Definition.

By apolynomialwe understand any function of the form

p(x) =a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n},where

a_{k}are real numbers.

Consider a polynomialWe define the p(x) =a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}.degreeofp, denoteddeg( as the largest indexp),ksuch thata_{k}≠ 0.

The fact that *a*_{k} are real is sometimes emphasised by
using the name "real polynomial". There are also other polynomials, for
instance "complex polynomials" arise when we allow
*a*_{k} to be complex numbers. However, here we will only
work with real functions, and so we just say "polynomials" and mean real
polynomials.

The numbers *a*_{k} are called coefficients. Some of them
may be zero, which means that the expression above is not unique for a given
polynomial. For instance, we can have
*p*(*x*) = 1 + 2*x*^{2} = 1 + 0*x* + 2*x*^{2} = 1 + 2*x*^{2} + 0*x*^{5}.*leading power*. Therefore people often write polynomials
in the opposite order, from the highest to the lowest power. Note that
polynomials of degree 0 are exactly the non-zero **constant functions**,
*p*(*x*) = *a*_{0}.

What is the degree of the zero polynomial (or constant zero function)? The
most popular choices are
*k* be the
smallest non-negative integer such that *k*-th*p* is zero. Then *k* − 1*p*.

We often use the phrase "let
*p*(*x*) = *a*_{n}*x*^{n} + *a*_{n−1}*x*^{n−1} +...+ *a*_{1}*x* + *a*_{0}*n*"; then it follows that the coefficient
*a*_{n}

The degree of polynomial satisfies the following properties:

(i) *p*⋅*q*) = deg(*p*) + deg(*q*);

(ii) *p* + *q*) ≤ max(deg(*p*),deg(*q*));

(iii) if *p*) > deg(*q*),*p* + *q*) = deg(*p*).

These formulas are definitely true for non-zero polynomials. Their general
validity dependas on how we define the degree of the zero polynomial.
If we set it to

In most situations it is enough to consider only polynomials with
*a*_{n} = 1,

*p*(*x*) =
*a*_{n}*x*^{n} + *a*_{n−1}*x*^{n−1} +...+ *a*_{1}*x* + *a*_{0} =
*a*_{n}[*x*^{n} + (*a*_{n−1}/*a*_{n})*x*^{n−1} +...+ (*a*_{1}/*a*_{n})*x* + *a*_{0}/*a*_{n}] =
*a*_{n}⋅*q*(*x*).

The polynomial *q* is of the same degree as *p* and the most
important properties of polynomials are the same for *p* and *q*,
so it is often enough to investigate *q* (and it has the coefficient at
the highest power equal to 1).

**Shape of a polynomial.**

For understanding polynomials it helps to know that when *x* gets close
to infinity (or negative infinity), then the leading term will win over the
other terms - they become unimportant. Thus a polynomial of degree *n*
has shape very similar to
*a*_{n}*x*^{n}*x*.
This can be very useful in many situations. Since we know how powers
*x*^{n} look like (see the
previous section), we can make a
good guess about shapes of polynomials.

These are the typical shapes for polynomials of even degree (on the left) and odd degree (on the right), assuming that there is a positive coefficient at the highest power.

If there is a negative coefficient (like in the polynomial
*x*^{3} + 5*x* + 1),*x*-axis:

How many "bends" are there for a polynomial *p*? At most
*p*) − 1.**parabola**.

From the typical shape it also follows that a polynomial of odd degree must have an even number of bends, since its "ends" at infinity and negative infinity always point in the opposite directions. On the other end, polynomials of even degree have "ends" pointing the same way, so they must have an odd number of bends. Thus for instance a polynomial of degree five can have these basic shapes:

**Linear functions.**

The name **linear function** denotes any function of the form
*f* (*x*) = *k**x* + *b*,*k*,*b* are arbitrarily chosen fixed constants. Linear functions
are therefore polynomials of degree at most 1.
They are exactly those functions whose graphs are lines. If
*k* = 0,*b*, in this way we get constant functions. Otherwise we get lines that
go up or down, depending on *k*. The coefficient *k* is called the
**slope** of the corresponging line and it gives the steepness of its
rise/descent. One popular value is *k* = 1,*b* gives the *y*-intercept.

The angle at which the line goes up/down is given by the formula
*α* = arctan(*k*).*k*,
*α*) = *k*,*k*
using triangles. Take an arbitrary triangle as indicated in the picture, find
its height *y* and base *x*, then
*k* = *y*/*x*.

The ability to quickly pass from a linear function to its graph and from a picture of a line to the formula that gives it is very useful, since linear functions belong among the most useful functions and appear very often.

**Roots of a polynomial.**

How many **roots** can we have for a non-zero polynomial *p*? Recall that by
a root we mean a number *c* such that
*p*(*c*) = 0.*p*)

We see that it can have between 1 and 5 roots. In fact, polynomials of odd degree have always at least one real root.

Sidenote: The claim about number of roots would be true for all polynomials only if we decided that the zero polynomial has degree infinity, but that would play such a havoc with other properties that people do not do it.

In complex numbers, polynomials
are better behaved. We will therefore briefly look at complex things, but we
will quickly get back to reals. Note that real polynomials are also complex
polynomials, since real numbers are a subset of complex numbers. Thus given a
real polynomial, one can substitute complex numbers for *x* and pretend
that it is a complex polynomial, drawing on some properties. We start right
away with the main result.

Theorem.

Every polynomial (real or complex) of degree at least 1 has a complex root.

Fact(factorization).

Letcbe a root of a polynomialp. Then there is a polynomialqwithdeg( such thatq) = deg(p) − 1

p(x) = (x−c)⋅q(x).

This Fact works in general for complex polynomials and complex roots, but if
we start with a real polynomial and a real root, then the new polynomial
*q* will be also real.

If we work in complex numbers, then we start with the Theorem, apply the Fact
on factorization, and if the new polynomial *q* is not a constant, we
can return to the Theorem and factor it further and so on, so we get

Corollary(factorization of polynomial to linear factors).

Letbe a polynomial of degree p(x) =a_{n}x^{n}+a_{n−1}x^{n−1}+...+a_{1}x+a_{0}Then there are complex numbers n> 0.which are all roots of c_{1},c_{2},...,c_{n}pand satisfy

p(x) =a_{n}(x−c_{1})⋅(x−c_{2})⋅...⋅(x−c_{n}).

Thus every non-zero polynomial has exactly as many complex roots as its degree (this also fits for degree zero), but some of these roots may be the same. When we put the corresponding roots together, we can rearrange the above decomposition into

*p*(*x*) =
*a*_{n}(*x* − *c*_{1})⋅(*x* − *c*_{2})⋅...⋅(*x* − *c*_{n})
=
*a*_{n}(*x* − *d*_{1})^{n(1)}⋅...⋅(*x* − *d*_{N})^{n(N)}.

Now *d*_{i} are distinct roots of *p*. Each
corresponding power *n*(*i*) is called the *multiplicity* of
the root *d*_{i} (see
this note). For example, we may have

*p*(*x*) = (*x* − 1)⋅(*x* − 1)⋅(*x* − 3)
= (*x* − 1)^{2}⋅(*x* − 3).

We would say that 1 is a root of multiplicity 2, while 3 is a root of multiplicity 1, also called a simple root.

Note that having such a nice decomposition with real coefficients is a matter of luck, since in general we are only guaranteed complex roots. Since we are only interested in real polynomials, we will now try to deduce some useful information that would not use complex numbers.

Fact.

Ifcis a complex root of a real polynomialp, then its complex conjugate is also a root ofp.

This is wonderful, because it means that when we start factoring the
polynomials, for every term *x* − *c*)*x* − *c**).

Fact.

For a complex numbercand its conjugatec* we have( wherex−c)⋅(x−c*) =x^{2}+ex+f,eandfare real numbers.

Thus when we start with a real polynomial, we may get complex roots, but their factors can be put together in pairs to create real quadratic polynomials. Therefore we get

Corollary.

Every real polynomial can be written as

p(x) =a_{n}(x−d_{1})^{n(1)}⋅...⋅(x−d_{N})^{n(N)}⋅(x^{2}+e_{1}x+f_{1})^{m(1)}⋅...⋅(x^{2}+e_{M}x+f_{M})^{m(M)},where

d_{i},e_{j}, andf_{j}are real numbers.

In this decomposition, the linear terms
*x* − *d*_{i})*p*,
the quadratic terms
*x*^{2} + *e*_{j}*x* + *f*_{j})**irreducible**, which means that they
cannot be further factored into
*x* − *a*)(*x* − *b*)*a*, which is impossible for a polynomial of
degree two.

Such factorization can be very useful, and we see that having it is
equivalent to knowing the roots of *p*. Unfortunately, finding such a
factorization is generally next to impossible. Everybody and his dog knows
the famous quadratic formula for finding roots of quadratic polynomials.
There are also formulas for finding roots of cubic polynomials (degree 3),
but they are not very pleasant and nobody remembers then anyway. Formulas for
roots of polynomials of degree 4 are so ugly that most books actually skip
them so as not to traumatize the more delicate reader. To top it off, one of
major results of general algebra is that for polynomials of degree 5 and
more there cannot exist any formula for roots.

Finding roots is thus a matter of chance and luck. Usually we try to guess
one root, then take out the corresponding factor
*x* − *c*)

**Example:**
We try to factor
*p*(*x*) = *x*^{6} − 3*x*^{4} + 2*x*^{3}.

**Solution:** We see that

*p*(*x*) =
*x*^{3}⋅[*x*^{3} − 3*x* + 2]
=
(*x* − 0)^{3}⋅[*x*^{3} − 3*x* + 2].

Thus we see that 0 is a root of *p* of multiplicity 3. What about
*q*(*x*) = *x*^{3} − 3*x* + 2?

*x*^{3} − 3*x* + 2)/(*x* − 1) =
*x*^{2} + *x* − 2.

We obtained a quadratic polynomial, whose roots are 1 and −2 by the quadratic formula. Thus 1 is actually a root of multiplicity 2 and −2 is a simple root. We get

*p*(*x*) =
(*x* − 0)^{3}⋅(*x* − 1)^{2}⋅(*x* − (−2)) =
*x*^{3}⋅(*x* − 1)^{2}⋅(*x* + 2).

**Note:** Instead of using the quadratic formula, one can also try to
guess the factorization, see
this remark.

**Example:**
We try to factor
*p*(*x*) = *x*^{3} + 3*x*^{2} + 3*x* + 2.

**Solution:** We try to guess, after a while we find that
*p*(−2) = 0.

*x*^{3} + 3*x*^{2} + 3*x* + 2)/(*x* − (−2)) =
(*x*^{3} + 3*x*^{2} + 3*x* + 2)/(*x* + 2) =
*x*^{2} + *x* + 1.

The quadratic formula shows that
*x*^{2} + *x* + 1

*p*(*x*) =
(*x* + 2)⋅(*x*^{2} + *x* + 1).

**Example:**
We try to factor
*p*(*x*) = *x*^{4} + *x*^{2} + 1.

**Solution:** We try to substitute 0, 1, −1, 2, −2,..., but after a while
we give up. What can we do? Since guessing did not help, this problem is not
solvable by tools we know.

Note that the only irreducible polynomials are those of degree 1 (linear) and
2 (quadratic), so there must be some factorization of *p*. Perhaps there
are real roots, but they are not integers so we can't guess them. Since the
polynomial is of even degree, it might also happen that there are no real
roots. In fact, that is exactly what is happening here, we have

*p*(*x*) =
(*x*^{2} + *x* + 1)⋅(*x*^{2} − *x* + 1).

How did we get it? We cheated, we created this example by starting with two irreducible quadratic polynomials and multiplied them together. If we did not know the answer from the very start, we would not know how to solve it.

Now we look at a different topic, related to factoring. We used the long
division of polynomials, in general one gets a remainder when dividing. To
show some motivation we recall the following: When we try to divide 17 by 5,
we get answer: It is 3, with remainder 2. We could write it as follows:
*r* such that there is an integer *s* with
*s*⋅5 + *r**r* < 5.

Remarkably, one can do the same for polynomials, instead of the magnitude of number we will use the degree of polynomial.

Fact(division with remainder).

Letp,qbe polynomials. Then there are unique polynomialss,rsuch thatand

p(x) =s(x)⋅q(x) +r(x)deg( r) < deg(q).

Again, one can express it as
*p*/*q* = *s* + *r*/*q*.

Sidenote: We have no restrictions on *p* and *q* in our theorem,
so one or both may be the zero polynomial. This fact remains true as long as
we define the degree of zero polynomial to be negative, so the popular
values

By a **rational function** we mean any ratio
*p*/*q**q*(*x*).**proper** if
*p*) < deg(*q*).

Every rational function can be written as a sum of a polynomial and a proper rational function with the same denominator. This is just a reformulation of the Fact on division with remainder. In practice we do it using the algorithm for long division.

Since polynomials are relatively easy to handle, we see that in order to work with rational functions, in many situations it is enough to know how to work with proper rational functions. This is particular applies to the method of partial fractions decomposition. Since this method is very important, we dedicated a special section to it, and since the main use (at least for us) is for integration, we put that section there, so for partial fractions check out Integrals - Theory - Integration methods - Partial fractions.

Briefly, the method first calls for breaking the denominator *q* into
linear and quadratic (if needed) factors as we discussed above, the proper
rational function can be then broken into a sum of simple so-called
*partial fractions*, very simple fractions whose denominators come from
the linear and quadratic factors of *q*. A practical point of view is
here.

One last note: When talking about polynomials we remarked that when *x*
is close to infinity (or negative infinity), then a polynomial is
essentially equal to its leading term. Therefore when
*p*(*x*) = *a*_{n}*x*^{n} +...*n* and
*q*(*x*) = *b*_{m}*x*^{m} +...*m*, then the rational function
*p*/*q**x* pretty much equal
to
*a*_{n}/*b*_{m})*x*^{n-m}.

Exponentials, logarithms

Back to Theory - Elementary
functions