Chapter 10  Polynomials and polynomial equations
This chapter discusses polynomials. It contains the following sections: section 10.1  In this section we introduce
polynomials, polynomial functions and polynomial equations.
 section 10.2  In this section we
describe two methods for factoring polynomials.
 section 10.3  In this section we show how to solve polynomial equations.
10.1  Introduction to Polynomials
Polynomials are an extension of quadratics so you may it useful to review quadratics before reading this section. Also many of the ideas discussed here involve complex numbers so you may want to review those as well.To create a polynomial, imagine carrying out the following steps:
 Start with a variable x.
 Multiply x by itself repeatedly.
That is, raise x to various powers,
starting with the power 0 and ending with the power n
(where n is some positive integer):
1, x, x^{ 2}, x^{ 3}, …, x^{ n}.
 Multiply each power of x by a coefficient.
Let a_{3} denote the coefficient of x^{ 3}, and so on.
 Add all the terms together.
Definition: A polynomial is an expression of the form: a_{n} · x^{ n} + a_{n −1} · x^{ n −1} + … + a_{2} · x^{ 2} + a_{1} · x + a_{0},where x is a variable, n is a positive integer and a_{0}, a_{1}, … , a_{n −1}, a_{n} are constants. The highest power of x that occurs is called the degree of the polynomial. The terms are usually written in order from highest power of x to lowest power. 
Examples:
 Any quadratic is a polynomial of degree 2.
 x^{ 4} − 8 x^{ 2}
is a polynomial of degree 4.
 x^{ 5} − 8 x^{ 3} + 10 x + 6 is a polynomial of degree 5.
Graph of a polynomial
We can create a polynomial function, called say f, whose input is x and whose output, f (x), is the polynomial evaluated at x:a_{n} x^{ n} + a_{n −1} x^{ n −1} + … + a_{2} x^{ 2} + a_{1} x + a_{0}
We can then call the output of the function “y” and make a graph of y versus x. We will get a curve like the two curves shown to the right. The graph of a polynomial function oscillates smoothly up and down several times before finally “taking off for good” in either the up or down direction.
The degree of the polynomial gives the maximum number of “ups and downs” that the graph of the polynomial can have. It also gives the maximum number of crossings of the x axis that the polynomial can have.
Polynomial equation
If we set the polynomial equal to zero or if we set y = 0 or f (x) = 0 then we get a socalled polynomial equation:a_{n} x^{ n} + a_{n −1} x^{ n −1} + … + a_{2} x^{ 2} + a_{1} x + a_{0} = 0.(Note that setting y = 0 in the polynomial’s graph means that we are looking at points where the graph crosses the x axis, and setting f (x) = 0 in the polynomial function means that we are looking for values of x for which the output of the polynomial function is zero.
There is a close connection between:
 The values of x that cause a polynomial to equal zero.
 The places where a polynomial function’s graph crosses the x axis.
 The solutions of a polynomial equation.
 The factors of a polynomial.
The factor theoremLet f (x) be a polynomial.

The fundamental theorem of algebraOver the complex numbers, a polynomial equation of degree n has exactly n roots. Over the real numbers it may have less than n.

Notes on the Factor theorem and the Fundamental theorem of algebra:
 The factor theorem essentially says that finding the roots of a polynomial equation amounts to
finding the factors of the polynomial and vice versa.
 If a is a real root, then x − a is a factor of the polynomial and
the graph of the polynomial crosses the x axis at x = a.
 If a is a double root, then (x − a)^{ 2}
is a factor of the polynomial and the graph of the polynomial just touches
the x axis at x = a rather than crosses it.
 Over the real numbers, complex roots are not allowed and must be excluded from the count of roots.
That is why there may be less than n solutions over the real numbers.
 If a + b i is a complex root then so is a − b i and the expression (x − a + b i) (x − a − b i) is a factor of the polynomial. Over the real numbers this complex expression is not allowed and is replaced by the real quadratic expression x^{ 2} − 2 a x + (a^{ 2} + b^{ 2}) (which is the previous expression, but in unfactored form). A complex root a + b i does not correspond to a graph crossing of the x axis.
Example: This example is meant to illustrate the various quantities related to polynomials that were defined above as well as these two theorems.
 x^{ 5} − 8 x^{ 3} +
10 x + 6 is a polynomial.
 f (x) = x^{ 5} − 8 x^{ 3} +
10 x + 6 is a polynomial function. Here are a few values of this function:
f (0) = 6
You should check these. Note that the last three are approximate. Values of x that cause the value of the polynomial to equal zero are called zeros of the polynomial. Thus −2.62, 1.6 and 2.4 are zeros of this polynomial.
f (1) = 9
f (−2.62) = 0
f (1.6) = 0
f (2.4) = 0  y = x^{ 5} − 8 x^{ 3} +
10 x + 6 is a polynomial relation between y and x
that can be plotted in
a graph. It is the blue curve shown above.
Any point on that curve satisfies the relation.
Notice that the curve intercepts the x axis at −2.62, 1.6 and 2.4.
 0 = x^{ 5} − 8 x^{ 3} +
10 x + 6 is a polynomial equation that can be solved for x.
Values of x that satisfy this equation are called roots or solutions
of the equation. There are 3 solutions over the real numbers:
x = {−2.62, 1.6, 2.4},
or 5 solutions over the complex numbers:x = {−2.62, 1.6, 2.4, −0.69 − 0.34 i, −0.69 + 0.34 i}.
Notice that the two complex number solutions are complex conjugates.
 A common task is to factor a polynomial.
It turns out that this polynomial factors into 3 linear factors and
1 quadratic factor over the real numbers:
(x + 2.62) (x − 1.6) (x − 2.4) (x^{ 2} + 1.3x + 0.59),
or into 5 linear factors over the complex numbers:(x + 2.62) (x − 1.6) (x − 2.4) (x + 0.69 + 0.34 i) (x + 0.69 − 0.34 i).
Notice that letting x = −2.62, 1.6, 2.4, −0.69 − 0.34 i, or −0.69 + 0.34 i causes each factor in turn to become zero, and thus causes the entire product to become zero. These values of x are, of course, just the solutions of the polynomial equation discussed in the previous bullet.
Example: Consider the polynomial equation x^{ 4} − 8 x^{ 2} = 0. The lefthandside can be factored as x^{ 2} (x + 2.83) (x − 2.83). If we write the equation like this:
(x − 0)^{ 2} (x + 2.83) (x − 2.83) = 0,then we see that it has roots at x = 2.83 and x = −2.83, as well as a double root at x = 0. This means that the graph of the polynomial function y = x^{ 4} − 8 x^{ 2} should cross the x axis at x = −2.83 and x = 2.83 and it should touch the x axis at x = 0. This can be verified by looking at the red curve shown above.
10.2  Factoring polynomials
You should read section 10.1, Introduction to polynomials, before reading this section.Factoring polynomials using the deflation method
Let f (x) denote the polynomial of degree n that we wish to factor. Also, let f (x) = 0 be the corresponding polynomial equation and let y = f (x) be the corresponding graph of the polynomial function.The deflation method starts by either:
 using trial and error (trying simple values like x = 0, 1, −1, 2,
−2, … ) and
finding a value of x that causes the polynomial to vanish
(i.e. causes f (x) to equal zero), or
 inspecting the graph and finding a value of x where the graph crosses or touches the x axis.
polynomial = (x − r) · (other factor).Then we find the other factor by dividing the polynomial by (x − r). This is called deflating the polynomial. (We saw how to divide polynomials in section 3.5.)
The process is then repeated with the deflated polynomial. The process stops (fails) when no more zeros of the deflated polynomial can be found by trial and error and the graph of the deflated polynomial has no more crossings of the x axis.
Note on graph touching the x axis: Suppose that the graph touches, rather than crosses, the x axis at x = 3. Then x = 3 is a double root and (x − 3)^{ 2} is a factor. In order to divide this factor into the polynomial to deflate it, we must write it in the expanded form x^{ 2} − 6 x + 9.
Example: Factor the polynomial 16 x^{ 3} − 13 x − 3.
Solution: It is not hard to see that x = 1 is a zero of this polynomial. This means that x − 1 is a factor. Divide this factor into the polynomial: This means that the other factor (the deflated polynomial) is the quadratic 16 x^{ 2} + 16 x + 3. It can be factored further by the deflation method but it is easier to use the method of section 8.3. As a result of that factoring, we get this final result:
(x − 1) (4x + 1) (4 x + 3).
Factoring polynomials numerically
Once the deflation method fails, the deflated polynomial can theoretically still be further factored over the real numbers into quadratic factors or over the complex numbers into linear factors containing complex numbers.Unfortunately there is no way to do this using algebra. However there is a numerical method, due to Laguerre, to do it. The Algebra Coach uses his method when the deflation method fails.
10.3  Polynomial equations
Before reading this section you may want to review the following topics: To solve a polynomial equation, follow these steps: Ensure that the polynomial equation is in standard form,
which is to have a polynomial on the lefthandside of the equation and zero on
the righthandside.
 Factor the polynomial on the lefthandside
completely using the methods of section 10.2.
 In section 4.1, we saw that if
a · b = 0 then a = 0 or b = 0.
Use this fact to set each factor equal to zero. This replaces a polynomial equation
with a set of new equations that are either linear or quadratic.
 Solve each of the new equations. The solution set for the polynomial equation is the collection of all the solutions of the new equations.
Example: Solve the polynomial equation 16 x^{ 3} − 13 x = 3.
Solution: First put the polynomial equation into standard form:
16 x^{ 3} − 13 x − 3 = 0.Factor the lefthandside using the deflation method:
(x − 1) (4x + 1) (4 x + 3) = 0.Set each factor equal to zero. This gives a set of equations:
x − 1 = 0Solve them. This gives a set of solutions:
4x + 1 = 0
4 x + 3 = 0
x = 1The solution set is x = {1, −¼, −¾}.
x = −¼
x = −¾