 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
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
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
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
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
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
tackle that significant digits issue with you. In this calculation, I’m just
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
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
we were told is 3 centimeters. Let’s go ahead and find the meter
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
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 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
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
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
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
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
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
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]