12.4: Polynomials

Expressions like $3 - 5x$ and $1 + 3x - 2x^{2}$ are examples of polynomials. In general, a polynomial is an expression of the form

$$f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_n x^n \nonumber$$

where the $a_i$ are numbers, called the coefficients of the polynomial, and $x$ is a variable called an indeterminate. The number $a_{0}$ is called the constant coefficient of the polynomial. The polynomial with every coefficient zero is called the zero polynomial, and is denoted simply as $0$.

If $f(x) \neq 0$, the coefficient of the highest power of $x$ appearing in $f(x)$ is called the leading coefficient of $f(x)$, and the highest power itself is called the degree of the polynomial and is denoted $\text{deg} (f(x))$. Hence

$$\begin{array}{ll} -1+5x+3x^2 & \mbox{ has constant coefficient } -1, \mbox{ leading coefficient } 3, \mbox{ and degree } 2, \\ 7 & \mbox{ has constant coefficient } 7, \mbox{ leading coefficient } 7, \mbox{ and degree } 0, \\ 6x - 3x^3 + x^4 - x^5 & \mbox{ has constant coefficient } 0, \mbox{ leading coefficient } -1, \mbox{ and degree } 5. \\ \end{array} \nonumber$$

We do not define the degree of the zero polynomial.

Two polynomials $f(x)$ and $g(x)$ are called equal if every coefficient of $f(x)$ is the same as the corresponding coefficient of $g(x)$. More precisely, if

$$f(x) = a_0 + a_1x + a_2x^2 + \cdots \quad \mbox{ and } \quad g(x) = b_0 + b_1x + b_2x^2 + \cdots \nonumber$$

are polynomials, then

$$f(x) = g(x) \quad \mbox{ if and only if } \quad a_0 = b_0, a_1 = b_1, a_2 = b_2, \dots \nonumber$$

In particular, this means that

$$f(x) = 0 \mbox{ is the zero polynomial if and only if } a_0 = 0, a_1 = 0, a_2 = 0, \dots \nonumber$$

This is the reason for calling $x$ an indeterminate.

Let $f(x)$ and $g(x)$ denote nonzero polynomials of degrees $n$ and $m$ respectively, say

$$f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \quad \mbox{ and } \quad g(x) = b_0 + b_1x + b_2x^2 + \cdots + b_mx^m \nonumber$$

where $a_n \neq 0$ and $b_m \neq 0$. If these expressions are multiplied, the result is

$$f(x)g(x) = a_0 b_0 + (a_0 b_1 + a_1 b_0)x + (a_0 b_2 + a_1 b_1 + a_2 b_0)x^2 + \cdots + a_nb_m x^{n+m} \nonumber$$

Since $a_{n}$ and $b_{m}$ are nonzero numbers, their product $a_{n}b_{m} \neq 0$ and we have

035227 If $f(x)$ and $g(x)$ are nonzero polynomials of degrees $n$ and $m$ respectively, their product $f(x)g(x)$ is also nonzero and

$$\text{deg} [f(x)g(x)] = n+m \nonumber$$

035231 $(2-x+3x^2)(3+x^2-5x^3)=6 - 3x+11x^2-11x^3+8x^4-15x^5.$

If $f(x)$ is any polynomial, the next theorem shows that $f(x) - f(a)$ is a multiple of the polynomial $x - a$. In fact we have

Remainder Theorem035236 If $f(x)$ is a polynomial of degree $n \geq 1$ and $a$ is any number, then there exists a polynomial $q(x)$ such that

$$f(x) = (x-a)q(x)+f(a) \nonumber$$

where $\text{deg} (q(x)) = n - 1$.

Write $f(x) = a_{0} + a_{1}x + a_{2}x^{2} + \dots + a_{n}x^{n}$ where the $a_i$ are numbers, so that

$$f(a) = a_{0} + a_{1}a + a_{2}a^{2} + \dots + a_{n}a^{n} \nonumber$$

If these expressions are subtracted, the constant terms cancel and we obtain

$$f(x)-f(a) = a_1(x-a)+a_2(x^2-a^2)+ \cdots + a_n (x^n-a^n). \nonumber$$

Hence it suffices to show that, for each $k \geq 1$, $x^{k}- a^{k} = (x - a)p(x)$ for some polynomial $p(x)$ of degree $k - 1$. This is clear if $k = 1$. If it holds for some value $k$, the fact that

$$x^{k+1} - a^{k+1} = (x-a)x^k + a(x^k-a^k) \nonumber$$

shows that it holds for $k + 1$. Hence the proof is complete by induction.

There is a systematic procedure for finding the polynomial $q(x)$ in the remainder theorem. It is illustrated below for $f(x) = x^{3} - 3x^{2} + x - 1$ and $a = 2$. The polynomial $q(x)$ is generated on the top line one term at a time as follows: First $x^{2}$ is chosen because $x^{2}(x - 2)$ has the same $x^{3}$-term as $f(x)$, and this is subtracted from $f(x)$ to leave a “remainder” of $-x^{2} + x - 1$. Next, the second term on top is $-x$ because $-x(x - 2)$ has the same $x^{2}$-term, and this is subtracted to leave $-x - 1$. Finally, the third term on top is $-1$, and the process ends with a “remainder” of $-3$.

$$\def\arraystretch{1.2} \begin{array}{rr @{\hskip\arraycolsep}c@{\hskip\arraycolsep} rrrrrr} & & & & x^2 & - & x & - & 1 \\ \cline{2-9} x-2 & \Big) & x^3 & - & 3x^2 & + & x & - & 1 \\ & & x^3 & - & 2x^2 & & & & \\ \cline{3-9} & & & & -x^2 & + & x & - & 1 \\ & & & & -x^2 & + & 2x & & \\ \cline{4-9} & & & & & & -x& - & 1 \\ & & & & & & -x& + & 2 \\ \cline{6-9} & & & & & & & - & 3 \end{array} \nonumber$$

Hence $x^{3} - 3x^{2} + x - 1 = (x - 2)(x^{2} - x - 1) + (-3)$. The final remainder is $-3 = f(2)$ as is easily verified. This procedure is called the division algorithm.¹

A real number $a$ is called a root of the polynomial $f(x)$ if

$$f(a) = 0 \nonumber$$

Hence for example, $1$ is a root of $f(x) = 2 - x + 3x^{2} - 4x^{3}$, but $-1$ is not a root because $f(-1) = 10 \neq 0$. If $f(x)$ is a multiple of $x - a$, we say that $x - a$ is a factor of $f(x)$. Hence the remainder theorem shows immediately that if $a$ is root of $f(x)$, then $x - a$ is factor of $f(x)$. But the converse is also true: If $x - a$ is a factor of $f(x)$, say $f(x) = (x - a) q(x)$, then $f(a) = (a - a)q(a) = 0$. This proves the

Factor Theorem035282 If $f(x)$ is a polynomial and $a$ is a number, then $x - a$ is a factor of $f(x)$ if and only if $a$ is a root of $f(x)$.

035287 If $f(x) = x^{3} - 2x^{2} - 6x + 4$, then $f(-2) = 0$, so $x - (-2) = x + 2$ is a factor of $f(x)$. In fact, the division algorithm gives $f(x) = (x + 2)(x^{2} - 4x + 2)$.

Consider the polynomial $f(x) = x^{3} - 3x + 2$. Then $1$ is clearly a root of $f(x)$, and the division algorithm gives $f(x) = (x - 1)(x^{2} + x - 2)$. But $1$ is also a root of $x^{2} + x - 2$; in fact, $x^{2} + x - 2 = (x - 1)(x + 2)$. Hence

$$f(x) = (x-1)^2(x+2) \nonumber$$

and we say that the root $1$ has multiplicity $2$.

Note that non-zero constant polynomials $f(x) = b \neq 0$ have no roots. However, there do exist nonconstant polynomials with no roots. For example, if $g(x) = x^{2} + 1$, then $g(a) = a^{2} + 1 \geq 1$ for every real number $a$, so $a$ is not a root. However the complex number $i$ is a root of $g(x)$; we return to this below.

Now suppose that $f(x)$ is any nonzero polynomial. We claim that it can be factored in the following form:

$$f(x) = (x-a_1)(x-a_2) \cdots (x-a_m)g(x) \nonumber$$

where $a_{1}, a_{2}, \dots, a_{m}$ are the roots of $f(x)$ and $g(x)$ has no root (where the $a_i$ may have repetitions, and may not appear at all if $f(x)$ has no real root).

By the above calculation $f(x) = x^{3} - 3x + 2 = (x - 1)^2(x + 2)$ has roots $1$ and $-2$, with $1$ of multiplicity two (and $g(x) = 1$). Counting the root $-2$ once, we say that $f(x)$ has three roots counting multiplicities. The next theorem shows that no polynomial can have more roots than its degree even if multiplicities are counted.

035311 If $f(x)$ is a nonzero polynomial of degree $n$, then $f(x)$ has at most $n$ roots counting multiplicities.

If $n = 0$, then $f(x)$ is a constant and has no roots. So the theorem is true if $n = 0$. (It also holds for $n = 1$ because, if $f(x) = a + bx$ where $b \neq 0$, then the only root is $-\frac{a}{b}$.) In general, suppose inductively that the theorem holds for some value of $n \geq 0$, and let $f(x)$ have degree $n + 1$. We must show that $f(x)$ has at most $n + 1$ roots counting multiplicities. This is certainly true if $f(x)$ has no root. On the other hand, if $a$ is a root of $f(x)$, the factor theorem shows that $f(x) = (x - a) q(x)$ for some polynomial $q(x)$, and $q(x)$ has degree $n$ by Theorem [thm:035227]. By induction, $q(x)$ has at most $n$ roots. But if $b$ is any root of $f(x)$, then

$$(b-a)q(b)=f(b)=0 \nonumber$$

so either $b = a$ or $b$ is a root of $q(x)$. It follows that $f(x)$ has at most $n$ roots. This completes the induction and so proves Theorem [thm:035311].

As we have seen, a polynomial may have no root, for example $f(x) = x^{2} + 1$. Of course $f(x)$ has complex roots $i$ and $-i$, where $i$ is the complex number such that $i^{2} = -1$. But Theorem [thm:035311] even holds for complex roots: the number of complex roots (counting multiplicities) cannot exceed the degree of the polynomial. Moreover, the fundamental theorem of algebra asserts that the only nonzero polynomials with no complex root are the non-zero constant polynomials. This is discussed more in Appendix [chap:appacomplexnumbers], Theorems [thm:034196] and [thm:034210].

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1. This procedure can be used to divide $f(x)$ by any nonzero polynomial $d(x)$ in place of $x - a$; the remainder then is a polynomial that is either zero or of degree less than the degree of $d(x)$.↩