb{I}$ and $a_n = 0$, then $p$ is a divisor of $a_0$ and $q$ that of $a_n$.

Proof: Since $\frac{p}{q}$ is a root, we have $$a_n\left(\frac{p}{q}\right)^n + a_{n - 1}\left(\frac{p}{q}\right)^{n - 1} + \cdots + a_1\frac{p}{q} + a_0 = 0$$ $\Rightarrow a_np^n + a_{n - 1}p^{n - 1}q + \cdots + a_1q^{n - 1}p + a_0q^n = 0$

$\Rightarrow a_{n - 1}p^{n - 1} + a_{n - 1}p^{n - 2}q + \cdots + a_1pq^{n - 2} + a_0q^{n - 1} = -a_n\frac{p^n}{q}$

Everything on L.H.S. is integer and $p, q$ are relatively prime therefore $q$ must divide $a_n$. Similalry, it can be proven that $a_0$ is divisible by $q$.

Corollary(Integer Root Theorem): If roots of $x^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 = 0$, where $0\leq i\leq n- 1$ are integers and coefficients are also integer, are integer then all the roots divide $a_0$.

Proof: This corollary is a direct result from previous corollary.

Vieta's Relations

If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are $n$ roots of the equation $a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0 = 0$, then $\displaystyle\sum_{i=0}^n\alpha_i = -\frac{a_{n - 1}}{a_n}, \sum_{1\leq i\leq j\leq n}\alpha_i\alpha_j = \frac{a_{n - 2}}{a_n}, \sum_{1\leq i\leq j\leq k\leq n}\alpha_i\alpha_j\alpha_n = -\frac{a_{n - 3}}{a_n}, \cdots, \alpha_1\alpha_2\ldots\alpha_n = (-1)^n\frac{a_0}{a_n}$.

These relations are denoted as $\sigma_1, \sigma_2, \ldots, \sigma_n$ as well. These relations are known as Vieta's relations.

Symmetric Functions

Consider functions $a + b + c, a^2 + b^2 + c^2, (a - b)^2 + (b - c)^2 + (c - a)^2,$ and $(a + b)(b + c)(c + a)$ in which the terms can be interchanged without changing the overall function. Functions demonstrating such behavior are known as {\it symmetric} functions.

In general, if a function is of $n$ variables then this definition warrants that any two variable can be interchanged without changing the function. Thus, we see that Vieta's relations are symmetric functions.

Common Roots of Polynomial Equations

If $\alpha$ is a common root of the polynomial equations $f(x) = 0$ and $g(x) = 0$, if and only if, it is a root of the HCF of the polynomilas $f(x)$ and $g(x)$. The HCF of two polynomials can be found exactly like HCF of two integers using Euclid's method.

Irreducibility of Polynomials

When we talk of irreducability we talk in terms of set to which the coefficients of the polynomial belong. The set could be $\mathbb{Q}, \mathbb{Z}, \mathbb{R}$ or $\mathbb{C}$.

An irreducible polynomial is, a non-constant polynomial which cannot have non-constant factors in the same set as coefficients of the polynomial itself.

Consider following example:

1. $x^2 - 5x + 6 = (x - 2)(x - 3)$
2. $x^2 - \frac{4}{9} = \left(x - \frac{2}{3}\right)\left(x + \frac{2}{3}\right)$
3. $x^2 - 5 = (x - \sqrt{5})(x + \sqrt{5})$
4. $x^2 + 9 = (x + 3i)(x - 3i)$

Over $\mathbb{I}$, first is reducible while other are irreducible, over $\mathbb{Q}$ first two are reducible bit last two are not, over $\mathbb{R}$, first three are reducible but last one is not and over $\mathbb{C}$ all are reducible.

Gausss' Lemma

If a polynomial with integer coefficients is reducible over $\mathbb{Q}$, then it is reducible over $\mathbb{Z}$.

Eisenstein's Irreducibility Criterion Theorem

Consider the polynomial $f(x) = a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0$ with integer coefficients. If there exists a prime $p$ such that the following three conditions apply

1. $p$ divides each $a_i$ for $0\leq i < n$,
2. $p$ does not divide $a_n$, and
3. $p^2$ does not divide $a_0$,

then $f(x)$ is irreducible over rational numbers and integers.

Proof: If possible, let us assume that $f(x) = g(x).h(x)$ such that $g(x) = b_kx^k + b_{k - 1}x^{k - 1} + \cdots + b_1x + b_0$ and $h(x) = c_lx^l + c_{l - 1}x^{l - 1} + \cdots + c_1 + c_0$, where $b_i, c_i\in\mathbb{Z}\;\forall\;i = 0, 1, 2, \ldots; b_k\neq =, c_l\neq 0; 1\leq k, l\leq n - 1$.

Comparing leading coefficient on both sides, we have $a_n = b_kc_l$. As $p\nmid a_n\Rightarrow p\nmid b_kc_l\Rightarrow p\nmid b_k$ and $p\nmid c_l$.

Similarly, $a_0 = b_0c_0$. As $p|a_0$ and $p^2\nmid a_0\Rightarrow p|b_0c_0$, but both $b_0$ and $c_0$ cannot be divided by $p$. Without loss of generality, we suppose $p|b_0$ and $p\nmid c_0$. Suppose $i$ be the smallest index such that $b_i$ is not divisible by $p$. There is such an index $i$ since $p\nmid b_k$, where $1\leq i\leq k$. Depending on $i$ and $k$, for $i\leq k, a_i = b_ic_0 + b_{i-1}c_1 + \cdots + b_0c_i$ and for $i > k, a_i = b_ic_0 + b_{i-1}c_1 + \cdots + b_{i-k}c_k$.

We have $p|a_i$ and by supposition $p$ divides each one of $b_0, b_1, \ldots, b_{i - 1}\Rightarrow p|b_ic_0$. But $p\nmid c_0 \Rightarrow p|b_i$, which is a contradiction, and therefore, $f(x)$ is irreducible.