So let’s talk about some

astronomical calculations. The average radius of

each of the eight planets is shown in the table

on the next screen. Let’s state the radii

of Mercury and Jupiter to two significant digits

using scientific notation and the units of meters. Now before I move

on to the problem, I do want to address a couple things

that you’re seeing on the screen. One is this weirdo

word R-A-D-I-I, radii, that’s the plural

of the word radius. The other thing is

you might be thinking, whoa, Steve, there’s nine planets. Actually, the latest thinking

amongst the astronomical community is that Pluto is not really

worthy of a planet designation, so the majority of the

astronomical community now agrees that there’s only

eight planets in our solar system, and I’m just a go with

the majority kind of guy. So here’s a table showing our eight

planets and their average radius in the right column. Notice that the radii

are given in kilometers, and one thing we want to

remember about the question that was posed to us is that we want to

state our final radii in meters, so we’ll have to keep that in mind. Now clearly we don’t have

any room to work the problem. So let’s highlight the two

rows that go with the planets that we’re asked to work

with, Mercury and Jupiter, and let’s get rid of

all the other clutter so we have some room to work. So let’s first begin with

the radius of Mercury. Well, one of the things

it asked us to do is to write our radii using

two significant digits. When you use scientific notation,

the number of significant digits is the same as the number of numbers

that you write in your mantissa. Now if I use all of the digits

2, 4, 3, and 9 in my mantissa, I would have four

significant digits. I only want two

significant digits, so I have to think how does

2 round to two digits. Well, it rounds down because

three is less than five. So my mantissa is going to be 2.4. And 2,439, I would need to move

the decimal point one, two, three places to the right. So this is 2.4 times

10 to the third. But that’s how many

kilometers are in the radius. We want to know how many

meters are in the radius. Now, you might know that there’s

1,000 meters in a kilometer. If you didn’t already know

that, you know it now, because I just told you. A thousand written in scientific

notation is 10 to the third power. So let’s go ahead and multiply

by the conversion fraction, 10 to the third meters in 1 kilometer. Notice now that our

units of kilometers are going to divide

to one, and I’m just going to be left with 2.4

times 10 to the sixth, with a unit of meters. Let’s talk about the

radius of Jupiter now. So the radius of

Jupiter is about– Well, let’s see, we know that

the radius of Jupiter to the number of significant

digits given in whatever table I got this information was 71,492. I want my mantissa to only

have two significant digits. Four is below five, so when I round

714 down to two significant digits, my mantissa is just going to be 7.1. This is times– Well, let’s

look at the number here. We need to move the decimal point

one, two, three, four places to the right. So this is times 10 to the fourth. And we already talked

about the fact that there’s 10 cubed metres in a

kilometer, so we also need to multiply by 10 cubed. Now we’re in meters. So all together this is 7.1

times 10 to the seventh meters. Notice that the power on

10 in the radius of Jupiter is seven, whereas the power of 10 on

the radius of Mercury is only six. Sometimes you’ll hear people

talk about magnitudes, and the magnitude of

the number has something to do with the power of 10. Since the power on 10 is one

more for the radius of Jupiter than for the radius

of Mercury, you’ll sometimes hear people say

that the radius of Jupiter is one magnitude larger

than the radius of Mercury. Now let’s use a

proportion to kind of get a sense of how much bigger

Jupiter is than Mercury. Suppose that Marge wants to paint

a scale drawing of the planets. If she uses a diameter

of 1 inch for Mercury, what diameter should

Marge use for Jupiter? So I’ve gone ahead and

created a new table, and I’ve gone ahead

and put into the table these meter measured

radii that we just computed for Mercury and Jupiter. Now, I’m about to perform some

calculations using these rounded figures for the radii. And if there’s any science teacher

sitting out in the audience right now, they’re probably

freaking out a little bit. They’re probably saying I can’t

believe he’s using these rounded numbers in future calculations. That’s because when

you round numbers before your final calculation,

you very frequently get something called propagated error. And I think I have a

pretty easy example for you to understand what this

propagated error is. Let’s pretend that a

baseball weighs 1.44 pounds. And suppose we were asked to

find to two significant digits the weight of two baseballs. Well, suppose that we first

rounded the weight of one baseball to two significant digits. That would be 1.4 pounds. And when you multiply 1.4 pounds

by two, you end up with 2.8 pounds. But notice that if we use

all three significant digits of the weight 1.44 pounds

and multiply by two, we end up with 2.88 pounds,

which in fact rounds up to 2.9 pounds versus the 2.8 pounds

that we got when we first rounded the weight before the calculation. That’s called propagated error. The error we introduced by

first rounding the weight was magnified when we

performed future calculations. So when you take science

classes, very early in the term they’re going to start talking

about significant digits and how calculations

affect significant digits. Lucky for you, I’m not

too concerned about that right now at this point

in the math class, because I just want you to

get some really vague sense of the relationships

between these numbers, but do be prepared to

get a pretty hardcore lesson on significant digits when

you do take your science courses. So using these rounded

numbers to just get a vague sense of the

dimensions that Marge would have to use in her

drawings, what we want to do is we want to define d to

be the diameter of Jupiter in Marge’s scale drawing. Remember Marge is going to draw a

scale drawing of the solar system, and she’s going to draw Mercury

with a diameter of one inch, and we want to figure out what the

diameter of Jupiter needs to be. Well, this is just a proportion. Let’s take the diameter of

Jupiter in Marge’s scale drawing, and let’s divide by the

real diameter of Jupiter, which is 2 times 7.1 times

10 to the seventh meters. This ratio needs to be the same

as the ratio of the diameter of Mercury in Marge’s scale

drawing, which is 1 inch divided by the actual diameter of Mercury,

which is 2 times 2.4 times 10 to the sixth meters. This is called a proportion. Whenever you have two

ratios equal to one another, you get a proportion. Let’s go ahead and solve for d. Multiplying both sides of my

equation by 2 times 7.1 times 10 to the seventh meters, I get

that d is equal to 2 times 7.1 times 10 to the seventh meters

over 2 times 2.4 times 10 to the sixth meters. All of this times 1 inch. Notice that my units of meters

are going to divide to one, and I am going to be left

with units of inches. Notice too that I have a factor

of 2 in the numerator, which divides with the factor

of 2 in the denominator. So to get the numerical part of

the diameter, all I need to do is take 7.1 times 10 to the

seventh and divide by 2.4 times 10 to the sixth. Let’s do that on a calculator. Now, when I was

preparing this lesson, I looked at several

different calculators to see how scientific notation works

on several different calculators, and I noticed that if I picked

up eight different calculators at random, chances are they

did scientific notation eight different ways. So all I can do is show you

how scientific notation works on the calculator

that I’m working with. Hopefully, you still have the

manual for your calculator so you can read the

directions to see how it works on your calculator. If you don’t, or even if you

do and you can’t figure it out by yourself, get some help, either

from your teacher or a math tutor, or maybe one of your friends can

help you figure out how it works. But do make sure you figure out

how it works on your calculator. Now on this calculator

that I’m working with, I have to tell the calculator

that I want it to work in scientific notation mode. The way I do that on this computer

calculator is I right click and I go to what is

called Options, and then I need to go to Display Options

and put it in scientific display. Obviously, that’s not the way it’s

going to work on your calculator, but there might be a

Mode button, or something like that, that will give

you an access to a menu that will allow you to change your

calculator to scientific notation mode. Now that I’m in

scientific notation mode, let’s see what happens when

I enter a number like 400. So I’m entering 400, and

let’s see what happens when I press the equal sign. Holy smokes! An e shows up. 4.e2. What on Earth could that mean? Well, in scientific notation

400 is the same as 4 times 10 to the second power. So that e must just be some short

calculator speak for times 10 to the power. Almost all calculators use e

or EE in scientific notation, and that is always

standing for times 10 to the power of whatever

number follows the e or the EE. So let’s do the calculations

that we’re asked to perform. We need to take 7.1 times 10 to the

seventh and divide by 2.4 times 10 to the sixth. So I start with my mantissa

in the numerator 7.1. On this calculator,

the way I indicate that I’m about to use scientific

notation is I press this EXP key. I saw several calculators in my

stack that had that exact same key. Several other calculators

had an EE key. You just need to find the

proper key on your calculator. So I press EXP. There’s my e that’s saying that

I’m going to do a power of 10. The power of 10 is seven. Let’s divide that by 2.4. Press my scientific notation key

and tell it that the power of 10 is six. And let’s go ahead and press equals. And we see that we get about

2.9583 times 10 to the first. Since I was working in two

significant digits on my last grid, I’m going to go ahead and write

down two significant digits here as well, even though

technically I don’t really have two significant digits

of accuracy in my result. I’m going to let your science

teachers explain that to you. When I round 2.9582 to just

two significant digits, that rounds up to 3.0. So I’m going to go ahead and

say that d is about 3.0 times 10 to the first inches,

which is just 30 inches. So the diameter of

Jupiter in Marge’s scale drawing of the solar system

needs to be about 30 inches. So Mercury has a diameter

of 1 inch, and Jupiter has a diameter of around 30 inches. How much bigger is that? Well, I had some

free time yesterday, so I went ahead and cut out

a little one-inch Mercury. I don’t know if you can

see it there or not. I also cut out a Jupiter

with a 30 inch diameter. And maybe this will help you see

Mercury and just see how much smaller Mercury is than Jupiter. Well, guess what. This is just a cross section of the

difference between the two planets. It turns out that

Jupiter, really, if you think about it in the correct

way is that much more bigger than Mercury, which we’re going

to see in our next example. So the average radii

of Mercury and Jupiter are shown on the next screen. Let’s compute and compare the

volumes of the two planets. This is the way that

we’re really going to get a sense of just how much

bigger Jupiter is than Mercury. So when I rounded the two radii

to two significant digits, Mercury was 2.4 times

10 to the sixth meters, and Jupiter was 7.1 times

10 to the seventh meters. Now, if we’re going

to compare the volumes I think we’re going to need to

figure out what the volumes are, so we need a volume formula. Assuming that the two

planets are perfect spheres, the volume formula for a sphere is

that the volume equals 4/3 times pi times the radius cubed. So let’s apply that

formula two times. The volume of Mercury is

about 4/3 pi times 2.4 times 10 to the sixth meters cubed. So let’s get out our

calculator and calculate that. So I’m going to go ahead and take my

4/3 and put it inside parentheses. And then I’m going

to multiply by pi. And then I’m going to multiply

by 2.4, hit my power of 10 key. My power of 10 is six. Hit my exponent key. Cube that number and press equal. And I’m getting about

5.8 times 10 to the 19th. How about the volume of Jupiter? Well, this is going to be

4/3 times pi times 7.1 times 10 to the seventh meters cubed. So getting out my calculator,

clearing away my last calculation, putting the 4/3 in parentheses. Now I need to multiply by pi. Now I need to multiply by

7.1 times 10 to the seventh. That number cubed. And this is rounding to about

1.5 times 10 to the 24th. But these volumes have

units attached to them. What are the units? Well, in both cases I had

meters inside parentheses where the expression inside

the parentheses is cubed. So the unit of meters is also cubed. The unit on each of these

volumes is meters cubed. And the unit is just as

important as the number, because there’s a

great big difference between 1.5 times 10 to

the 24th cubic meters, and, for example, 1.5 times

10 to the 24th cubic inches. The units are really important. Now, this question,

what we want to do is we want to compare

these two volumes, and the appropriate way to compare

the two volumes is with a ratio. So let’s make a ratio. Let’s compare the volume of

Jupiter with the volume of Mercury by dividing the volume of Mercury

into the volume of Jupiter. Now again, I’ve gone ahead and

put my rounded volume values into my table. So in terms of significant

digits, I don’t really have two significant

digits in this ratio here, but I’m just trying to

get some sort of sense about how much bigger

Jupiter is than Mercury. And like I promise you, when

you get to your science classes, your science teachers will

tackle that significant digits issue with you. In this calculation, I’m just

going to go ahead and say that this ratio is about

the ratio of 1.5 times 10 to the 24th cubic meters to 5.8

times 10 to the 19th cubic meters. Looks like calculator time to me. So I need to take my 1.5 and put

in my scientific notation, EXP. That’s times 10 to the 24th. And divide by 5.8. Use my scientific notation key. That’s times 10 to the 19th. And I’m getting about 2.6

times 10 to the fourth. What’s the unit? Well, in my calculation,

I had cubic meters in the numerator and cubic

meters in the denominator, so they divide to one. There is no unit on this ratio. What does 2.6 times 10 to the

fourth mean in standard notation? Well, I need to move that decimal

point four digits to the right. Two, six, there’s one

digit to the right. Two digits to the right,

three digits to the right, four digits to the right. This ratio is about 26,000. The volume of Jupiter is about

26,000 times the volume of Mercury. Good golly, what does that mean? Well, let’s think about it. I’ve gone ahead and made

up a little animation with a Jupiter-like planet

and a Mercury-like planet, but hopefully you’re saying,

you know what that just doesn’t seem very proportional to me. Doesn’t look to me

like you’ve got Jupiter 26,000 times as big as Mercury. Well, on our little

TV screen, there’s no way that I could get you

to be able to see Mercury, if I really made it only 126

thousandths the size of Jupiter. But I can take this

animation and help you understand what it

means for Jupiter to be 26,000 times the size of Mercury. What I’m going to do

is I’m going to think of Mercury as a little marble. And I’m going to think of Jupiter

as a great big spherical shaped fish tank. And I’m going to see how

many marbles it’s going to take to fill up this fish tank. Before I do that, I better

put the fish in a safe place so it doesn’t get

clobbered with the marbles. So now let’s take our

Mercury-sized marbles and start placing them in

the Jupiter-sized fish tank. How many marbles is

it going to take? Well, Jupiter is 26,000

times as big as Mercury. So it would take around

26,000 Mercury-sized marbles to fill a Jupiter-sized fish bowl. And that’s a lot of marbles! Now, I want to talk about an

example that’s rooted in chemistry. The radius of a water molecule

is about 0.14 nanometers. Suppose that a spherical snowball

with a radius of 3 centimeters was blown up to the

size of the Earth. If each water molecule

in the snowball grew proportionally

with the snowball, what would be the resultant

radii of the water molecules? The reason I’m working this example

is because, first thing, you probably didn’t know what nm meant. It’s a nanometer. The second thing is 0.14 nanometers

is such a teeny tiny little length that it’s really

impossible to imagine it, but you still can get a

sense of how teeny tiny it is using a proportion. What I have here is

a little table that has several common metric names. We’re using the unit of

nanometer in our problem. A nanometer is 10 to the

negative ninth meters. Some other common units

you’ll see are a micrometer, which is 10 to the negative

sixth meters, a millimeter, which is 10 to the negative

third meters, a centimeter, which is 10 to the

negative second meters, a kilometer, which

is 10 cubed meters. These [? prefixes, ?] they also

apply to other metric units. For example, 1,000

liters is a kiloliter. 1,000 watts is a kilowatt. Look at the next word, a megameter. That’s 10 to the sixth meters. You might have heard mega in the

context of a computer discussion. Megabyte, that’s a million bytes. Similarly, a gigameter

is a billion meters. So a gigabyte is a billion bytes. I included this last row,

not because you see 10 to the 24th meters

all that frequently, I included it just because

I have a special fondness for the name of that. That’s a yottameter. And if you want to

have some fun, see if you can come up with

several different ways to pronounce the word yottameter. It’s kind of a fun little game. But in our problem, we’re not

playing they yottameter game. We’re playing a game involving

nanometers, kilometers, and centimeters. So let’s put those

three rows up, and let’s define r to be the radius

of the molecule in meters after the snowball is blown

up to the size of the Earth. So let’s review the information

we were given in the problem. We were told that we were

starting with a snowball that has a radius of 3 centimeters. I looked it up on the

internet, and 3 centimeters is a radius that’s just

a little bit smaller than the radius of a tennis ball. So we have a snowball that’s just

a little bit smaller than a tennis ball. The radius of a water

molecule is 0.14 nanometers. And the radius of the Earth

is about 6,400 kilometers. Now we’re going to get a

sense of what this 0.14 nanometers is by making a ratio. When we make our ratio, we

want all our units to go away, and that’s not going to happen

if we have mismatched units. So the first thing

I’m going to do is I’m going to take our

three radii, and I’m going to find their equivalent

radii in terms of meters. So the radius of the snowball

we were told is 3 centimeters. Let’s go ahead and find the meter

equivalent of 3 centimeters. So let’s start with

our 3 centimeters, and looking at the

table above, we see that 1 centimeter is the same as

10 to the negative second meters. So 3 centimeters is

the same as 3 times 10 to the negative second meters. All righty. Let’s take our water molecule

radius and convert it to meters. So we have 0.14 nanometers. Since I want to express this radius

using proper scientific notation, there’s two things

I need to attend to. One thing is I need

to make my mantissa a number between one and ten. I can’t use 0.14, because

that’s not between one and ten. So I’m going to make

my mantissa 1.4. Now, what do I need to do to

get from 1.4 back to 0.14? I need to move the

decimal point one digit to the left, which I

can do by multiplying by 10 to the negative first. Now looking up at

my table at the top, I see that 1 nanometer is equivalent

to 10 to the negative ninth meters. So I also have times 10 to

the negative ninth meters. So when I simplify, I see

that in terms of meters, the radius of a water

molecule is 1.4 times 10 to the negative tenth meters. How about the radius of Earth? Well, I’m starting

with 6,400 kilometers. 6,400 in proper scientific

notation is 6.4 times 10 cubed, because 6,400 is 6,400

and 1,000 is 10 cubed. How about a kilometer? Well, my table up top

tells me that one kilometer is equivalent to 10 cubed meters. So in simplified form using meters

and proper scientific notation, the radius of Earth is about 6.4

times 10 to the sixth meters. So what we want to do is we want

to blow the snowball up so it’s as big as the earth,

and we want to imagine the molecules growing at the same

ratio as the snowball is growing. That’s called a proportion. So I’ve gone ahead and

typed up a proportion that we could solve

to determine what the radius of the enlarged

water molecule would be. There’s a lot of numbers

and letters here, so let me kind of make it real

clear what’s playing what role here. That r in the numerator is

the new water molecule radius. It’s the radius of

the water molecule after the snowball blows up. The 1.4 times 10 to the

negative tenth meters, that’s the original real

actual water molecule radius. The 6.4 times 10 to the sixth,

that’s the new snowball radius. The snowball is blowing

up so its radius is as big as the radius of the Earth. And our snowball originated

with a 3.0 times 10 to the negative second meter radius. So new radius over old radius

equals new radius over old radius. So let’s go ahead and solve for r. r is going to be 6.4 times 10

to the sixth over 3.0 times 10 to the negative second. I don’t need to write

any units there, because I can see that

on the right-hand side of my original equation the

fraction has units of meters over meters, which divide to one. Now, I also need to

multiply by my 1.4 times 10 to the negative tenth meters. I’m doing that because one of my

principles of solving equations says if I multiply both sides of

the equation by the same number, I’ll get an equivalent equation. So let’s get out our calculator,

and perform this calculation. So let’s see, we’ve got 6.4

times 10 to the sixth power and that needs to be

divided by 3.0, which I can just enter as 3 times 10

to the negative second power. Something curious

about this calculator is that when you want

to negate a number, you hit the plus-minus

key after you enter the number, with one

exception, and that’s when you’re using scientific notation. For whatever reason, I need

to hit the plus-minus key before I hit the two. The only reason I’m

telling you that is because your calculator might

have some fluky operation just like this calculator does. So I wanted to give

you a heads-up in case you start banging your

head against the wall when you try doing scientific

calculations on your calculator. Now I need to multiply by 1.4 times

10 to the negative tenth power. And all together I get about 2.9867

times 10 to the negative second. If I round that 2.9867 to

two significant digits, that rounds up to 3.0. So something really

curious is happening. This is about 3.0 times 10 to

the negative second meters, but remember that 10 to

the negative second meters is equivalent to one centimeter. So it turns out that the new

radius of our water molecule is about 3.0 centimeters. Now why is that so curious? That’s so curious because

that was the original radius of the snowball. This is a fluke. It just happened. There was nothing natural

about this happening. It just turns out that if

you have a snowball that has a radius of 3

centimeters, and you blow it up to the size of the

Earth, coincidentally when the water molecules

grow proportionately, they end up being

the exact same size that the snowball was

in the first place. So if the water molecule grows

proportionally with the snowball, the radius of the water

molecule is about 3 centimeters when the radius of the snowball

is the same as Earth’s radius, and an implication of that

is that the ratio of a water molecule to a snowball with

the radius of 3 centimeters is the same as the ratio

of a snowball to the Earth. If you think about how a snowball

with the radius of 3 centimeters compares to the Earth,

that’s the same ratio you get if you take one

of those little water molecules inside the snowball

and compare it to the snowball. So I kind of thought about

how I could convey just how big a difference it is

between the size of a snowball and the size of the Earth. And I thought back to our

Jupiter and Mercury example, and a way to really get a sense

of how much bigger Jupiter is than Mercury, I ended up

dividing the volume of Mercury into the volume of

Jupiter, and I saw that Jupiter was about 26,000

times larger than Mercury. Well, before I prepared

this lesson or while I was preparing this lesson, I

performed a similar calculation comparing the volume of Earth

with the volume of the snowball. And it turns out that it would

take roughly 9.7 times 10 to the 24th snowballs

to fill up the Earth. And that’s a lot of snowballs. But it’s more than just

a lot of snowballs. Look at this 10 to the 24th. So that’s not only a lot of

snowballs, that’s a yotta snowball. And I think with that example,

I yotta turn it over to Candace, so that she can work

her part of the lesson. For the last part

of this lesson, I’m going to review lesson four

through what Anne and Steve talked about today in lesson eight. I’m going to start by

going over the rules and definitions of

exponents one last time and then work several examples. Then I’m going to review

polynomials again and work examples, because as I work examples

multiplying and dividing polynomials, I use some

of the rules of exponents. So let’s go over our rules

and definitions of exponents. If m and n are integers and

a, b, and c are real numbers, then first we have the

product rule, which tells us when our bases are the

same and we’re multiplying, we can add the exponents. We have our power to a power rule. This is the case where

we multiply exponents. a to the m to the n-th

power is a to the m times n. We have our power

over product rule that tells us the power of the product

equals the product of the powers. Another way to remember

this is the exponent distributes over multiplication. We have our quotient

rule, and this is the rule we can subtract exponents. a to the m divided by a to the

n equals a to the m minus n. The power of the

quotient rule is the rule where we remember that exponents

distribute over addition. For zero exponents, as long

as our base is not zero, when we raise a number to

the zero power, we get one. Our definition of

negative exponents says we can simplify the

expression a to the negative n by writing it as one

over a to the positive n. Then we had some consequences of

exponent rules and definitions. Remember that you have

to remember the rules and definitions of exponents. If you happen to forget

one of these consequences, that’s not quite as important

because you can go back to the rules and definitions. What these consequences help

with is saving some steps. The consequences of exponent

rules and definitions allow us to skip several

steps in multi-step problems. One of these

consequences is a over b raised to the negative m equals b

over a raised to the positive m. We just need to take

the reciprocal of what’s inside parentheses to change

the sign of the exponent outside the parentheses. This next one might look like

the quotient rule to you. Notice I put a little parenthetical

comment here that we usually use this when m is less than n. So when the exponent in the

numerator, m in this case, is smaller than the

exponent in the denominator, we can move everything

to the denominator. b over m divided by b over n equals

one over b to the n-th minus m-th. One One over a to

the negative n-th can be simplified most

quickly by writing it as a to the positive n-th. So let’s look at some

examples using some of these rules and definitions. Let’s say we want to simplify

4 to the negative 1 plus 4 to the negative 2. Here we have to think about

the order of operations. The order of operations tells me

to apply exponents before adding. So 4 to the negative 1 is 1/4 plus

4 to the negative second is 1 over 4 to the positive second. So I have 1/4 plus 1/16. And to add these fractions,

I need a common denominator. 1/4 can be written as 4/16, and

when I add 1/16, I have 5/16. What if I want to

simplify 12r to the 17th t to the 15th over negative

3r to the 7th p to the 10th? Well, here I can start by figuring

out what my coefficient is by dividing 12 by negative 3. And then I can deal with the

factors of r and the factors of t. 12 divided by negative

3 is negative 4. I have 17 factors of

r in the numerator and 7 factors of r

in the denominator. This reduces to r to the 10th. I have 15 factors of

t in the numerator and 10 factors of t

in the denominator. This reduces to t to the fifth. Now let’s simplify the opposite of

2 to the zero minus 3 to the zero. In this problem we have to think

about the order of operations again. The order of operations

tells us we need to apply exponents

before we can subtract, but also before we can negate. So if I have the opposite

of 2 to the zero, I need to find 2 to the

zero first, which is 1. And then negate that. And then I’m subtracting 3

to the zero, which is also 1. Negative 1 minus 1 is negative 2. Let’s simplify negative 1

over x to the negative fifth. Well, this is the same

as negative 1 times 1 over x to the negative fifth,

which is negative 1 times x to the positive fifth. That’s negative x to the fifth. In a problem like this, we

need to be careful and remember that a negative exponent does not

ever affect the sign of its base. Some people might

look at this problem and see a couple of

negative signs and say, do those negatives

cancel one another? And you’ll remember that

they don’t if you always remember a negative exponent does

not affect the sign of its base. Now let’s simplify a to the

fifth b over a to the eighth b to the negative first all raised

to the power negative third. There are several properties of

exponents that I can apply here. Remember when we’re applying

the rules of exponents that there are different

orders that we can apply them, and as long as we

apply them correctly, we should all come to

the same final result. In this problem, something

that I’m noticing is that I have factors of a in

both the numerator and denominator. So I think I’m going to simplify

what’s in parentheses first, and deal with that exponent

of negative 3 later. So I have a to the fifth in the

numerator and a to the eighth in the denominator. If I simplify that, that leaves

a cubed in the denominator. I have b in the numerator

and b to the negative first in the denominator. If I move that b to the

negative first to the numerator, that gives me another factor

of b in the numerator. So I have two factors of

b in the numerator, which I write as b squared. All of this is still raised

to the power negative third. Remember the quickest

way to simplify this, to make that negative 3

exponent into a positive 3, is to take the reciprocal

of what’s in parentheses. So in parentheses we change

this to a cubed over b squared. And now we have this raised

to the power positive 3. Exponents distribute over division. So I have a to the third cubed,

which is a to the ninth over b squared cubed, which

is b to the sixth. Now, let’s simplify, negative

3x to the negative first y cubed all raised to the

negative second power over xy to the negative third all raised

to the negative third power. This time I have different

exponents on the bases in the numerator and denominator. So I’m going to attack this

problem a little bit differently. What I think I’ll do in

this problem is go ahead and distribute the exponent through

the numerator and the exponent through the denominator. So in this case I have negative

3 to the negative second in the numerator. x to the positive second

y to the negative sixth. In the denominator I have

x to the negative third y to the positive ninth. Now let’s concentrate on

making our exponents positive. I can make that negative

3 to the negative second have an exponent of

positive 2 by moving the factor to the denominator. So it’s negative 3 squared

in the denominator. I have x squared in

the numerator and x to the negative third

in the denominator. Simplifying gives me

x to the second minus negative 3, which is x

to the second plus 3. x to the fifth in the numerator. I have y to the negative

sixth in the numerator and y to the ninth in the denominator. So I’m going to move that

y to the negative sixth to the denominator, which

gives me a total of 15 factors of y in the denominator. So I have x to the fifth

over 9y to the 15th. Now let’s write the 87,000

in scientific notation. This is one of our

applications of exponents. So I know in scientific

notation, I need my mantissa to be between one and ten, so

I need my decimal point moved between the eight and

seven, so I have 8.7. Now I need to think

about how many places I need to move the decimal point

to get from 8.7 back to the 87,000, I’d have to move the

decimal one, two, three, four places, so I need to

multiply 8.7 by 10 to the fourth. Let’s go the other direction. How do I write 7.3 times

10 to the negative fourth in standard notation? Well, that negative 4 exponent

tells me to move the decimal four places to the left. So the first place to the left

takes me to the left of the seven and then I need three

zeroes for placeholders to move the entire four

places to the left. So I have 0.00073 in

standard notation. Now, let’s review polynomials. Let’s start with the

definition of polynomials. A polynomial is an

expression containing one or more terms added together. The exponents on the variables in

each term must be whole numbers. Zero, one, two, three, and so on. We had a second definition

of a polynomial. We also said that a polynomial

in x is a finite sum of terms of the form ax to the n-th,

where a is a real number and n is a whole number. So we have some real number a

times some whole number power of x. Is this a polynomial? 3x cubed plus x squared

plus 6x minus 7. Since all of my

terms are of the form some real number times x to a whole

number power, even the constant, remember constants have degree

zero, this negative 7 at the end can be thought of as

negative 7x to the zero. This is a polynomial

with four terms. It’s a third degree polynomial,

since our highest power here is three. What about 2x to the negative third

plus x squared plus x minus 1? Well, you should recognize quickly

that this is not a polynomial, because of this

exponent of negative 3. Negative 3 is not a whole number. What about x to the

1/2 plus 3x minus 4? Again, you should

quickly recognize this is not a polynomial, because 1/2 as

an exponent is not a whole number. What about negative 1/2

x cubed plus x squared y. It’s OK to have negative

1/2 as a coefficient. We just want real

number coefficients. This is a polynomial

in two variables. Both terms have degree 3.

x cubed and x squared to y. Remember when we have

more than one variable, we add the exponents,

2 plus 1, to find the degree of the term, which is 3. This polynomial is a binomial,

because there are two terms. What about the constant 9? Well, remember with constants,

we can think of our constant as 9x to the zero, for instance,

it’s a zero degree term. So this is a polynomial. It happens to be a monomial,

since there’s only one term. What about 3x squared

plus 2 over x squared? At first glance, you might think

this is a polynomial, because you see whole number exponents. But remember, you can’t divide

by a variable in a polynomial. And in fact, another

way we can think of this is 3x squared plus 2x

to the negative second. When we write this as 2x

to the negative second, we see the exponent of negative

2 is not a whole number, and this is not a polynomial. Now, let’s add the trinomial

x squared plus 2x minus 5 and the trinomial 3x cubed

minus 5x squared plus 2. Remember, when we’re

adding polynomials, we’re just adding like terms,

like we did in Math 60. So I have my first polynomial,

a second degree polynomial, x squared plus 2x minus 5 plus

my trinomial, 3x cubed minus 5x squared plus 2, which is

a second degree trinomial. The parentheses here

are just for emphasis. I don’t need to write parentheses. The only reason I did

is to show that I’m adding two trinomials together. All I need to do here

is look for like terms. I’m going to start with

my highest degree term. So I can write my simplified

polynomial in descending order. So 3x cubed has an exponent of 3. That’s my highest degree term. There aren’t any other third

degree terms to add to this. Now I’m going to look

for second degree terms. In the first polynomial,

I have x squared. In the second polynomial,

I have minus 5x squared. Adding these together,

I have minus 4x squared. Now, I’m going to look

for my first degree terms. In the first polynomial, I have 2x. In my second polynomial, I don’t

have any first degree terms, so I’m just going to write down plus 2x. Now, I’m down to my zero degree

terms, which are my constants. Negative 5 in the first polynomial,

and 2 in the second polynomial. When I add those together,

I get negative 3. Now, let’s subtract x

squared plus 2x minus 5 from 3x cubed minus

5x squared plus 2. The trickiest part about

setting up this problem is deciding which order to

write our polynomials in. If I ask you to subtract

2 from 5, pretty quickly you’d probably tell me that was 3. What you do there? 5 minus 2. So we need to start with the

polynomial we’re subtracting from, which is the 3x cubed minus

5x squared plus 2 minus, because we’re subtracting

the polynomial x squared plus 2x minus 5. This time my first set of

parentheses aren’t really needed. My second set of

parentheses are essential. I need the parentheses

around the polynomial I’m subtracting to remember that

each term in that polynomial needs to be subtracted. So to simplify this, I’m going

to start by removing parentheses. The first set of parentheses

is not essential. I can just remove them. To remove that second

set of parentheses, I need to distribute the

negative through to each term, so I change the sign of each term. Minus x squared minus 2x plus 5. Now, I can go ahead and look

for like terms to combine. Starting with my

highest degree term, 3x cubed, there’s only

one third degree term. Looking for second degree terms,

I have minus 5x squared and minus x squared. All together I have

minus 6x squared. Looking for first degree terms

I have minus 2x, and that’s it. So I bring down minus 2x. And in my constants, I

have 2 plus 5, which is 7. Now, I want to talk a little bit

about multiplying polynomials. I’m going to review some of

the things that we went over. I’m going to start by looking

at a plus b times a minus b. Do you remember what

these are called? The quantity a plus b and

the quantity a minus b, these are conjugates. We have a binomial times a binomial. A two term polynomial multiplied

by a two term polynomial. When we’re multiplying two

binomials together, we can use FOIL. Let’s quickly review FOIL. In FOIL, we take the

product of the first terms. In this case, that’s a

times a, which is a squared. Then we take the product

of the outer terms. In this case, that’s a times

negative b, which is negative ab. We take the product

of the inner terms, in this case that’s b times a. So plus ba. We finally take the

product of the last terms. In this case, that’s b times

negative b, which is negative b squared. Remember that when we are

multiplying polynomials, it’s a good idea to be in the habit

of always writing your variables in the terms in alphabetical order. So if we look at our product

of our inner terms ba, we see this isn’t in

alphabetical order. It helps us to locate like

terms if we write the variables in a term in alphabetical order. So let’s go ahead and do that. Let’s rewrite this ba as ab. When we do that, we see we

have like terms, negative ab and positive ab, which add to zero. So when we add these

together, we’re left with a squared minus b squared. Notice this is one of

our special products. When we multiply two

conjugates together, we get a difference of squares. In this case, a squared

minus b squared. a squared and b squared

are perfect squares. We’re subtracting them, thus

the difference of squares. A couple of other special

products that we looked at. a plus b quantity squared gives

us a perfect square trinomial, a squared plus 2ab plus b squared. a minus b quantity squared,

another perfect square trinomial. a squared minus 2ab plus b squared. Let’s square r cubed plus t cubed. This is one of those

special products. Here we should get a squared, in

this case a is r cubed, plus 2ab. So 2 times r cubed. And in this case, b is t cubed. Plus b squared. So t cubed squared. This simplifies to r to

the sixth plus 2r cubed t cubed plus t to the sixth. When you first looked

at this problem, did you think about distributing

that two to the r cubed and the t cubed? Hopefully you didn’t, because

exponents do not distribute over addition or subtraction. That’s one of the most important

things you need to remember. However, exponents do distribute

over multiplication and division. Let’s look at an

example of exponents distributing over multiplication. Let’s say we want to square

r cubed times t cubed. Now, since we have multiplication

inside the parentheses, we can distribute the exponent. That gives us r cubed

squared t cubed squared, which simplifies to r to

the sixth t to the sixth. If you don’t need to show that

step in between, that’s great. I was just making a point

here that the exponent distributes over multiplication. If it helps to show

that step, go ahead and show that step just like I did. Now let’s multiply k plus

1/3 m by k minus 1/3 m. You might recognize these

as being conjugates. When we multiply two

conjugates together, we get a difference of squares. In this case, k squared

minus 1/3 m quantity squared. Simplify, and that is k

squared minus 1/9 m squared. Now, let’s multiply a cubed plus

11 by a to the fourth minus 2. Here I have a binomial

times a binomial, but it doesn’t look like

any of the special products. So I can just go

ahead and apply FOIL. The product of my first terms,

a cubed times a to the fourth, is a to the seventh. The product of my outer terms,

a to the third times negative 2, is negative 2a to the third. The product of my inner

terms 11 and a to the fourth is 11a to the fourth. And the product of the last terms,

11 and negative 2, is negative 22. I don’t see any like

terms here to combine, but I am going to take

one more step and rewrite this in descending order. That’s the way answers are generally

written in the back of the book, and it’s important to remember that

a to the seventh minus 2a cubed plus 11a to the fourth

minus 22 is the same as a to the seventh plus 11a to the

fourth minus 2a cubed minus 22. Now, let’s multiply x plus 2 by

x to the fourth minus 3x plus 5. This time I have a binomial, x plus

2, multiplied by the trinomial, x to the fourth minus 3x plus 5. FOIL doesn’t work here. FOIL only works when you have

a binomial times a binomial. What I need to do here is

multiply the x by x to the fourth by negative 3x and by 5. Then I need to multiply 2 by x to

the fourth by negative 3x and by 5 and combine my like terms. So distributing the x through

the second set of parentheses, I have x to the fifth

minus 3x squared plus 5x. Distributing the 2 through

the second set of parentheses, I have plus 2x to the

fourth minus 6x plus 10. Now, I need to look for like terms. I’m going to start with

my highest degree terms. It looks like I just have one

fifth degree term, x to the fifth. So now I’ll look for

fourth degree terms. I have 2x to the fourth. I don’t see any third degree terms. I have just the one second

degree term, negative 3x squared. I have two first degree terms,

5x and negative 6x adds up to negative 1x, which we

write simply as negative x. And then just the

one constant plus 10. For my last example, I want to write

an algebraic expression in reduced form for the shaded region. So looking at my region here,

I have this big periwinkle rectangle, which has a length of 3x

plus 2, and a width of 2x minus 1. From the periwinkle rectangle, we

have a little white square cut out, which has sides of length x. So if I want to find the area

of this periwinkle region, I need to take the

area of the rectangle minus the area of the square. So the area of the rectangle,

length times width, is 3x plus 2 times 2x minus 1. Minus the area of the square

would be minus x squared. I’m asked to write my algebraic

expression in reduced form, so I need to go ahead

and do some algebra here. At the beginning of

this expression I have a binomial times a

binomial that gives me the area of the large rectangle. Since this is a binomial times

a binomial, I’ll apply FOIL. 3x times 2x is 6x squared. 3x times negative 1 is negative 3x. 2 times 2x is plus 4x. 2 times negative 1 is negative 2. And finally I have that minus

x squared, which represents taking out the area of that square. Combining like terms, 6x squared

minus x squared is 5x squared. Negative 3x plus 4x

is plus x minus 2. So the area of the periwinkle

region is 5x squared plus x minus 2. This concludes lesson eight. [MUSIC PLAYING]