# Polynomial Equations

In Chapter 1, just like integers, we discovered we could multiply polynomials by each other using the distributive property.

 ( 3x + 5 )( 2x – 1 )  =  3x( 2x – 1 ) + 5( 2x – 1 )  =  6x2 – 3x + 10x – 5  =  6x2 + 7x – 5

In Chapter 2, we discovered we could reverse this process and write polynomials in factored form as well.

 x2 – 6x + 8  =  ( x – 4 )( x – 2 )

In both chapters, the focus was on algebraic representations of polynomials and their coefficients, the values which give each polynomial their uniqueness.  In Chapter 3, we will discover how using the transformations, which we explored in chapter 0, can provide information about the variables of polynomials.

Like the examples above, we’re going to continue to focus on univariate polynomials, or polynomials in one variable (no matter how many times the variable is written).

 univariate polynomial (polynomial in one variable): 2x3 + 2x2– 5x + 1 bivariate polynomial (polynomial in two variables): 2x – 3y multivariate polynomial (polynomial in three or more variables): x2 + y2 + z2

Using function notation, makes it easy to tell if you have a univariate, bivariate, or multivariate polynomial.

 f( x )  =  2x3 + 2x2– 5x + 1 g( x , y )  =  2x – 3y h( x , y , z )  =  x2 + y2 + z2 Can you figure out what the output value of  g( 5 , 2 )  would be?

Polynomials can also be written using a different variable for the output values instead.

 2x3 + 2x2– 5x + 1  =  y 2x – 3y  =  z x2 + y2 + z2  =  w

In either case, as soon as we use an equal sign, we have turned the polynomial expression into a polynomial equation.  This reminds us that polynomials have solutions, but what exactly is a solution to a polynomial? By the way, g(5, 2)  =  2(5) – 3(2)  =  10-6 = 4.

To answer the question, “what is a solution to a polynomial?”, we’ll use the common strategy of looking at simpler examples first, such as degree 1 polynomials.  Below is a snapshot of an applet which will let you explore this question.

Run the applet  “solution.html”  , and move the input/output value sliders, until you can answer the 2 given questions.

Use a copy of the pdf appletnotes1g to record your answers and any discoveries.

After recording and defending your answers to the 2 questions, go on to the next page.

To answer the questions, “what is a solution to a polynomial?”, and “how many solutions does a polynomial have?”, we will have to look at higher degree polynomials as well.  It’s possible our answers for degree 1 polynomials are good for all polynomials, but it’s also possible our answers are only good for degree 1 polynomials.  In other words, the more examples we use, the more convincing our arguments will be (even if we can’t prove them).

Below are snapshots of an applet which will let you explore polynomials up to degree 4.

Run the applet  “tracer.html”  , and move the input/output slider, until you can answer the 2 given questions.

Use a copy of the pdf  appletnotes1g  to record your answers and any discoveries.

After recording and defending your answers to the 2 given questions, go on to the next page.

From exploring the applets, we discovered the solutions of polynomials, were represented by points on the graphs. In other words, a solution to a univariate polynomial equation, is any input value and the resulting output value.

Another way of saying this is, solutionsto polynomials, are the sets of values for the input and output variables, which make polynomial equations true.

Since polynomials are functions, there will always be exactly one output value for every input value.  Since there are an infinite number of possible input values, polynomials have an infinite number of solutions.  That’s one of the reasons for graphing polynomials, it’s a useful way of visually representing an infinite list of solutions.

Since every polynomial has an infinite number of solutions, it’s easy to find one to explore.  If we are interested in a particular solution though, we have to be more careful.  When the input value of a particular solution is known, the number of possible output values for the solution, shrinks from an unlimited number, down to just one.

Below are snapshots of an applet which will challenge you to find the missing output values of polynomial solutions.

Run the applet  “traceroutput.html”  , and move the sliders until you can answer the 2 questions for any example.

Use a copy of the pdf  appletnotes1g  to record your answers and any discoveries.

After recording and defending your answers to the 2 given questions, for any example, go on to the next page.

By using the applet, it wasn’t too much trouble finding the missing output value to a particular solution.  Since polynomials are functions, each input value has exactly one resulting output value.  So all we had to do, was go to the point on the graph which had the given input value, and whatever the output value of that point was, had to be the one and only possible missing output value for that particular solution.

The problem is, the applet is only good for showing solutions with input values that happen to only go to the tenths place.  Then again, due to limits on size and precision, no graph could possibly show every solution.  We can always find the missing output value to a solution however, by simply letting the input value do its thing, and plug it into the polynomial.

x3 – 4x2 + x + 3

If we are interested in a particular solution, when the input value is missing, things get a little trickier.

Below are snapshots of an applet which will challenge you to find the missing input values of polynomial solutions.

Run the applet  “tracerinput.html”  , and move the tracer slider until you can answer the 4 questions for any example.

Use a copy of the pdf  appletnotes1g to record your answers and any discoveries.

After recording and defending your answers to the 4 given questions, for any example, go to the next page.

If we explore enough examples, we will notice polynomials with an even degree have a maximum output value or a minimum output value, while polynomials with an odd degree don’t have any output value limits at all.  For polynomials with an odd degree, we will always be able to find at least one missing input value.  For polynomials with an even degree, we will only be able to find at least one missing input value, if the output value is within the given limit.

Show that the above 3 graphs can have 0 to 2, 0 to 4, and 1 to 3 input values, respectively, for a given output value.

Use a copy of the pdf  booknotes2g, to sketch the above three graphs, and draw horizontal lines to represent output values, with 0 to 2, 0 to 4, and 1 to 3 input values, respectively.

After sketching the 3 graphs with 3, 5, and 3 horizontal lines, respectively, go on to the next page.

Using the applet to find missing input values to solutions of polynomials, was pretty straight forward.  All we had to do was go to the point, or points, on the graph with the given output value.

The problem is, for polynomials of degree 2 or higher, the chances of the missing input values being “nice”, is extremely small.  Most likely, the applet will only be able to give you estimates.

Even when the missing input values are “nice”, as in small fractions or integers, we can be left with a pretty tough puzzle.

 For example, without using the applet, what are the different missing input values to the solutions with an output value of 3, for the degree 4 polynomial 12x4+32x3-17x2-13x+9 ?

The good news is, the puzzle of finding  missing input values to polynomial solutions, can be expressed as equations.

12x4 + 32x3 – 17x2 – 13x + 9  =  3

The bad news is, the above example is still a bit too much to solve at this point.

In case you were wondering though, there are 3 different input values for x,  which will make the above equation true, -3 , -2/3 , and 1/2 (see graph to the right).

For now, we’ll focus on degree 1 polynomial equations, where there will always be  just one missing input value for every given output value.  Another reason for starting with degree 1 polynomial equations is, it will provide an opportunity to discover the connection between equations and transformations.  Knowing this connection will be helpful later in the book for solving more complicated degree 1 polynomial equations, using mathematical formulas, applying physics formulas, and exploring degree 2 polynomial equations in the next chapter.

In other words, being able to do something is nice, but understanding that something, leads to making new connections and further learning.

Below are snapshots of an applet which will allow you to solve degree 1 polynomial equations geometrically, using vertical translations and vertical scalings.

Run the applet  “transmag.html”  , and move the translation slider, and then the scaling slider, all the way to the right. Hit the refresh button and repeat, until you can answer the 2 questions at the bottom of the applet.

Use a copy of the pdf  appletnotes1g  to record your answers, and to defend them with at least one sketch.

After recording both answers, and defending them with at least one sketch, go on to the next page.

Since solutions are just points on the graphs of polynomials, when the line was translated down and scaled vertically, the output value of the solution was translated down and scaled vertically as well.

In other words, the transformations never changed the input value of the solution.

After being translated and scaled, the original polynomial was always transformed into the polynomial x, which is a 45º diagonal line going through the origin.  The polynomial x is unique, because the input and output values are always the same.  This makes finding the input value of a solution quite easy, because it will be the same as the given output value.

In other words, finding the input value of the solution to 2x + 6 = 8 , is the same as finding the input value of the solution to x = 1 .

In either case, x, the input value of the solution, is a 1, but it’s a lot easier to see in the second case.

Understanding how to solve degree 1 polynomial equations geometrically with transformations, explains the steps behind solving them algebraically.

Run  program 301  until you can correctly solve at least 5 out of the 7 given equations.

Run  program 302  until you can correctly solve at least 5 out of the 7 given equations.

After solving, reading, and observing, at least 5 equations from each program, go on to the next page.

If we look closely, we’ll notice in all of the applets and programs so far, the coefficient of the x-term has been positive.  If the x-term coefficient was negative, as in the polynomial equation  -2x + 3 = 11 , we could translate the polynomial  -2x + 3  and the output value  11  down  3  without any trouble, giving you  -2x = 8  .  However, we wouldn’t be able to vertically scale the new polynomial  -2x  and the new output value  8  by  -1/2  .  This is because, there is no such thing as a negative scaling.  When objects are scaled by a scale factor greater than one, something gets bigger and/or farther away from something else.  When objects are scaled by a scale factor of one, nothing changes.  When objects are scaled by a scale factor between zero and one, something gets smaller and/or closer to something else.  If objects could be scaled by a negative scale factor, they would have to have areas less than zero or be a distance less than zero away from something else, which is impossible.

Run the applet  “scale.html”  , if it would help to explore scaling again.

Most likely though, you already have experience solving degree 1 polynomial equations, where the x-term coefficient was negative.  How were you able to get away with this?

To the right are snapshots of an applet, which will help  us answer that question.

Run the applet  “reflect.html”  , until you can answer the 2 questions which will appear at the bottom.

Use a copy of the pdf  appletnotes1g, to record your answers, and defend them with at least one example.

After recording your answers, and defending them with at least one example, go on to the next page.

So it turns out, multiplying or dividing both sides of an equation by a negative number, isn’t a “negative scaling”.  it’s a reflection and then a scaling (or a scaling and then a reflection), where the negative is the reflection, and the number is the scaling.

In the following example, the step numbers below, are also the line labels in the graphic on the right.

1. -2x + 3 = 9 original polynomial equation
2. -2x = 6 after translating down 3 (subtracting 3 from both sides)
3.  2x = -6 after reflecting about the x-axis (multiplying both sides by -1)
4.  x = -3 after scaling vertically by 1/2 (dividing both sides by 2)

Since all of the transformations are vertical, the  missing input value for x always stays the same.

Algebraically, this is why all of the above steps are equivalent equations.  Since solutions are just points on a line, the output value automatically gets transformed along with the rest of the line.  Algebraically, this is why we have to remember to do the same thing to both sides of the equation.  Since the transformations turn the polynomial into x, the missing input value will be the same as the output value.  Algebraically, this is why we solve for x.

Knowing that multiplications are scalings, and negatives are reflections, explains multiplying with negatives.

3 • 2 = 6, is just the point 3 on a number line, being scaled by 2 (made twice as far from the origin).

-3 • 2 = -6, is just the reflection of 3 on a number line, being scaled by 2 (made twice as far from the origin).

3 • -2 = -6, is just the point 3, being reflected and scaled by 2 (turned into a -3 and made twice as far from the origin).

-3 • -2 = 6, is just the reflection of point 3, being reflected and scaled by 2 (turned into a 3 and put twice as far away).

To explore more examples, run the applet  “multiply.html”.

Knowing that additions are translations, and negatives are reflections, explains addition (and subtraction) with negatives.

5 + 2 = 7 ( and 5 – -2 = 7 ), is just a 2 unit translation from the point 5.

-5 + 2 = -3 ( and -5 – -2 = -3 ), is just a 2 unit translation from the point -5.

5 + -2 = 3 ( and 5 – 2 = 3 ) , is just a 2 unit translation from the point 5, reflected about the point 5.

-5 + -2 = -7 ( and -5 – 2 = -7 ) , is just a 2 unit translation from the point -5, reflected about the point -5.

To explore more examples, run the applet  “add.html”.

Remembering that additions are vertical translations, negatives are reflections about the x-axis, and multiplications are vertical scalings, gives a clearer picture of solving degree 1 polynomial equations.

Below is a snapshot of a program, which will randomly generate 10 degree 1 polynomial equations for you to solve.

Run  program 303  , until you can successfully solve at least 8 of the 10 degree 1 polynomial equations.

Use a copy of the pdf  programnotes, to record the original equations, and their 2 resulting equivalent forms.

After solving at least 8 equations, and recording them along with their 2 equivalent forms, go on to the next page.

Understanding the connection between solving degree 1 polynomial equations, geometrically and algebraically, will be helpful when we have to simplify the equations first.

Below are snapshots of programs which will each randomly generate 3 unsimplified polynomial equations to be solved.

Run each of the above programs, until you can successfully solve at least 2 out of the 3 equations given.

Use a copy of the pdf  programnotes, to record the original equations, and to neatly show your steps.

After solving at least 2 equations from each program, and neatly showing all of your steps, go on.

Understanding that negatives are reflections about the x-axis, will also help us better understand degree 1 polynomial inequalities, and degree 1 polynomial equations/inequalities with absolute values.

Below are snapshots of applets and programs, which will allow you to explore degree 1 inequalities and absolute values.

To explore the above applets and programs, open the pdf  inequalities and/or the pdf  absolutevalues.

We have made many discoveries about polynomial functions already.  There is still plenty left to explore in the coming chapters however, and after finishing this book, you will be well prepared to continue with both your formal and your independent mathematical learning.  It should be pointed out though, finishing this book, does not mean you will have learned everything there is to know about polynomials.  In general, it’s impossible to know everything about a particular branch of mathematics, because new discoveries and new math is being created every day.  In particular, it’s impossible to know everything about polynomials, because they also belong to a branch of mathematics called dynamical systems.  A dynamical system is something that changes over time, and quite often can’t be predicted, even when following very simple rules.

Treating polynomial functions as dynamical systems, means choosing an input value, calculating the resulting output value, using this output value as the new input value, and repeating.  The question in dynamical systems is, what happens as the process continues over time?  This process of using outputs as new inputs and repeating, is called iteration.  The initial input values are called seeds, and the eventual pattern of the outputs are called orbits.  In the example below, function notation is used to represent the seed, and superscripts are used to show the number of iterations.

f( x ) = 2x + 1

f1( 1 ) = 2( 1 ) + 1 = 2 + 1 = 3

f2( 1 ) = 2( 3 ) + 1 = 6 + 1 = 7

f3( 1 ) = 2( 7 ) + 1 = 14 + 1 = 15

f4( 1 ) = 2( 15 ) + 1 = 30 + 1 = 31

After 4 iterations, we know the orbit of the seed 1, tends to infinity (the outputs will keep getting bigger forever).

The idea of iteration has been around for a long time. Around 2,000 years ago, a polynomial rational expression was iterated to calculate square roots.

In the polynomial rational expression below, the number we want the square root of, is represented by n.

For a positive seed, or initial input value for x, the orbit, or pattern that results, will end up being the square root of n.

x2 + n

f( x ) = ______

2x

Although clever, a lot of computation is needed to iterate the above polynomial rational expression.

With computers however, this isn’t a problem.

To see that iterating the above polynomial rational expression works, run  program 314  .

A modern example of iterating, is the degree 1 polynomial function, which lets you know how much is left on a mortgage. The seed, or initial input value for x, is the amount of the loan, the r is the interest rate, and m is the monthly payment.  The outputs let you know how much of the loan is still left.

r

f( x ) = ( 1 + ___ )x – m

12

Since interest is charged monthly, the above function would be iterated 360 times, before a 30 year mortgage is paid off.  Again, thanks to computers, it isn’t a problem to find out what a financial commitment having a mortgage is.

To explore what it would be like to take out a mortgage (a loan to buy a house), run  program 315.

Below are examples of degree 1 polynomials being iterated.

Below are examples of degree 2 polynomials being iterated.

To explore orbits of iterated degree 1 and degree 2 polynomial functions, run  program 316.

Earlier in the chapter we discovered that the polynomial x, is unique, because in all if its solutions, the input and the output values were always the same.  This special property of the polynomial x, allows us to graphically find orbits.

Below are snapshots of a program which will graphically determine the orbits for seeds of degree 1 and 2 polynomials.

Can you figure out how the above example, used the polynomial x, to determine the orbit of  .5x + 1  with seed  -1 ?

Run program 317  to explore finding the orbits of iterated degree 1 and 2 polynomials, graphically.

Below are 7 questions related to graphically determining the orbits of degree 1 and degree 2 polynomial functions.

1.  What kind of degree 1 polynomials have orbits that tend to infinity?
2.  What kind of degree 1 polynomials have orbits that eventually become fixed points?
3.  What kind of degree 1 polynomials have orbits that graphically spiral?
4.   What kind of degree 1 polynomials have orbits that graphically staricase?
5.  How many fixed point orbits can a degree 1 polynomial have?
6.  How can you algebraically determine the value of a fixed point orbit of a degree 1 polynomial (without iterating)?
7.  How can you algebraically determing the values of fixed point orbits of degree 2 polynomials (without iterating)?

Run  program 317  to see how many of the above questions you can answer.

Use a copy of the pdf  booknotes, to record and defend your answers, along with any other discoveries.

If you found the Julia set at the beginning of the chapter interesting, run  program 318  , and try to creating one yourself.

Below is a snapshot of the chapter 3 programs menu program.

Run the program  “3menu.bas”   , or its alias to access every program used in chapter 3.

Below is a snapshot of the chapter 3 Julia set programs menu program.  Run  program 323  to acces these programs.

“The mathematician does not study pure mathematics because it is useful; he or she studies it because he or she delights in it, and he or she delights in it because it is beautiful.” — Henri Poincare —

“it may be well doubted whether, in all the range of science, there is any field so fascinating to the explorer – so rich in hidden treasures – so fruitful in delightful surprises – as pure mathematics” — Lewis Carroll —

“The science of mathematics presents the most brilliant example of how pure reason may successfully enlarge its domain without the aid of experience” — Emmanuel Kant —

“[Mathematics] is an independent world created out of pure intelligence” — William Wordsworth —

“Pure mathematics is, in its way, the poetry of logical ideas.” — Albert Einstein —