Talk:Algebra/ToInclude

This are lecture notes that I wrote for the University of Phoenix class on algebra. They are under the GFDL. --Roadrunner 19:54, 17 June 2006 (UTC)

Lecture 1
Let me start by talking about ferrets.

When I was in graduate school, I had an office mate who wrote her doctoral disseration on how fluids behaved in magnetic fields. The equations that she had to deal with were really long and nasty polynomial equations. She also was a ferret lover. So everytime I see a polynomial, I imagine the equation being a small furry creature.

Thinking of a polynomial as a small furry creature isn't a bad way of approaching this chapter. Basically this chapter can be entitled the care and feeding of polynomials. In this chapter, we are going to learn the different parts of these small furry equations. We are going to learn how to what happens when you feed a polynomial some numbers and lets it digest it. We are going to learn how to make these small furry polynomial do tricks (which might be useful), and finally we are going to learn how to breed polynomials. We'll put two polynomials in a cage and let them give birth to new polynomials.

This is all going to be useful because you find polynomials all over the place. Equations which describe how big something is (i.e. Length, volumes, and surface areas) tend to end up to be polynomials. Polynomials are found in equations which describe temperatures, pressures, how long it takes for a computer program to run. Just as a quick example of the use of polynomials is the Bezier curve. I draw some dots have have the computer to connect them with a line

The line that it connects them by is a third-degree polynomial. Notice that I just used a special word. What is a third-degree polynomial you might ask? Well that's what section one is about. Section 5.1 defines some terms for polynomials. Just as ferret lovers have special terms to describe different types and different parts of ferrets, polynomial lovers have different words to describe different types and different parts of polynomials.

The next section is on adding polynomials. Why would you want to add polynomials? Well lets take a trip to the carnival and buy some ice cream.........

As I mentioned earlier, polynomials are often found in equations involving volume. So let be go through a problem that involves adding, subtracting, and multiplying polynomials.

So suppose we have an ice cream cone. How much ice cream is in it. You can figure this out by splitting the ice cream cone into the cap (which is a half a sphere) and the cone.


 * (volume of ice cream) = ? * (sphere) + (cone)

now figure this out you need to add two polynominal expressions and you end up with something like where r is the radius of the cone and h is the height of the cone. But I'm hungry, I want a double scoop.


 * (volume of ice cream) = ? * (sphere) + (sphere) + (cone)

Or since I don't know how many scoops of ice cream I want. I just have n scoops.


 * (volume of ice cream) = (n ? ?) * (sphere) + (cone)

So we can figure out the volume of ice cream. Well since there are a lot of business people out there, lets think about costs. Let p be the cost of ice cream per cubic centimeter. So we have


 * (cost of ice cream) = (price per volume) * (volume)

But we've forgotten something. We have the price of the ice cream, but what about the price of the cone......


 * (price of ice cream cone) = (price of ice cream) + (price of cone)

Now the price of the cone is going to be


 * (price of cone) = (price per area) * (area of cone)

We have get more complicated. We can take into account inflation and so make p and q variables which change year to year. Or we could try to figure out what fraction of the price of the ice cream cone is cone.


 * (fraction of price of cone) = (price of cone) / ((price of cone) + (price of ice cream))

Now I have actually put in algebraic equations for any of these things, because you haven't read chapter 5 yet, and I wanted to show you what the point of chapter 5 was before actually going through the content. But once you know how to add, subtract, multiply, and divide polynomials, you should be able to start with the equations for the volume of a sphere and cone, and the area of a cone and work to the end and get the (fraction of the price of cone) in an algebraic equation. I've worked through the equation in an appendix which you should read after you've gotten half way through chapter 5.

Now as far how do approach chapter 5. Think of the ability to add, subtract, multiple, and divide polynomials as a skill, like riding a bicycle. The best way of learning how to do it is to just keep practicing until its second nature. Something that I find useful is to make up problems and start with really easy problems, and then make them progressively harder. For example, an easy problem with adding polynomials is
 * 2x + x

which is of course
 * 3x

You can then make it a bit harder
 * (x+y) + 2x

which is
 * 3x + y

and just work your way up until you can deal with longer and longer polynomials without your mind going numb (and given a large enough polynomial, your mind will go numb).

Something that has always helped me is to just think of the equations as squiggles on a page, and the lesson as rules for turning one set of squiggles into another set of squiggles.

One of the more important things that you will learn in this chapter are the product and quotient rules for exponents. The way I think about them is that they turn multiplication and division into addition and subtraction. Section 9.1 starts with radicals which is a different way of writing things.

Now I've saved what I think is the most important section for the end of the lecture. Section 5.7 is about negative exponents and scientific notation, and I don't think that the author of the text quite realizes how important that section is. Let me give an example of how important section 5.7 is with a money example. A good salary for an experienced programmer is about $80,000 per year. A good salary for a CEO is about $2,000,000 per year. The revenue for the University of Phoenix is $300,000,000 per year. The revenue for a state government is $10,000,000,000 per year. The revenue for the federal government is about $2,000,000,000,000 per year. Now if you are like me, your mind started to go numb somewhere along the way, and all you were seeing by the end were big numbers with lots of zeros. Now lets write them in scientific notation.


 * (salary for programmer) = $8 x 10^4
 * (salary for CEO) = $2 x 10^ 6
 * (revenue for UOP) = $3 x 10^ 8
 * (state government budget) = $1 x 10^10
 * (federal budget) = $2 x 10^12

Now I'm hoping that you notice that the last set of numbers is just easier to think about. The federal budget rather than being a number with lots of zeros is now a number with 12 zeros as opposed to one's salary which has 4 zeroes.

With the exponent rule we can do some more magic. For example, what do you suppose is the average tax burden per person in the United States. We'll we look up the population of the United States which is 200,000,000 or in scientific notation 2x10^7.


 * (taxes per person) = (federal budget) / (people in the US)
 * (taxes per person) = $2x10^12 / 2x10^7

with the magic of the quotient rule, this equals


 * (taxes per person) = $2/2 times 10^(12-7)
 * (taxes per person) = $10^5.

Which means that on the average a person pays about $10,000 per year in taxes.

We can play some more games. For example suppose UoP spend 10% of its budget on IT, how many programmers is that......


 * (UoP IT budget) = 10% times $3 x 10^8
 * (UoP IT budget) = $3 x 10^7

Now each programmers salary is about $10^5, and
 * = $3 x 10^7 / $10^5
 * = 3 x 10^2

which says that UoP probably has around about 100 ? 500 people in IT. Notice that I did that with very little math. I've found this to be really useful in reading corporate reports. For example, if I know that a vacuum cleaner manufacturer made $10^6 a year last year, I can guess that each vaccuum cost between $10^2 and $10^3 and so they manufactured between 10^3 and 10^4 vacuums.

Rather than thinking in term of standard numbers and having my mind blank out from zeros, I rewrite the numbers in scientific notation, and then with the exponental and quotient rules can start making some quick calculations, like how many employees does the company have, what is its market capitalization, how does the size of this company compare with other companies, how many widgets did the company sell.

So rather than being called ¡°Negative Exponents and Scientific Notation¡± it should be called how ¡°You Can Handle Really Big and Really Small Numbers Without Going Crazy¡±

Hope that was useful.

Types of Polynomials
In one of the questions last week, I asked about what to do if you have a large polynomial. One of the important things to look at the polynomial and look at how big the largest exponent (or order) is. In MTH 208, you dealt with order-1 polynomials, which were linear equations. These involved straight lines. Since we are in MTH 209, we now learn how to deal with more complex polynomials, mainly those of order 2. Where you see those polynomials can be shown in the following chart.

Factoring
Now in order to solve these more complex equations, we need to take a short detour into factoring. Factoring is cool, and section 6.1 didn't explain how cool it is.

The reason factoring is cool is that multiplication is easy but factoring is hard. Multiplying two numbers together is like scrambling eggs, and factoring is like trying to unscramble eggs.

For example suppose I have two numbers say...
 * 3490529510847650949147849619903898133417764638493387843990820577

and
 * 32769132993266709549961988190834461413177642967992942539798288533

If I multiply them together, I get this one big number...
 * 114381625757888867669235779976146612010218296721242362562561842935706935245733897830597123563958705058989075147599290026879543541

Now the hard part comes if I give you this big number and then offer you money to figure out what two numbers I choose to get that big number. We can use the techniques we learned last week to figure out how long that will take. We want to factor a 110-digit number, and suppose we know that there are two factors. We then know that one of the numbers must be less than 55 digits. So we try all the numbers between 2 and 10**55.

Suppose we have a computer that can process a billion (10**9) numbers per second. So to look at all the numbers
 * 10 ** 55 numbers / 10 ** 9 numbers per second = 10 ** 46 seconds.

Now there are 3 x 10 ** 7 seconds per year. Now 3 is almost the square root of 10 so it is about 10 ** (1/2). So we can express this number as 10 ** (7.5).

So
 * 10 ** 46 seconds / 10 ** 7.5 seconds per year = 10 ** 39.5 years

Which poses a small problem since there has been 10 ** 10 years since the big bang. So factoring this number brute force will take 10 ** 29 times the lifetime of the universe. Note that getting a faster computer won't help you very much. If your computer is a million times faster, the time it will take to solve this problem is only 10 ** 23 times the lifetime of the universe. Now you can get somewhere if you find some mathematical shortcuts that identify possible factors, and that how they managed to factor that number, but without a basic breakthrough in mathematics, you can overwhelm those shortcuts by choosing an even bigger number to factor.

But there is actually a use for this, and you use it every time you buy something from ebay or send your credit card number over the internet. I can create a mathematical lock box and write a big number on it. Only the person who has a factor of that number can open the box, and the cool part is that you can create that big number without knowing the factor. This is what your web browser does when it sends a credit card number. It creates a big number that only Ebay or Microsoft or someone else knows how to factor, and mixes that number with your credit card number.

Needless to say, there is some very big money involved in being able to factor big numbers. To give you some idea how big, take a look at rsasecurity.

The reason it useful to factor polynomials is that it lets you solve polynomial equations. Suppose you have an equation
 * x ** 2 + (something) * x + (something else) = 0

if you can figure out a way of rewriting it so that it looks like
 * (x - a) (x - b) = 0

Then you know that a and b are solutions of the equation. One thing about polynomials is that if the highest term in the polynomial is x ** n, you should be able to split it into n factors, which means that there are n solutions to the equation. You can sort of see how this works by going the other way.
 * (x - a)(x - b)(x - c)

is going to end up with a polynomial with x ** 3 in front.
 * (x - a)(x - b)(x - c)(x - d)

is going to end up with a polynomial with x ** 4 in front and so forth.

Where this would be useful is if you have a business problem with an x**2 in it. Typically this be the case when you have a problem that involves area. One example of this would be a situation where you are running a pizza parlor. The price of a pizza is typically a quadratic equation
 * ($1.00 per square foot of dough) * (area of pizza) = (cost of pizza)
 * $1.00 * pi * r ** 2 = (cost)
 * $3.00 * r ** 2 = (cost)

So how big does the pizza need to be to cost $12.00?


 * $3.00 * r ** 2 = $12.00
 * r ** 2 = $4.00
 * r**2 - $4.00 = 0
 * (r + $2) (r - $2) = 0 (Where did this come from?)


 * r = $2 or -$2

Note that you end up with an extra root. This is because you are dealing with a second degree polynomial and so the solutions to it will usually have two answers.

The good news is that if you can factor a polynomial you can solve it. The bad news is that as we have seen above factoring is sometimes hard. Now scrambling an egg is straightforward. Unscrambling it is much harder. Chapter 6 has some techniques for unscrambling a two egg polynomial omelet. Once you put another egg in there, it gets much harder.

Incidentally in terms of general usefulness, the formulas for squaring (a+b) and for reducing a**2 - b**2 are worth remembering. Somehow they manage to turn up in all sorts of places. Actually solving a quadratic equation by factoring is something that is much less useful. The problem is if someone hands you a quadratic equation in the real world, the odds that it will separate cleanly into nice numbers is small.

In chapter 7, we are looking at rational expressions, which is another fancy way of saying fractions with polynomials. The first section is on simplifying rational expressions. The result for this is that suppose you have a fraction 1/4 and someone else has a fraction 2/8. They are actually the same thing, and so it's nice to put all of them into one common format. Then there is more stuff with multiplying, adding, and subtracting polynomials (i.e. slicing and dicing numbers).

I don't know why I have some many food metaphors this week. Maybe I can order online for some pizza and have Domino's factor some of my numbers.

Introduction
This week's assignment there are three threads in this weeks assignment which all build on the work we did last week. In the last week, you learned to work with rational expressions. This week's work uses those rational expressions to solve word problems involving rates. Also in last week's lesson you learned to solve equations using factoring. This week we will learn another way to solve quadratic equations using a single formula. Finally, we are going to learn out to solve power equations.

Rational Expression Word Problems
There are two parts of solving a problem. The first is writing the problem down as a equation, the second is solving the equation. Generally the first part of the problem is the hardest part. One you have an equation, solving it is usually easy using the techniques we've learned in the previous weeks. The hard part is setting up the problem. So I'd like into a bit more detail about how that works. Also, I'd like to use as an example, a problem that is more relevant to business than the standard algebra textbook problem.

So let me work through a problem.The V Corporation is a dot-com with huge problems. It has $40 million in the bank and it spending it at a rate of $8 million per month. The venture capitalist is willing to fund $15 million if it can prove that the V Corporation will be around in two years.

The question that faces the executives of the V Corporation how much do they have to cut their burn rate in order to survive two years when hopefully the economy will improve.

The first thing to do is to just write down, what you know and what you want to find out.

I know
 * $40 million in the bank now
 * $65 million if VC funding goes through
 * $8 million burn rat
 * goal is 24 month survival

I want to find
 * How much do I want to cut? Call this x

The next thing to do is to draw a chart

Since Burn = Time * Money
 * 65 = (8-x) * 24

I'll let you grind through the numbers and break the bad news to the employees of the V Corporation.

Note that this is just a simple linear question, not a fancy rational one. Let try something more complex.

I have two plants that are produces 40 cars per month. I know that one of the plants is producing 1 car every two days, and I want to see how this will impact production if I half the amount of time it takes to produce a car at the other plant.

I know
 * total production 30 cars per month
 * production at Plant #1 is 2 cars per day

I want to find out
 * how long it takes to build a car at Plant #2 (?)

So I draw my chart

Since we know that cars = rate * times, we can fill out some more stuff

We then can add together the number of cars we have each month to get the equation
 * 15 + 30 / x = 40

Multiplying each side by x we get
 * 15 x + 30 = 40
 * 15 x = 25
 * x = 25 / 15
 * = 5 / 3

So it takes a little more than a day to create a car at plant 2

We can then play some what if games. Suppose we double production at plant #2

Filling in stuff in we get

We can play some more games
 * What happens if we add a third plant that produces a car every two days?
 * We want to increase production to 100 cars a month and have two choices, we can increase production at plant one or build a new plant? How much do we need to increase production at plant #1 to build 100 cars per month, plant #2, what about building a third plant?
 * Alternatively the economy is bad, and there is now demand for only 30 cars a month. We can either leave plant #1 idle for part of the month or leave plant #2 idle for part of the month?  How much do you have to idle plant #1 and how much to you have to idle plant #2?
 * Union rules say that you have to idle plant #1 and plant #2 the same  amount.  How many days do you have to idle both plants to cut production?

As an exercise you can try to go through each of the above scenarios and see if you can handle it. If you don't run an auto company, you can try replacing cars and plants with something in your own experience.

One final thing. You notice that I've put everything in nice neat rows. Mathematicians don't do problems that way, but you see these sorts of tables everywhere in business. Part of the reason to put these sorts of problems in table format is that it helps you to also put those sorts of tables in Excel. For example, with the above scenarios is something that you probably would use Excel to help you with.

The Quadratic Equation
And now for for the second of the three threads in this weeks lesson.

You've seen three ways of solving quadratic equations
 * factoring
 * completing the square
 * the quadratic equation

One thing that will be useful for you do to know is that under normal circumstances, the quadratic equation is the only one that you will ever use. Factoring only works if you happen to have a problem that can be factored and those tend to occur only in math textbooks :-). Completing the square will work for everything, but you won't remember how to do it. The quadratic equation, is easy to use because you just put the numbers in the formula and out pops the answer.  It's short enough to memorize and even if you don't memorize it, its something that you can know exists and look up if you ever need it.

One thing that helps a lot in dealing with quadratic equations is to visualize what a quadratic function looks like.

The equation
 * a x**2 + b x + c

is going to be a U shaped parabola if a is positive and an upside down parabola if a is negative. Changing a changes how narrow the parabola is. Changing b moves it left and right and changing c moves it up and down. If you imagine an intersection between a U and a straight line, you can see that some of the time, you get two intersections, some of the time, you get one intersection, and some of the time you get no intersections.

One equation you might have at this point is the quadratic equation is the most useful way to deal with quadratic equations, then why did we spend most of last week talking about them.

The reason that mathematicians are interested in factoring has to do mainly with higher order equations. There are formulas for the cubic and quartic equations, but they take up two pages and no one ever uses them. What gets even more interesting are quintic and higher equations. A person named Galois proved in the early 19th century that there are *no* general formulas for 5th degree and higher polynomials. He was 21 at the time, and shortly (i.e. days) after proving this and inventing an entire branch of mathematicians, he lost a pistol duel over a woman of questionable morals. And his great legacy to the world was leaving the world the knowledge that solving the equation
 * <span style="color: rgb(204, 0, 0);">0 = a + b x + c x ** 2 + d x ** 3 + ... + x ** n

is really, really hard when n is larger or equal to five. But there is good news, which comes in the section on power equations ....

Power equations
The bad news is that general polynomials are hard to solve. The good news you usually won't see a general polynomial often and that a lot of the time you will see a subset that is much easier to solve. If you see the equation
 * <span style="color: rgb(204, 0, 0);">x ** n = b

You can solve it by taking the n-th root of both sides, giving you
 * <span style="color: rgb(204, 0, 0);">x = b ** (1/n)

There is a subtlity though. If you multiply a number by itself, the negative signs disappear. So the solution to
 * <span style="color: rgb(204, 0, 0);">x ** 2 = 4

is both 2 and -2. By contrast,
 * <span style="color: rgb(204, 0, 0);">2 ** 3 = 8

but
 * <span style="color: rgb(204, 0, 0);">(-2) ** 3 = -8.

The consequence to this is that there are two solutions to a power equation if n is even and b is positive, and no real solutions if n is even and b is negative. By contrast if n is odd, then you have one solution regardless of whether b is positive or negative.

Conclusion
If the three threads in this weeks lesson look unconnected, its because they are. At this point, you've mastered the basics of manipulating equations, and what is left now is to teaching you mathematical tools which you can use to solve different classes of equations. These tools are specialized, which means what works with one type of equation will completely fail for another type, but the more tools you have in your toolbox, the more likely it is that something will fit the equation you are trying to solve.

Lecture 4
My how time flies. We have one more week of new material and then a final week of review. This week there are three new concepts, the first is graphing quadratic equations, the second is doing quadratic inequalities, and the third is doing rational inequalities.

Graphing Quadratic Functions
The first new concept is graphing quadratic functions. I've generally found graphing functions to be useful because it gives you some visual idea of how that function behaves. So lets start with the quadratic equation
 * y = a x ** 2 + b x + c

There are examples of what you get if you graph that equation in section 10.3. The thing that these graphs, particularly figure 10.4, should remind you of is a ball flying through the air. This is not a coincidence as the equation that describes a ball flying through air is a quadratic equation. Now if you were to draw a picture of a ball flying through the air, you could do this if someone gave you three points: So lets try to figure out where the ball hits the ground. The equation that describes where the ball hits the ground is
 * where the ball starts from the ground.,
 * the highest point of the ball, and
 * where the ball hits the ground again
 * 0 = a x ** 2 + b x + c

and you can find these points via the quadratic equation or by factor. Notice that there are usually two such points. Now the highest point of the curve is midway between the points that the curve hits the ground. And that is at
 * x = -b / a

You can find the y value of the equation at that point by substituting back into the quadratic equation. This gives you three points, where the ball leaves the ground, where the ball is at its highest point, and where the ball hits the ground again. You can then connect the dots and draw a quadratic equation.

And now for the usual question. Why would you want to do this?

During last weeks discussion, we talked about a situation where a vendor would want to give a discount on their software depending on how many licenses one buys. Suppose the base price for a license is 200, but the vendor is willing to discount $5 for each license. The equation that describes the total cost is
 * y = x * (200 - 5x)
 * = -5 x **2 + 200 * x

Now one might want to know how the cost of the software varies with the number of licenses, and this is a good example of where graphing this on a sheet of graph paper might be useful. I'll leave this as an exercise for you, and once you graph it, you should realize that this is probably not a good deal for the vendor if you want to buy a really large number of licenses.

Quadratic Inequalities
This leads to the next topic. You are a purchaser and you will consider the deal only when the total cost is less that $1000. This creates an equation
 * 1000 &lt; x * (200 - 5x)

or
 * 0 &lt; -5 x ** 2 + 200 * x - 1000

There are two ways of solving this equation. If you can factor the equation, then you can find the signs of the factors. If they are the same, then the equation is greater than zero. If they are different, then the equation is less then zero. You can draw things on a number line similar to 10.15, and figure out where your equation is satisfied. The other way is to find the zero-crossings using the quadratic equation, you then pick a test point between the zero-crossings, and some test points outside the zero-crossings. This will tell you the sign of the equation, and you can see where your inequality is satisfied.

Rational Inequalities
The final topic for this week involves rational inequalities. An example where this would be the case is if you had a situation in which the total cost producing an item is $250 plus $5 per item. The unit cost of the item would be
 * <span style="color: rgb(255, 0, 0);">250 + 5 x
 * <span style="color: rgb(255, 0, 0);">--
 * <span style="color: rgb(255, 0, 0);">     x

You want to know howmany items you need to produce so that the cost of each item is less than $10.
 * <span style="color: rgb(255, 0, 0);">250 + 5 x
 * <span style="color: rgb(255, 0, 0);">--   &lt; 10
 * <span style="color: rgb(255, 0, 0);">      x

The rule for doing rational inequalities is that you shouldn't multiply factors on both sides of the inequality, just add and subtract them. This gives you
 * <span style="color: rgb(255, 0, 0);">250 + 5x
 * <span style="color: rgb(255, 0, 0);">-- - 10 &lt; 0
 * <span style="color: rgb(255, 0, 0);">     x
 * <span style="color: rgb(255, 0, 0);">250 + 5x       10x
 * <span style="color: rgb(255, 0, 0);">-- -  -- &lt; 0

<div style="color: rgb(255, 0, 0);">
 * x                 x
 * 250 - 5x
 * -- &lt; 0
 * x

The numerator is positive if x &lt; 50 and negative if x &gt; 50. The denominator is negative if x &lt;0 and positive if x &gt; 0. Thus the signs are different if x &lt; 0, the same if 0 &lt; x &lt; 50, and different if x &gt; 50. Thus the inequality is satisfied if x &gt; 50.

Conclusion
I mentioned last week that you can think of algebra as a set of mathematical tools in your toolbox. This week we've added three more tools in your toolbox. That's all of the tools for this class. Once you get past this week, the next thing to do is to review all of what we have covered.