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Very early in these notes we found that the natural numbers are inadequate for dealing with many situations encountered in everyday life (e.g., bank overdrafts), and our solution to that problem was to enlarge that collection to the system of integers. But we subsequently discovered deficiencies in this new collection (e.g., one cannot always divide), so we again added on, this time achieving the rationals. Low and behold, problems still occurred (e.g., very few square roots), and as a result we created the real numbers (in which, at least, all positive reals have square roots).

Even the reals have problems, the most obvious being that negative numbers still do not have square roots. This particular deficiency can be overcome by extending the number system yet one more time, in this instance achieving the complex number system, but we are not going to worry about the specifics of that construction43. What we would like the reader to remember is that when some mathematical deficiency is found in a ``number system'', the problem can often be overcome by enlarging the system.

From our point of view the collection of polynomials amounts to an enlargement of the real number system for the purpose of giving a very simple way to describe many physical phenomena which ``change over time'', e.g., the height of a balloon in flight, or the amount of carbon 14 in a dinosaur bone44. (In particular, the enlargement is for reasons other than to solve the ``no square root'' problem, and in that sense the construction of polynomials is an extension of the real numbers in a completely different direction than is the construction of the complex numbers.)

The definition of a polynomial found in standard elementary texts is usually along the following lines: a polynomial is an expression of the form

\begin{displaymath}c_nx^n + c_{n-1}x^{n-1} + \cdots + c_1x + c_0, \end{displaymath}

where $\,c_0,\dots, c_{n-1}\,$ and $\,c_n\,$ are real numbers, e.g., $\,8x^4-6x^3 + \sqrt{2}x^2 - \pi x + \frac{14}{3}$. (If you are comfortable with this approach to polynomials, fine! Skip the next two paragraphs, the subsequent sample test question, and begin again with the paragraph beginning ``As was the case ... ''.) So what is the problem? Answer: it is a ``definition in terms of notation'', i.e., it attempts to define something not by specifying the appropriate characteristics but by describing how ``it'' (whatever ``it'' is) is to be represented notationally. That's analogous to defining the number three (which lives in the untouchable mathematical world) to be the numeral 3 (which lives in the real world of death and taxes). Or, to dredge up a prior analogy, it's akin to confusing a picture of George Washington with George Washington. Until we are told the meaning of ``$x$'' this is not a definition; it is slight-of-hand.

We intend to give a definition of ``polynomial'' more in the spirit late twentieth century mathematics. But you'll need to put your thinking caps on45.

What is ``$x$''?

Let A denote the system46 of integers, or the system of rational numbers, or the system of real numbers and let B denote a system larger than A, e.g., the system of rational numbers, real numbers or complex numbers respectively. (We will illustrate parenthetically with the following example: A is the collection of rational numbers; B is the collection of real numbers.) Choose an element $\,b\,$ of $\,{\bf B}\,$ and consider the collection ${\bf A}[b]\,$ (read ``A square brackets B'') of all non negative integer powers $\,b^n\,$ of $\,b$; all multiples $\,c\cdot b^n\,$ of such powers by numbers $\,c\,$ contained within A; and all (finite) sums and differences of such multiples. Elements of $\,{\bf A}[b]\,$ are called
polynomials in $\,b\,$ with coefficients in A. (Example: Let $\,b = \sqrt[3]{5}$. Then $\,\frac{1}{2} (\sqrt[3]{5})^7 - 17(\sqrt[3]{5})^5 + \frac{228}{19}$, $\, 84 (\sqrt[3]{5})^2 + 1\,$ and $\,14\,$ ( $\,=
14(\sqrt[3]{5})^0\,$) would be
typical elements of $\,{\bf A}[\sqrt[3]{5}]\,$, and therefore would be called ``polynomials in $\,\sqrt[3]{5}\,$ with rational coefficients''. The coefficients of the first polynomial are $\,\frac{1}{2},\ 17\,$ and $\,\frac{228}{19}$, those of the second are $\,84\,$ and $\,1$, and that of the third is $\,14$.) We say that $\,b\,$ is algebraic over A if at least one of these polynomials, with at least one non zero coefficient, is zero47. (Example: $\,\sqrt[3]{5}\,$ is algebraic over the integers, because $\,(\sqrt[3]{5})^3-5 = 0$.) When $\,b\,$ is not algebraic over A it is (a) transcendental ( element) over A, e.g., $\,\pi\,$ is transcendental over the integers, and is also transcendental over the rationals48. Transcendental elements (over A) are also called indeterminants (over A) or variables49. When one speaks of a polynomial in an indeterminant (or variable) $\,x\,$ (with coefficients in A) one means a polynomial in $\,x\,$ (with coefficients in A) in the sense just described, always with the understanding that $\,x\,$ is a symbol representing some otherwise unspecified50

transcendental element over A. Examples: $\,x^4 +3x + 5\,$ is a polynomial over the integers, i.e., a polynomial with integer coefficients; $\,x^2 +
\frac{1}{4}x + 3\,$ is a polynomial over the rationals, i.e., a polynomial with rational coefficients; and $\,x + \pi\,$ is a polynomial over the reals, i.e., a polynomial with real coefficients. Note that $\,x^4 +3x + 5\,$ may also be considered a polynomial over the rationals or reals, and that $\,\frac{1}{4}x
+ 3\,$ may also be considered a polynomial over the reals. On the other hand, $\,x + \pi\,$ is neither a polynomial over the integers nor a polynomial over the rationals.

Sample Test Question : What is $\,x\,$?

Sample Responses :

$\,x\,$ is the ``unknown''.
$\,x\,$ is a symbol which can be viewed as representing any particular transcendental element over the system of numbers under discussion.

The first answer fails to satisfy my need to have mathematical symbols represent specific entities in the mathematical world51. (Moreover, the ``unknown'' terminology is a bit frightening, conjuring up images of ``secret initiation rites'' which must be attended before one can understand $\,x$.) The second answer does fill the need, because I have the freedom to view $\,x\,$ as representing a number if I so desire, e.g., I could let $\,x\,$ represent $\,\pi$.

As was the case with real numbers we define $\,x^0\,$ to be $\,1$. An integer (rational number, real number) $\,r\,$ may then be imagined as the polynomial $\,rx^0\,$ with integer (rational, real) coefficients; polynomials of this form are said to be constant. With this ``think of $\,rx^0\,$ as $\,r$'' interpretation as our justification we henceforth regard the collection of polynomials in $\,x\,$ with integer (rational, real) coefficients as an enlargement of the system of integers (rational numbers, real numbers).

The reader needs to be aware of certain terminology associated with polynomials.

Polynomials of the form $\,c x^k$, where $\,c\,$ is a constant and $\,k\,$ is a non negative integer, are called monomials. Examples are: $\,3x^2; \ -\frac{3}{\pi}x^{17}; \ 5 \
$($k=0$); and $\,x^2\,$($c = 1$).

Notice that $\,0\,$ is a monomial: take $\,c=0\,$ in the definition. All other monomials are called non zero monomials.

For examples of polynomials which are not monomials consider $\,x^2 +
2x\,$ and $\,-x +3$.

The exponent $\,k\,$ in a non zero monomial $\,cx^k\,$ is the degree of that monomial. (The degree of the zero monomial is not defined52.) Example: $\,-\frac{27}{5}x^9\,$ has degree $\,9$; $\,3\,$ has degree $\,0\,$ (since $\,3 = 3x^0$).

In all subsequent definitions relating to non zero polynomials we always regard such a polynomial as having been rearranged (if necessary) into a sum of monomials of different degrees. An example will be seen in the very next item, wherein we think of $\,x^2 +
(x-7x^2)\,$ as the polynomial $\,-6x^2 + x$.

The leading coefficient of a non zero polynomial is defined to be the constant $\,c\,$ which appears in the non zero monomial of that polynomial of highest degree, and that degree is called the degree of the polynomial. When the polynomial is denoted $\,p(x)\,$ the degree is denoted $\,deg(p(x))$. Examples: The leading coefficient of $\,p(x) = 5x^2 + 2x + 8\,$ is $\,5$, and $\,deg(p(x)) = 2$; the leading coefficient of $\,q(x) = x^2 + (x -
7x^2)\,$ is $\,-6\,$ (because $\,x^2 + (x-7x^2) = -6x^2 + x$), and $\,deg(q(x)) = 2\,$; the leading coefficient of $\,3x^8 - 3(x^4 + 2)(x^4+x+1)\,$ is $\,-3\,$ (because $\,3x^8 - 3(x^4 + 2)(x^4+x+1) = -3x^5 - 9x^4-6x-6$), and that polynomial has degree $\,5$; the leading coefficient of $\,r(x) = x^3 - (x^3+
2)\,$ is $\,-2\,$ and $\,deg(r(x)) = 0$.

A non constant non zero polynomial with leading coefficient $\,1\,$ is monic. Examples are: $\,x^2; \, x-2;\, x^7 + 5x^4 - \pi x +
17$; and $\,-9x^8 + (3x^4 + 9)(3x^3 + \frac{1}{3}x^2+x)\,$ (because $\,-9x^8 + (3x^4 + 9)(3x^3 + \frac{1}{3}x^2+x) = x^6 + 3x^5+ 27x^4 +
3x^2 + 9x$). In contrast, the polynomials $\,3x^7 - 5\,$ and $\,3x^2 - (3x +
1)(x - \frac{1}{3})$ are not monic (the second because that polynomial is equal to $\,\frac{1}{3}$).
A polynomial of degree one (i.e., of the form $\,ax + b\,$ with $\,a \neq
0$) is linear; a polynomial of degree two is quadratic; a polynomial of degree three is cubic; a polynomial of degree four is quartic; a polynomial of degree five is quintic.

The reader is assumed familiar with the arithmetical manipulation of polynomials, i.e., with addition, subtraction and multiplication thereof, but perhaps not with formal statements of the properties being used. These are now listed: the similarity with the formal laws of addition and subtraction of natural numbers, integers, rational and real numbers should not be ignored. Here $\,p(x),\ q(x)\,$ are $\,r(x)\,$ denote typical polynomials.

The Commutative Law of Addition : $\,p(x) + q(x) = q(x) +
p(x)$. Example: $\,(3x^2
+ 5x+2) + (7x^3 + x) = 7x^3 + 3x^2 + 6x + 2 = (7x^3 + x) + (3x^2 + 5x
+ 2)$.
The Associative Law of Addition : $(p(x) + q(x)) + r(x) =
p(x) + (q(x) + r(x))$. Example: $\,((2x
+ 1) + (x^4-6x)) + (5x+17) = (x^4 - 4x + 1) + (5x+17) = x^4 + x + 18
= (2x + 1) + (x^4 - x + 17) = (2x + 1) + ((x^4-6x) + (5x +
The Commutative Law of Multiplication : $\,p(x)q(x) =
q(x)p(x)$. Example: $\,(4x^5 - \frac{1}{2}x +3)(8x-2) = 32x^6 -4x^2 +
24x - 8x^5 + x - 6 =
(8x-2)(4x^5 -\frac{1}{2}x + 3)$.
The Associative Law of Multiplication : $\,(p(x)q(x))r(x) =$
p(x)(q(x)r(x))$. Example: $\left((x^3-6x+2)(x^4 + x^2-7)\right)(x^2+5) =
(x^7-5x^5+2x^4-13x^3 +2x^2 + 42x ...
...210x-70 = (x^3-6x + 2)(x^6 + 6x^4 - 2x^2 -
35) = (x^3-6x+2)((x^4+x^2-7)(x^2+5))$.
The Distributive Law : $\,p(x)(q(x)+r(x)) =
p(x)q(x) + p(x)r(x)$. Example: $\,(\frac{1}{2}x + \frac{3}{5})((7x+9) + (x^8 + x -1)) = (x^4 +
2x)(x^8 + 8x + ...
...(\frac{1}{2}x +
\frac{3}{5})(7x+9) + (\frac{1}{2}x + \frac{3}{5})(x^8 + x

The Multiplicative Property of 1 : For any polynomial $\,p(x)\,$ we have $\,1 \cdot p(x) = p(x)$, e.g., $\,1\cdot (x^3 + \sqrt[5]{\pi}x + 17) = x^3 + \sqrt[5]{\pi}x +
The Additive Property of Zero : When $\,p(x)\,$ is any polynomial we have $\, p(x) + 0 = p(x)$, e.g., $\,(\sqrt{13}x^4 + 7x) + 0 = \sqrt{13}x^4 + 7x$.

These laws are so familiar in algebraic calculations that they are usually applied unconsciously. For example, ``factoring out'' the $\,3\,$ so as to justify writing $\,3x^2 + 15 = 3(x+5)\,$ is a straightforward application of the distributive law (e), although this fact is somewhat obscured by the omission of the central equality in the unabbreviated computation $\,3x^2
+ 15 = 3x^2 + 3\cdot 5 = 3(x^2 + 5)$.

We further illustrate the use of laws with a discussion of the example used in (d), which (without comment) made use of the equality

\begin{displaymath}(x^3-6x+2)(x^4 + x^2-7) =
x^7-5x^5+2x^4-13x^3 +2x^2 + 42x -14.\end{displaymath}

How do we know this is true? Here is an explanation with more detail than you probably wish to see. The lines are numbered for reference in the discussion which follows the calculations.

(i) & (x^3-6x+2)(x^4 + x^2-7) = ((x^3 -6x...
...ii) & = x^7 - 5x^5 + 2x^4 - 13x^3 + 2x^2 + 42x -14.

The calculations are justified as follows.

We can think of $\,x^3 - 6x + 2\,$ as a sum of three monomials (i.e., $x^3,\, -6x$, and $\,2$), and since we can only add two monomials at a time we choose to add $\,2\,$ to the sum of $\,x^3\,$ and $\,-6x$, which is precisely what the placement of parentheses in $\,(x^3-6x) + 2\,$ signifies. But there is nothing special about this choice. Writing the expression as $\,x^3 - (6x - 2)\,$ would have changed the subsequent steps, but not the final answer.

Similarly, we choose to think of $\,x^4 + x^2 - 7\,$ as $\,x^4\,$ added to the sum of $\,x^2\,$ and $\,-7$.

We are using the distributive law. Specifically, in (f) we are thinking of $\,p(x) = (x^3 - 6x) + 2, \
q(x) = x^4\,$ and $\,r(x) = x^2 - 7$.

Now we are using the distributive law of addition over multiplication.

Both distributive laws are being used.

As in item (iii).

This is a blend of one law of exponents a commutativity of multiplication.

The commutative law of addition.

Note that we have not bothered with parentheses in connection with addition, i.e., precision is somewhat lacking in our explanation53.

As a consequence of the fact that polynomials obey the same rules are real numbers we could also do the same calculation in the style of elementary arithmetic, i.e.,

\begin{array}{r} x^3 - 6x + 2 \\
x^4 + x^2 - 7 \\
...line{x^7 -5x^6 + 2x^4 -13x^3 + 2x^2 \ + \ 42x - 14}

Indeed, this is more likely the way the reader would have done it. But it is simply an alternate way of writing the same thing, although the placement of parentheses is a bit different. The points to remember are: the laws (a)-(h) are what guarantee the correctness of such calculations; wrong answers (which are not simply careless mistakes) are usually due to incorrect applications and/or misunderstandings of these laws54.

A Common Mistake : The assertion

\begin{displaymath}\frac{1}{2}\,x \neq \frac{x}{2}\end{displaymath}

is wrong. Or, what amounts to the same thing: the formula $\,\frac{1}{2}x = \frac{x}{2}\,$ is correct. Students who feel the expressions are ``different'' would probably not question the (correct) equality $\,\frac{1}{2} \cdot
7 = \frac{7}{2}$, but they do not seem to appreciate the fact that manipulations of the indeterminant $\,x\,$ are governed by the the same rules which govern the manipulation of real numbers. Here's another example which can ``trip'' students. Is $\,(.5)x = x/2\,$ a correct formula? It certainly is. Why? Because $\,.5 = 1/2\,$ and $\,(1/2)\cdot x = x/2$.

Here are some further examples of correct formulas: $\,\frac{2}{5}x =
\frac{2x}{5}$; $\,\frac{x^2\cdot 7}{14} = \frac{1}{2}x$; $\,\frac{4x^2
+ 2x}{2} = 2x^2 + x\,$ (because $\,\frac{4x^2+2x}{2} = \frac{ \! \not
\hspace{-.001in}2(2x^2 + x)}{ \! \not \hspace{-.001in}2} = 2x^2+x$).

When doing polynomial arithmetic there are two important properties of degrees to keep in mind:

The degree of the sum or difference of two non zero polynomials is no greater than the largest of the degrees of the given polynomials. Examples: The degree of $\,p(x) = x^6 + 8x + 5\,$ is $\,6$, the degree of $\,q(x) = x^5 + \pi x^4\,$ is $\,5\,$ (i.e., the degrees involved are $\,6\,$ and $\,5$), and the degree of the sum $\,p(x) + q(x) = x^6 + x^5 + \pi x^4 + 5\,$ is $\,6$; the degree of $\,x^7 + 6\,$ is $\,7$, the degree of $\,-x^7 + x\,$ is $\,7$, and the degree of the sum $\,(x^7 + 6)+(-x^7 +x) = -x + 6\,$ is $\,1\,$ (in particular, the degree of a sum of polynomials can be less than the degrees of the polynomials involved); the degree of $\,8x^4 +
2x^3\,$ is $\,4$, the degree of $\,-3x^4 + x^3 + \sqrt{3}x + \pi\,$ is $\,4$, and the degree of the sum $\,5x^4 +3x^3 + \sqrt{3}x +
\pi\,$ is $\,4$.
(The additive property of degrees under multiplication): The degree of the product of two non zero polynomials is the sum of the degrees of these polynomials. Example: $\,p(x) = x^6 + 8x + 5\,$ has degree $\,6$, $\,q(x) = x^5 + \pi x^4\,$ has degree $\,5$, and the product $\,p(x)q(x) = x^{11} + \pi x^{10} + 8x^6 + (5 + 8 \pi )x^5 +
5 \pi x^4\,$ has degree $\,11 = 6 + 5$.

Powers of polynomials are defined exactly as were powers of real numbers, e.g., $\,(x^3 + 7x^2-2x +4)^5\,$ means ``write down $\,x^3 +
7x^2 - 2x + 4\,$ five times and multiply all the entries together55''. And there is nothing new about the laws of exponents: they work exactly as before. Specifically, if $\,p(x)\,$ and $\,q(x)\,$ are polynomials, and if $\,k\,$ and $\,\ell\,$ are non negative integers, then

(6.1) \begin{displaymath}
\left\{ \begin{array}{lll}
p(x)^k p(x)^{\ell} & = & p(x)^{k ...
(p(x)q(x))^k & = & p(x)^kq(x)^k.
\end{array} \right.

Examples56: Let $\,p(x) =
x^3 - \sqrt[3]{2}x + 4$, $\,q(x) =
x^5 - \pi$, $\,k = 3\,$ and $\, \ell = 2$. Then

\begin{displaymath}\begin{array}{c} p(x)^kp(x)^{\ell} = ( x^2 - \sqrt[3]{2}x + 4...
- \sqrt[3]{2}x + 4)^5 = (p(x))^{k + \ell}\ ;

(p(x)^k)^{\ell} = (( x^2 - 2^{1/3}x + 4)^3)^...
...= ( x^2 - \sqrt[3]{2}x + 4)^6 = (p(x))^{k\ell} \ ;
\end{array} \end{displaymath}


\begin{displaymath}\begin{array}{c} (p(x)q(x))^k = (( x^2 - \sqrt[3]{2}x +
... - \sqrt[3]{2}x +
4)^3(x^5-\pi)^3 = p(x)^kq(x)^k.
\end{array} \end{displaymath}

The first and third laws of exponents also generalize as in (1.4) and (1.5), but for the sake of saving space we will omit examples.

In reading the next few paragraphs the reader should be reminded of a similar discussion in §1.

A polynomial $\,q(x)\,$ divides a polynomial $\,p(x)$, in which case we write $\,q(x)\vert p(x)$, if there is a polynomial $\,s(x)\,$ such that $\,p(x) = q(x)s(x)$. When this is not the case we say that $\,q(x)\,$ does not divide $\,p(x)\,$ and we write $\,q(x) \! \not
\hspace{-.01in} \vert \ p(x)$. Note from the additive property of degrees under multiplication that $\,q(x)\vert p(x)\,$ is impossible when the degree of $\,p(x)\,$ is less than that of $\,q(x)$. Examples: $(x^2-4)\vert(x^4-16)\,$ (because $\,x^4
-16 = (x^2-4)(x^2+4)$, i.e., $s(x) = x^2+4$); $5\vert(x^2 + 4x)\,$ (because $\,x^2 + 4x = 5 \cdot (\frac{1}{5}x^2 + \frac{4}{5})$, i.e., $\,s(x) = \frac{1}{5}x^2 + \frac{4}{5}$); $\,x^3 \! \not
\hspace{-.01in} \vert \ x^2$.

When $\,p(x) = q(x)s(x)\,$ as in the previous paragraph one writes $\,s(x) = \frac{p(x)}{q(x)}\,$ and refers to $\,s(x)\,$ as the result of dividing $\,p(x)\,$ by $\,q(x)$. The first two examples of the previous paragraph could thus take the form: $\,\frac{x^4-16}{x^2-4} = x^2 + 4\,$ and $\,\frac{x^2+4x}{5} = \frac{1}{5}x^2 + \frac{4}{5}x$, whereas expressing the third example in the form $\,\frac{x^2}{x^3}\,$ would make no sense (at this point) because we have yet to discuss quotients of polynomials57.

Notice that $\,\frac{p(x)}{q(x)}\,$ makes sense when $\,p(x)\,$ and $\,q(x)\,$ are monomials, say $\,p(x) = a \cdot
x^n\,$ and $\,q(x) = b \cdot x^m$, provided $\,deg(p(x)) = n \geq m =
deg(q(x))$. Indeed, under these assumptions we have

(6.2) \begin{displaymath}
\frac{a\cdot x^n}{b\cdot x^m} = \frac{a}{b}\cdot x^{n-m}. \vspace{.07in}

When a polynomial $\,p(x)\,$ is a product $\,q_1(x) \cdot q_2(x) \cdot
\, \cdots \, \cdot q_k(x)\,$ of other polynomials we call that collection a factorization of $\,p(x)\,$ and refer to the $\,q_j(x)\,$ as factors of $\,p(x)$. In particular, $\,q(x)\,$ is a factor of $\,p(x)\,$ if $\,q(x)\vert p(x)$, for in that case we have a factorization $\,p(x) = q(x)s(x)$. Here is a concrete example: $\,(x^2 + 2x + 4)(x^3-1)(x^3-1)(x^3-1)(x^6 + 2)(x^6+2)$, which would more likely be written $\,(x^2+2x+4)(x^3-1)^3(x^6+2)^2$, is a factorization of

\begin{displaymath}\begin{array}{c} x^{23} + 2x^{22} + 4x^{21}- 3x^{20} -6x^{19}...
...7 -64x^6 + 12x^5 +
24x^4 + 48x^3 - 4x^2 -8x - 16; \end{array} \end{displaymath}

the factors are $\,x^2 + 2x + 4, \,x^3-1\,$ and $\,x^6+2$.

A polynomial $\,p(x)\,$ is irreducible if in any factorization $\,p(x) = q(x)s(x)\,$ it must be the case that at least one of $\,q(x)\,$ and $\,s(x)\,$ is a real number; otherwise $\,p(x)\,$ is reducible, i.e., it admits a factorization $\,q(x)s(x)\,$ in which both $\,q(x)\,$ and $\,s(x)\,$ have positive degree58. Example: Any linear polynomial is irreducible. (Recall that a linear polynomial is one of the form $\,ax+b$, where $\,a \neq
0$. This can be factored as $\,a(x+b/a)$, or as $\,(-1)(-ax-b)$, or as $\,\frac{1}{2}(2ax + 2b)$, etc., but by the additive property of degrees it cannot be the case that each of two factors involves $\,x$.)

As will soon be seen, monic irreducible polynomials assume the role of prime numbers within the system of polynomials. But they are much easier to identify.

Proposition 6.2   : The monic irreducible polynomials are precisely:
the monic linear polynomials; and
the monic quadratic polynomials $\,x^2 + bx +c\,$ satisfying $\,b^2 - 4c < 0$.
In particular, any monic polynomial of degree three or greater, or any monic quadratic for which $\,b^2-4c \geq 0$, can be factored into a product of two polynomials each having positive degree.

Examples: $\,x,\,x-3$, and $\,x + \pi$ are irreducible, and the same is true of $\,x^2 +1\,$ (because $\,b^2-4c = 0-4 = -4 < 0\,$) and $\,x^2 + x
+ 1\,$ (because $\,b^2-4c = 1-4 = -3 < 0$).

The number $\,b^2-4c\,$ which arises in the proposition statement is called the discriminant of the polynomial $\,x^2 + bx + c$. More generally, the discriminant of a quadratic polynomial $\,ax^2 + bx +c\,$ is $\,\frac{1}{4a^2}(b^2-4ac)$, but in elementary courses the factor $\,\frac{1}{4a^2}\,$ is usually omitted, and we will follow this custom 59.

The proposition is a bit deceptive: it can tell you if a given polynomial can be factored, but when this is the case it is of no help in finding a factorization60. For example, it tells us immediately that $\,x^4 + 4\,$ can be factored, but it takes some fiddling61 to come up with

(6.3) \begin{displaymath}
x^4 + 4 = (x^2 + 2x + 2)(x^2-2x+2). \vspace{.07in}

An exception to the difficulties alluded to in the previous paragraph occurs with monic quadratic polynomials $\,x^2 + bx +c\,$: if Proposition 6.3(b) indicates that factorization is possible, i.e., if the discriminant $\,b^2-4c\,$ is non negative, then one can immediately write down that factorization. Indeed, simply check in that case that

(6.4) \begin{displaymath}
x^2 + bx + c = \left(x + \frac{b -
\sqrt{b^2-4c}\,}{2}\right)\left(x + \frac{b + \sqrt{b^2-4c}}{2}\right).\vspace{.07in}

Examples: (i) For $\,x^2 + 5x - 7\,$ we have $\,b =5, \ c = -7\,$ and therefore $\,b^2 - 4c = 25 + 28 = 53 > 0$. From the formula, or by straightforward verification, we see that $\,x^2 + 5x-7 = (x+
\frac{5 + \sqrt{53}}{2})(x + \frac{5 - \sqrt{53}}{2})$. (ii) $x^2 - 5x
+ 6 = (x-3)(x-2)$.

Example (ii) of the previous paragraph is usually viewed from a different perspective, i.e., as a particular case of the formula

(6.5) \begin{displaymath}
(x-u)(x-v) = x^2 - (u+v)x + uv. \vspace{.07in}

As an immediate consequence we see that to factor $\,x^2 + bx + c$, when it is known from Proposition 6.3(b) that factorization is possible, we simply look for two numbers which multiply to $\,c\,$ and add up to $\,-b$. For ``simple'' quadratic polynomials this method of factoring is generally faster than appealing to (6.5). Examples: (a) To factor $\,x^2 - 15x +
36\,$ look for numbers $\,u,\,v\ $ which provide a factorization of $\,36\,$ and sum to $\,15$. Since $\,36 = 4\cdot 9\,$ we might try $\,u = 4\,$ and $\,v = 9$, but then $\,u + v = 13 \neq 15$, and we thus move on to other combinations. Eventually we find that $\,u = 12\,, \ v = 3\,$ works, and therefore $\,x^2 - 15x + 36 = (x-12)(x-3)$. Of course this polynomial could also be factored using (6.5). (b) $x^2 + 5x - 6 =
(x+6)(x-1)\,$ (because $\,-6 + 1 = -5\,$ and $\,6 \cdot (-1) =
-6$). (c) $\,x^2 - 4x -12 = (x-6)(x+2)\,$ (because $\,6\cdot (-2) =
-12\,$ and $\,6 + -2 = 4$).

When $\,a \neq 0\,$ is a real number and $\,b^2 - 4ac \geq 0\,$ formula (6.5) generalizes to

(6.6) \begin{displaymath}
ax^2 + bx + c = a\left(x + \frac{b - \sqrt{b^2 -
...eft(x + \frac{b + \sqrt{b^2 - 4ac}}{2a}\right). \vspace{.07in}

Indeed, simply note that $\,ax^2 + bx + c = a(x^2 + (b/a)x +
(c/a))\,$ and then apply (6.5), with $\,b\,$ and $\,c\,$ in that formula replaced by $\,b/a\,$ and $\,c/a\,$ respectively. Example: $\,14x^2 +
6x - 2 = 14 \cdot (x + \frac{6 -
\sqrt{36 + 112}}{28}) (x + \frac{6...
...}{28}) = 14 \cdot (x + \frac{3 - \sqrt{37}}{14}) (x
+\frac{3 + \sqrt{37}}{14})$.

There are similar (but less well-known) methods for factoring third and fourth degree equations, and one can actually prove (but the methods are quite sophisticated) that there is no analogous method for factoring arbitrary fifth degree equations62.

But there are tricks that enable one to factor some high degree polynomials. To illustrate these we first point out that formulas (2.4) hold, precisely as written, when $\,a\,$ and $\,b\,$ are regarded as polynomials, e.g., in polynomial form (2.4c) gives

(6.7) \begin{displaymath}
(p(x))^2 - (q(x))^2 = (p(x) + q(x))(p(x) - q(x)). \vspace{.07in}

Taking $\,p(x) = x^4\,$ and $\,q(x) = x^3 - 6 \,$ it follows, for example, that

\begin{displaymath}x^8 - x^6 + 12x^3 - 36 = (x^4)^2 - (x^3-6)^2 = (x^4 + x^3-6)(x^4 -
x^3 + 6). \end{displaymath}

Of course Proposition 6.3(b) guarantees that each of the polynomials on the right-hand-side can in turn be factored, but none of the formulas (2.4) will do the job, and we therefore leave the task unfinished63.

Completing The Square

This phrase is used when a term of the form $\,(x^n+a)^2\,$ is introduced into an expression without changing the value (of the original expression). Perhaps the most common example is the replacement of $\,x^2 + bx\,$ within an equation by $\,(x+\frac{b}{2})^2 -
\frac{b^2}{4}$, which would usually be summarized by the statement: ``To complete the square divide the coefficient of $\,x\,$ by two, square it, and add it''. But keep in mind that you must also subtract that quantity to keep things in balance. Example: $\,x^2 + 4x + 3 =
x^2 + 4x + 4 - 4 + 3 = (x+2)^2 - 1$.

Why would we ever want to do this? Because it often enables us to ``see'' a factorization (``factoring by completing the square''). Indeed, note that $\,(x+2)^2 - 1 = (x+2)^2
- 1^2\,$ has the form $\,a^2-b^2\,$ of (2.4c), and therefore factors as $\,(a+b)(a-b)$, i.e., as $\,(x+2+1)(x-1) = (x+3)(x+1)$. Thus $\,x^2 + 4x + 3\,$ factors as $\, (x+3)(x+1)$. Of course this particular factorization is more easily achieved using (6.6), but the general technique applies to a much wider class of examples.

Here's a more sophisticated application of the idea. Notice that we introduce $\,(x^2 +
A)^2\,$ rather than $\,(x+A)^2$, and that we do so by adding and subtracting the quantity $\,4x^2\,$ rather than dividing some coefficient by two and then squaring and adding.

x^4 + 4 & = & x^4 + 4x^2 + 4 - 4x^2 \vspac...
\ & = & (x^2 + 2x + 2)(x^2-2x+2).

This is, in fact, how we obtained (6.4).

The reader should note that the last few examples have simply been variations on a theme, i.e., that of $\,a^2-b^2 = (a-b)(a+b)$, whereas (2.4) involves many other formulas. These are also available for use: ``completing the square'' is just the tip of the iceberg. Here is an example of ``completing the cube'' using (g) and (d) of (2.4):

x^3 + 3x^3 + 3x - 26 & = & x^3 +3x^2 + 3x ...
\ & = & (x-2)(x^2 + 5x + 13).
\end{array} \vspace{.07in}

We now present the polynomial analogue of the Fundamental Theorem of Arithmetic64 (Theorem 1.6).

Theorem 6.7   : Every monic polynomial $\,p(x)\,$ with real coefficients and positive degree has a factorization into irreducible polynomials, i.e., $\,p(x)\,$ can be written in the form $\,p(x) = (p_1(x))^{k_1} \cdot
(p_2(x))^{k_2} \cdot \, \cdots \, \cdot (p_r(x))^{k_r}\,$ where the monic polynomials $\,p_j(x)\,$ are irreducible and the $\,k_j\,$ are exponents. Moreover, the factorization is unique in the sense that any other factorization into irreducible monic polynomials would agree with this one, or at worst would be a reordering65.

Example: $\,(x^2 + x + 5)^3(x-3)^2(x^2+6)^5\,$ is the factorization of the monic polynomial

\begin{displaymath}x^{18} - 3x^{17} + 39x^{16} - 140x^{15} + 696x^{14}-2766x^{13} +
7865x^{12} \vspace{-.07in}\end{displaymath}

\begin{displaymath}- \ 30,135x^{11} + 65,355x^{10} - 196,650x^9 +
417,006x^8 -
776,088x^7 \end{displaymath}

\begin{displaymath}+ \ 1,950,264x^6 - 1,756,080x^5 + 6,086,016x^4 \end{displaymath}

+ 11,061,360x^2 - 583,200x + 8,748,000\end{displaymath}

into irreducible monic polynomials. Here $\,r = 3\,$ ($r$ is simply the number of irreducible polynomials involved in the factorization), $\,p_1(x) = x^2 + x + 5,\, k_1 = 3, \,p_2(x) =
x-3,\,k_2 = 2, \,p_3(x) = x^2+6, \,k_3 = 5$. Readers without computer access are may not wish to verify the example, but they should certainly recognize the analogy with the examples following Theorem 1.6.

Continuing with analogies from §1 we define the greatest common divisor or ``g.c.d.'' of a collection of monic polynomials to be the monic polynomial which divides all these polynomials, and is divisible by any other monic polynomial which has this property. Example: $\,x^2-4\,$ is the greatest common divisor of the three monic polynomials $\,x^4 - 16,\
x^5-3x^3 - 2x^2 - 4x + 8\,$ and $\,x^3 + 3x^2 - 4x - 12$. (To see that it divides all these polynomials simply note that $\,x^4-16 = (x^2-4)(x^2+4), \ x^5 - 3x^3 -2x^2 - 4x + 8 =
(x^2-4)(x^3+x-2)\,$ and $\,x^3 + 3x^2 - 4x - 12 =
(x^2-4)(x+3)$. That it is the greatest common divisor can be established by comparing degrees.) Notice that $\,x+2\,$ and $\,x-2\,$ also divide each of these polynomials, and that both also divide $\,x^2-4$. Two monic polynomials are relatively prime when their g.c.d.$\!$ is $\,1$, e.g., $\,x^2 + x
+ 1\,$ and $\,x^2 + 5x + 6$. (Remember that real numbers also qualify as polynomials.) The least common multiple or ``l.c.m.'' of a collection of monic polynomials is the the polynomial which is divisible by all these polynomials and which divides any other polynomial with this property. Example: The l.c.m.$\,$ of the monic polynomials $\,x^4 - 16,\
x^5-3x^3 - 2x^2 - 4x + 8\,$ and $\,x^3 + 3x^2 - 4x - 12\,$ is $\,x^6
+ 3x^5 - 3x^4 - 11x^3 - 10x^2 - 4x + 24$, i.e., the product $\,(x^2-4)(x^3 + x - 2)(x+3)\,$.

When $\,p(x)\,$ and $\,q(x)\,$ are monic polynomials and $\,q(x)\! \!
\not \!
\vert\ p(x)\,$ the best one can do is the following66.

Theorem 6.7   (The Division Algorithm) : Suppose $\,q(x)\,$ and $\,p(x)\,$ are polynomials, that $\,deg(q(x)) < deg(p(x))$, and that $\,q(x)\! \! \not \! \vert\ p(x)$. Then there are unique polynomials $\,s(x)\,$ and $\,r(x)\,$ such that

\begin{displaymath}p(x) = s(x)q(x) + r(x) \end{displaymath}

and $\,deg(r(x))< deg(q(x))$.

The polynomial $\,r(x)\,$ in the statement is called the remainder when dividing $\,p(x)\,$ by $\,q(x)$. Examples: For $\,p(x) = x^3 - 7x + 2\,$ and $\,q(x) = 3x^2 + 1$ we have $\,s(x) =
\frac{1}{3}\,x\,$ and $\,r(x) = -\frac{22}{3}\,x + 2\,$; for $\,p(x)
= -7x^5 + 3x^4 + x^3- \frac{8}{3}x^2 + 12x + 3\,$ and $\,q(x) = 9x^3
+ 4x + 1\,$ we have $\,s(x) = -\frac{7}{9}\,x^2 + \frac{1}{3}\,x +
\frac{37}{81}$ and $\,r(x) = -\frac{29}{9}\,x^2 + \frac{797}{81}\,x
+ \frac{206}{81}$. The reader can verify both examples simply by working out $\,s(x)q(x) + r(x)$. But to see how they were computed we need to review long division.

Polynomial Long Division

The calculations involved in working out the last example of the previous paragraph would usually appear on paper as

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x + \frac{3...
...9}\,x^2 \, + \frac{797}{81}\,x +

wherein $\,q(x)\,$ and $\,p(x)\,$ appear on the second line and $\,s(x)\,$ and $\,r(x)\,$ appear in the first and last lines respectively. This is an example of (``polynomial'') ``long division'', which is assumed familiar to the reader. Nevertheless, recalling the procedure might be worthwhile.

Step I : Write down $\,q(x)\,$ and $\,p(x)\,$ as shown, i.e., the first step is to write out the second line of the calculation shown above.

9\,x^3 + 4\,x + 1\ \raisebox{.04in}{$\vert$}\overline{\ -7x^...
x^3 \ - \ \textstyle{\frac{8}{3}}\,x^2\ +\ 12\,x \ + \ 3} \end{displaymath}

Step II : Divide the highest degree term of $\,p(x)\,$ by the highest degree term of $\,q(x)\,$ (i.e., calculate $\,-7x^5/9x^3\,$) and write down the answer (i.e., $\,-\frac{7}{9}x^2\,$) above the highest order term of $\,p(x)\,$ as shown next.

-\frac{7}{9}\,x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ ...
... \,
x^3 \ - \ \frac{8}{3}\,x^2\ +\ 12\,x \ + \ 3} \end{array}\end{displaymath}

(The placement of $\,-\frac{7}{9}\,x^2\,$ is simply to remind us that we have now ``finished'' with the term it sits above.)

Step III : Multiply $\,q(x)\,$ by the term calculated in Step II (in this case $\,-\frac{7}{9}\,x^2\,$) and write the answer on Line 3 as shown. (Here and in all subsequent steps ``like powers'' of $\,x\,$ are aligned in columns as indicated.)

-\frac{7}{9}\,x^2 \, \ \ \ \ \ \ \ \ \ \ \ \...
...7}{9}\,x^2 \, \ \
\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \end{array}\end{displaymath}

Step IV : Subtract Line 3 from $\,p(x)$, obtaining

-\frac{7}{9}\,x^2 \, \ \ \ \ \ \ \ \ \ \ \ \...
...7}{9}\,x^2} \ \ \ \ \ \, \ \ \ \ \ \ \ \ \ \ \ \ \

Step V : Now repeat Steps II-IV, with $\,p(x)\,$ replaced by what is in Line 4, leaving the term already in Line 1 (i.e., $\,-\frac{7}{9}\,x^2\,$) exactly where it is, but otherwise ignoring it. Specifically:

Divide the highest degree term of Line 4 (i.e., $\,3x^4\,$) by the highest degree term of $\,q(x)\,$ (i.e., $\,9x^3\,$) and write down the answer (i.e., $\,\frac{1}{3}x\,$) on Line 1 as shown.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x \ \ \ \ \...
...^3 \, - \,
\frac{17}{9}\,x^2} \ + \ 12\,x \ + \ 3 \end{array} \end{displaymath}

Multiply $\,q(x)\,$ by the term just calculated (i.e., $\,\frac{1}{3}x\,$) and record the answer, as shown next, as Line 5.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x + \frac{3...
...frac{1}{3}\,x \, \, \ \ \ \ \
\, \ \vspace{.1in} \end{array} \end{displaymath}

Subtract Line 5 from Line 4 and enter the result67 as Line 6.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x \ \ \ \ \...
...c{29}{9}\,x^2 \ + \
\frac{35}{3}\,x} \, \ + \
3 \end{array} \end{displaymath}

Step VI : Again repeat Steps II-IV, this time with $\,p(x)\,$ replaced by what is in Line 6, leaving the terms already in Line 1 (i.e., $\,-\frac{7}{9}\,x^2 +
\frac{1}{3}\,x\,$) exactly where they are, but otherwise ignoring them.


Divide the highest degree term of Line 6 (i.e., $\,\frac{37}{9}x^3\,$) by the highest degree term of $\,q(x)\,$ (i.e., $\,9x^3\,$) and write down the answer (i.e., $\,\frac{37}{81}\,$) on Line 1 as shown.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x + \frac{3...
...c{29}{9}\,x^2 \ + \
\frac{35}{3}\,x} \, \ + \
3 \end{array} \end{displaymath}

Multiply $\,q(x)\,$ by the term just calculated (i.e., $\,\frac{37}{81}\,$) and record the answer as Line 7.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x + \frac{3...
...\frac{148}{81}\,x + \
\frac{37}{81} \vspace{.1in} \end{array} \end{displaymath}

Subtract Line 7 from Line 6 and enter the answer68 as Line 8.

-\frac{7}{9}\,x^2 + \frac{1}{3}\,x + \frac{3...
...9}\,x^2 \, + \frac{797}{81}\,x +

Step VII : Observe that we can no longer repeat Steps II-IV, with $\,p(x)\,$ replaced by Line 8, because we cannot divide the highest order term on Line $8$ by the highest order term of $\,q(x)$. We have thus finished: the bottom line must be the remainder. Indeed, we have achieved the schematic which opened the discussion of long division.

Synthetic Division

When $\,q(x)\,$ is a monic linear polynomial, i.e., has the form $\,x-r$, there is a much easier way to divide $\,p(x)\,$ by $\,q(x)$. An explicit example should suffice to convey this method of synthetic division. We choose $\,p(x) = 9x^5 - 3x^4 + \frac{11}{2}x^2 + \pi x
- 6\,$ and $\,q(x) = x-4$.

Step I : We begin by writing

9 \ \ -3 \ \ \ 0 \ \ \ \frac{11}{2} \ \ \ \...
...\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ }

which is obtained as follows.
The right-hand-entry in Line 1 (i.e., the $\,4\,$) is the negative of the constant term of $\,q(x)\,$. (If $\,q(x)\,$ had been $\,x+4$, this entry would have been $\,-4$.) The remaining entries of this line are the coefficients of the powers of $\,x\,$ in $\,p(x)\,$ written in decreasing order.
The first entry of Line 3 is written next; it always coincides with the first entry of Line 1.
The entry in Line 2 is the next to be calculated: it is the product of the last entry on Line 1 (i.e., $\,4\,$) and the first entry in Line 3 (i.e., $\,9\,$).
The $33$ is the sum of the two entries above it.

Step II : We augment the table with the addition of two additional terms, as indicated.

9 \ \ -3 \ \ \ \ \ \, 0 \ \ \ \frac{11}{2} ...
...\ \ 132 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ }

Were do these terms come from? The $\,132\,$ on Line 2 is the product of $\,4\,$ (from the far right in Line 1) and $\,33\,$ (the second entry in Line 3), and the $\,132\,$ on Line 3 is the sum of the two entries above it.

Step III : We enter two more terms, as shown.

9 \ \ -3 \ \ \ \ \ \, 0 \ \ \ \ \ \frac{11}...
...\ \ \ \frac{1067}{2}\ \ \ \ \ \ \ \
\ \ \ \ \ \ }

Why these terms? Because $\,528 = 4 \cdot 132\,$ and $\,\frac{1076}{2}
= \frac{11}{2} + 528$.

Step IV : By this point the reader should be catching on to the pattern. The next two entries bring us to

9 \ \ -3 \ \ \ \ \ \, 0 \ \ \ \ \ \frac{11}...
...rac{1067}{2}\ \ \ \pi + 2134 \ \ \
\ \ \ \ \ \ }

and the final step gives

9 \ \ -3 \ \ \ \ \ \, 0 \ \ \ \ \ \frac{11}{...
...}{2}\ \ \ \pi + 2134 \ \ \
4\pi + 8530 \ \ \ \ }

The final entry on the bottom line (i.e., $\,4\pi + 8530\,$) is the remainder $\,r(x)$ (which we note in this context is always a number); the other entries on that line represent the coefficients of the powers of $\,x\,$ in $\,s(x)\,$ (we are thinking of $\,p(x) =
s(x)(x-4) + r(x)\,$) written in decreasing order. The calculation shows that

\begin{displaymath}\textstyle{9x^5 -3x^4 + \frac{11}{2}\,x^2 + \pi x - 6} \end{displaymath}

\textstyle{ = (9x^4 + 33x^3 +
132x^2 + \frac{1067}{2}\,x + \pi + 2134)(x-4) + (4\pi + 8530).} \vspace{.07in}

We note that if the final entry in a synthetic division calculation is zero (i.e., the right entry on the bottom line), then the linear polynomial $\,q(x)\,$ ``evenly divides'' $\,p(x)$, i.e., there is no remainder, and therefore $\,p(x) = s(x)q(x)$. For example, from the calculation

1 \ \ 0 \ \ 0 \ \ 0 \ \ 0 \ \ 0 \ -1 \ \
... \, 1 \ \ 1 \ \ 1 \ \ 1 \ \ 1 \ \ \ \
0 \ \ \ \ }

we conclude that $\,x^6-1 = (x^5 + x^4 + x^3 + x^2 + x + 1)(x-1)$. In fact this formula generalizes: for any positive integer $\,n\,$ we have69

\begin{displaymath}x^n-1 = (x^{n-1} + x^{n-1} + \cdots + x + 1)(x-1).\end{displaymath}

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Next: Functions Up: arith Previous: Logarithms
Rob Thompson 2003-04-08