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The Factor Theorem and a corollary of the - UMass Blogs

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The Factor Theorem and a corollary of the - UMass Blogs 99584_6FactorTheoremEvaluated.pdf

Math 421 è Fall 2010

The Factor Theorem and a corollary of the Fundamental

Theorem of Algebra

27 August 2010

Copyright © 2006-2010 by Murray Eisenberg. All rights reserved.

Prerequisites

Mathematica

Aside from having a working Mathematica system at your disposal and knowing how to type input, how to evaluate

an Input cell, and how to navigate around a notebook, there are really no prerequisites.

In fact, working through this notebook is a good way to learn some Mathematica basics and even some more

advanced Mathematica techniques.

All the input shown uses the display form ä obtained by typing åiiå, but everything would work identically if

you typed I instead.

As you work through this notebook in Mathematica, if you come across a Mathematica function you don't under-

stand, try using the ? information command to find out something about it. (And follow the ‡ hyperlink in it, if any,

to Documentation Center help.) For example, to learn about PolynomialQuotient, used in the first section

below, evaluate:

In[1]:=?PolynomialQuotient

PolynomialQuotient@p,q,xDgivesthequotientof

pandq,treatedaspolynomialsinx,withanyremainderdropped. ‡ David Park's Presentations application is not needed for this notebook.

Mathematics

You need to know about adding and multiplying complex numbers.

Clearing a variable

Many Mathematica examples will use the variable z. Make sure no value has been assigned to z yet:

In[2]:=Clear@zD

The Division Theorem

You know from arithmetic that you can divide one positive integer a by another positive integer b to obtain an integer

quotient q and a remainder integer r. That is, given positive integers a and b, there are unique integers q and r with

a b=q+r b, or, equivalently, a=bq+r, in each case with

0£r the latter inequality says that the remainder r is less than the "divisor" b.

For example, if you use long division to divide 2356 by 14, you obtain a quotient of 168 and a remainder of 11, so

that

2356=14´168+11.

Mathematica: can verify that result for you:

In[3]:=2356Š14´168+3

Out[3]=False

And you can use Mathematica in the first place to calculate the quotient and remainder:

In[4]:=8Quotient@2356,14D,Mod@235,14D<

Out[4]=8168,11<

Likewise, you can use long division of polynomials to divide one polynomial AHzL in a variable z by another polyno-

mial BHzLin that variable so as to obtain a quotient QHzL and a remainder RHz), with the remainder having a smaller

degree than that of the "divisor" BHzL. More precisely: The Polynomial Division Theorem. Let AHzL and BHzL polynomials in the variable z with real or complex coefficients and with BHzL not the zero polynomial. Then there are unique polynomials QHzL and RHzL in z with real or complex coefficients, respectively, such that

AHzL=BHzLQHzL+RHzL

and degRHzL2 FactorTheorem.nb

The Polynomial Division Theorem. Let AHzL and BHzL polynomials in the variable z with real or complex coefficients and with BHzL not the zero polynomial. Then there are unique polynomials QHzL and RHzL in z with real or complex coefficients, respectively, such that

AHzL=BHzLQHzL+RHzL

and degRHzLFor example, z4-4z3+z+6 z3-2=Hz-4L+3z-2 z3-2, or equivalently, z4-4z3+z+6=Iz3-2MHz-4L+H3z-2L.

Again, Mathematica can verify that result...

In[5]:=z4-4z3+z+6ŠIz3-2MHz-4L+H3z-2LExpand

Out[5]=True

...and Mathematica can calculate the polynomial quotient and remainder for you:

In[6]:=aAz_E:=z4-4z3+z+6;

b@zD:=z3-2;

8PolynomialQuotient@a@zD,b@zD,zD,PolynomialRemainder@a@zD,b@zD,zD<

Out[8]=8-4+z,-2+3z<

Probably you've only used polynomial long division when both AHzL and BHzL have integer coefficients. In that case,

the result still holds - and both the quotient QHzL and remainder RHzL also have integer coefficients - provided that the

coefficient of the highest power of z in the divisor BHzL is 1.

A rigorous proof of the theorem, which is beyond the scope of this course, uses mathematical induction. The informal

idea of the proof is what happens in long division of polynomials: at each step the degree of the remainder at that step

has a lesser degree than does the remainder at the previous step, so that you can keep going until you reach a step

where the degree of the remainder at that step is less than the degree of the divisor. Exercise 1. With paper and pencil, carry out long division to express the given polynomial AHzL in the form BHzLQHzL+RHzL for the given polynomial BHzL. Then use Mathematica to verify the result. (a) PHzL=z3-5z2-4z+20 and QHzL=z-5. (b) PHzL=z3-5z2-4z+20 and QHzL=z-1. (b) PHzL=z5-12z4+48z3-62z2-33z+90 and QHzL=z2-3z+1.

FactorTheorem.nb 3

The Factor Theorem

The theorem is:

The Factor Theorem. Let PHzL be a polynomial in z (with real or complex coefficients) of degree n>0. Then a (real or complex) number z0 is a root of PHzL if and only if

PHzL=Hz-z0LQHzL

for some polynomial QHzL of degree n-1.

Proof. First assume that z0 is a root of PHzL. By long division, there is a quotient polynomial QHzL and a remainder

polynomial RHzL for which PHzL=Hz-z0LQHzL+RHzL.

Since the divisor z-z0 is of degree 1, the remainder polynomial RHzL is of degree 0, that is, a constant r. Then

PHzL=Hz-z0LQHzL+r.

Take z=z0 in both sides of this equation to obtain

0=PHz0L=0QHz0L+r,

so that r=0. Hence

PHzL=Hz-z0LQHzL.

Since PHzL has degree n whereas z-z0 has degree 1, then QHzL has degree n-1.

Conversely, assume that PHzL=Hz-z0LQHzL for a polynomial QHzL of degree n-1. Take z=z0 in this equation to

obtain PHz0L=0QHz0L=0. Hence z0 is a root of PHzL. ð

Example of the Factor Theorem

Make sure no value has been assigned to the variable z yet:

In[9]:=Clear@zD

Define a polynomial:

In[10]:=pAz_E:=z3-5z2+17z-13

By inspection, 1 is a root of this polynomial:

In[11]:=p@1D

Out[11]=0

4 FactorTheorem.nb

Give this root a name:

In[12]:=z0=1;

The question is how to form the quotient polynomial QHzL in Mathematica. (As usual, begin user-defined Mathemat-

ica names with use lower-case letters.)

One way to obtain the quotient:

In[13]:=qAz_E:=PolynomialQuotient@p@zD,z-z0,zD

q@zD

Out[14]=13-4z+z2

Another way to obtain the quotient:

In[15]:=Simplify@p@zDHz-z0LD

Out[15]=13-4z+z2

In turn find roots of the quotient:

In[16]:=quadraticRoots=Solve@q@zDŠ0,zD

Out[16]=88z®2-3ä<,8z®2+3ä<<

Factor the quotient:

In[17]:=Factor@q@zDD

Out[17]=13-4z+z2

Ugh! That does nothing. You have to tell Mathematica that you're looking for factoring "over the complex numbers".

And you may do that as follows, using the Extension option to Factor: In[18]:=factoredQuotient=Factor@q@zD,Extension®8äOut[18]=HH-2-3äL+zLHH-2+3äL+zL

It so happens that the two complex roots of the quadratic quotient polynomial have integers as real and complex parts;

in other words, these two complex roots belong to the set of Gaussian integers. Then you could also use Factor but

with the option GaussianIntegers ® True:

In[19]:=Factor@q@zD,GaussianIntegers®TrueD

Out[19]=HH-2-3äL+zLHH-2+3äL+zL

Factor the original cubic:

In[20]:=factoredCubic=Hz-z0LfactoredQuotient

Out[20]=HH-2-3äL+zLHH-2+3äL+zLH-1+zL

Exercise 2. Check that the product of three linear factors really is the original polynomial. Rather

than looking back at the original cubic polynomial p[z], you should do your checking by evaluating a suitable equation of the form p[z]==... and seeing that the result is True.

FactorTheorem.nb 5

Exercise 2. Check that the product of three linear factors really is the original polynomial. Rather

than looking back at the original cubic polynomial p[z], you should do your checking by evaluating a suitable equation of the form p[z]==... and seeing that the result is True. Exercise 3. Repeat the work that was done with the cubic polynomial z3-5z2+17z-13 but now doing all the calculations with paper and pencil. (You'll need to carry out a "long division" of the cubic by the linear polynomial z-1.) Exercise 4. Use Mathematica to factor the original cubic polynomial all at once. Exercise 5. With Mathematica, repeat for z3-3z2+z+5 what we did above for p[z]. Begin by observing that now -1 is a root. Exercise 6. Repeat the preceding exercise but for z3-äz2+z-ä. Begin by observing (and checking!) that ä is a root.

So far, the example and exercises used only half of the Factor Theorem - that if z0 is a root, then z-z0 is a factor of

the polynomial. But there is another half to the Factor Theorem, the converse: if z-z0 is a factor of the polynomial,

then z0 is a root. The following exercise illustrates that half with an example. Exercise 7. In Mathematica: Define the polynomial pHzL to be Hz+2LIz2-3z+5M. Multiply the linear factor and quadratic factors there to obtain an unfactored cubic polynomial. Finally, verify that -2 is a root of this unfactored cubic. Alternate way to factor, given the roots (advanced)

There is another way to form the (unexpanded) product of factors of a polynomial if you already know its roots:

In[21]:=quadraticProduct=Apply@Times,z-Hz.quadraticRootsLD

Out[21]=HH-2-3äL+zLHH-2+3äL+zL

Let's analyze how that worked, step-by-step. Start with:

6 FactorTheorem.nb

In[22]:=quadraticRoots

Out[22]=88z®2-3ä<,8z®2+3ä<<

Use the replacement rules in the preceding output to obtain a list of the actual roots:

In[23]:=z.quadraticRoots

Out[23]=82-3ä,2+3ä<

Form the corresponding linear polynomials:

In[24]:=z-Hz.quadraticRootsL

Out[24]=8H-2+3äL+z,H-2-3äL+z<

Finally, multiply those linear polynomials to obtain the factored quadratic:

In[25]:=Apply@Times,z-Hz.quadraticRootsLD

Out[25]=HH-2-3äL+zLHH-2+3äL+zL

That final step used the "functional programming" construct Apply[Times,...] in order to apply the Times

(multiplication) function to the list of two linear polynomials. Notice that Mathematica did not expand - multiply out - the product.

Corollary to the FTA

The following theorem is very important. Its proof is harder than you might at first suppose. One proof uses the theory

of integrating a complex-valued function of a complex variable around a closed curve in the complex plane.

The Fundamental Theorem of Algebra (FTA). Every non-constant polynomial with real or complex coefficients has at least one real or complex root.

Starting with that restatement of the FTA, a proof by mathematical induction establishes the following corollary.

Corollary. Let PHzL be a non-constant polynomial with complex coefficients, of degree n. Then PHzL has exactly n (not necessarily distinct) complex roots.

The easiest way to make sense for now of the "not necessarily distinct" part of the conclusion there is to rephrase the

corollary as follows:

FactorTheorem.nb 7

Corollary (restated). Let PHzL be a non-constant polynomial with complex coefficients, of degree n. Then there are n complex numbers z1,z2,...,zn, not necessarily different from one another, for which

PHzL=Hz-z0LHz-z1LºHz-znL.

The justification for that restatement is as follows. In view of the Factor Theorem and the FTA: If pHzL is a non-

constant polynomial, then there is a number z0 and a polynomial qHzL such that pHzL=Hz-z0LqHzL.

Example of the Corollary

Here is a polynomial of degree n=5:

In[26]:=pAz_E:=z5+H3-4äLz4-H1+12äLz3-H11+12äLz2-H12+4äLz-4;

Factor it:

In[27]:=factoredQuintic=Factor@p@zDD

Out[27]=H-2ä+zL2H1+zL3

(Notice that the Extension®{i} option to Factor was not needed because the polynomial already had complex

coefficients.)

Find the roots of the polynomial:

In[28]:=z.Solve@p@zDŠ0,zD

Out[28]=8-1,-1,-1,2ä,2ä<

There you see that the root -1 has "order 3", that is, appears three times in the list; similarly, the root 2ä has "order

2". On the other hand, the linear polynomial z-H-2L appears to order 3 in the factored polynomial; similarly,

z-H-2äL appears to order 2. This situation is precisely as predicted by the Corollary. Exercise 8. Carry out a similar confirmation of the Corollary with each of the following polynomials: (a) z6-6z5+11z4+4z3-29z2+10z+25. (b) z5-z4-9z3+5z2+16z-12. (c) z5-H4+äLz4+H7+2äLz3-H8+äLz2+6z-2.

8 FactorTheorem.nb


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