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Hi, class.
Today we're going to be multiplying polynomials.
And as we get into multiplying polynomials,
we're going to need a few exponent rules.
So, we're going to take a look at some problems
that we might have to deal with
and kind of come up with what our exponent rules are.
And then I'll get into multiplying.
So, for instance, let's say you had x squared [ x² ]
and you were going to multiply it by x cubed. [ x³ ]
In order to figure out what to do with these,
let's take a look at what x squared [ x² ] really is.
And it's really just x times x. [ x · x ]
And we're multiplying that by what x cubed [ x³ ] really is.
So let's take a look at that.
And x cubed [ x³ ],
the cube is telling us to multiply 3 of these together.
So that's x times x times x. [ x · x · x ]
If you take a look at this right here, [ x · x · x · x · x ]
you can see that we are multiplying 1, 2, 3, 4, 5 x's together.
So this is really x to the 5th. [ x² · x³ = x⁵ ]
Now, you could expand every problem out like we did right here,
in order to come up with the solution.
But it would be nice if we had some kind of a short cut
to get from here [ x² · x³ ] over to here [ x⁵ ]
So, let's examine these two things we multiplied
and what our final answer ended up being
based on what happened when we expanded things out.
Notice that we have a 2 here [ x² ] and a 3 here. [ x³ ]
And after multiplying, we ended up with a 5. [ x⁵ ]
A shorter way to do this problem would be
to possibly just add the 2 and 3 and you would get 5.
In fact, that is what
our first rule for exponents is going to be.
It's our product rule.
And the product rule says that if you have bases the same,
for instance, maybe this one's x and this one's also x,
and you are multiplying, all you have to do is add the exponents.
So, for instance, if I had an m to the 10th [ m¹°]
and I'm multiplying it by m to the 7th, [ m⁷ ]
the answer would be m to the 17th. [ m¹°m⁷ = m¹⁷ ]
So you just add things that are alike.
Now, let's expand on that idea a little bit farther
and see if maybe we can come up with another rule.
So, for instance, what if you had x squared [ x² ]
and maybe you were cubing it? [ (x²)³ ]
This cube up here [ (x²)³ ] is telling us
that we are multiplying three of these together.
So this would be the same as
x squared times x squared times x squared. [ x² · x² · x² ]
If you take a look at the rule that we just learned right here,
we need to add these exponents.
And 2 plus 2 plus 2 happens to be 6. [ 2 + 2 + 2 = 6 ]
Now, I don't want [ (x²)³ = x² · x² · x² = x⁶ ]
to have to do all this work every single time.
This was a lot to do.
It would be nice if we could go from here [ (x²)³ ]
straight to here. [ = x⁶ ]
If you examine our numbers, [ (x²)³ ]
when we had something raised to a power
and then raised to another power, the 2 and the 3 became a 6.
You could just multiply these together [ 2 · 3 = 6 ]
to get this. [ (x²)³ = x⁶ ]
So if you have a power and it's raised to another power,
all you have to do is multiply those powers.
This is our power to a power rule.
xⁿ = xⁿ
And we can expand on this a little bit farther.
For instance, let's say you have x squared, y cubed [ x² · y³ ]
and you are squaring it. [ (x²y³)² ]
Basically, this is the same as x squared, y cubed
times x squared y cubed. [ (x² · y³)²=(x² · y³)(x² · y³) ]
We can rearrange things since it's all multiply,
and move these around
so that like bases are next to each other
to get x squared times x squared times y cubed times y cubed.
[ = x²x²y³y³ ]
Multiplying things that have like bases,
we end up with a 2 plus a 2 on our x's for x to the 4th [ x⁴]
and a 3 plus a 3 on our y's for y to the 6th. [ x⁴·y⁶ ]
Notice, if you look at the original problem
and you go down to the answer, really all that happened
was this exponent was applied to everything inside.
So we could have done a short cut
of just 2 times 2 to get the 4
and 2 times 3 to get the 6
which is applying this rule here to everything inside.
So you can extend this rule to multiple things inside.
If you have a power outside, it gets applied to everything inside.
[ (xⁿ·y)^p = xⁿp·yp ]
Also, if you have a fraction,
which I'll give us a little more room here
so you can see that happen, the same kind of thing happens.
Let's say you have
x to the n over y to the m [ (xⁿ/y) ]
and it's raised to the p. [ (xⁿ/y)^p ]
Notice that when you multiply fractions like,
say x squared, y cubed times an x squared y cubed,
we multiply straight across.
Which means this times this [ x² · x² ] would be x⁴.
And this times this [ y³ · y³ ] would be y to the 6th [ = y⁶ ]
So the answer would be x to the 4th, y to the 6th. [ x⁴/y⁶]
And essentially what we did here because these are both the same,
is we really did x squared over y cubed squared. [ (x²/y³)²]
And essentially what happened was,
this power here [ ²] was applied to everything inside,
even though we had a fraction.
So, our power to a power rule here can be expanded
to whether you have a fraction.
If you have a fraction,
you just make sure that the exponent on the outside
gets applied to everything on the inside.
So those are your basic rules.
We're going to use those rules to do our multiplying.
So let's say that you had something like
5x squared, y cubed ( 5x²y³ )
and you're multiplying it by [ (7x⁵ · y⁴) ]
[ (5x²y³)(7x⁵ · y⁴) ]
Because it doesn't matter what order we multiply in,
we can multiply this in any order we would like.
So, we could multiply the numbers,
multiply the x's, multiply the y's.
We can, actually, drop our parentheses
since everything is multiplying, and rearrange things
because multiplication is commutative.
So we could actually rearrange this to 5 times 7 times x squared
times x to the 5th times y cubed times y to the 4th.
The commutative property of multiplication allows us
to rearrange things wherever we want them for multiplication.
[ 5 · 7 · x² · x⁵ · y³ × y⁴ ]
And then, since 5 and 7 are both numbers,
we just multiply them to get 35. [ 5 · 7 = 35 ]
The x squared times the x to the 5th, [ x² · x⁵ ]
since those have the bases the same,
we add our exponents to get an x to the 7th. [ x² · x⁵ = x⁷]
And our y to 3rd times y to the 4th, [ y³ · y⁴]
again we add our exponents
to get a y to the 7th. [ y³ · y⁴ = y⁷]
[ 35x⁷y⁷ ] And that's all there is to that.
Some people will need to write this interim step right here.
Some will be able to go from here straight to here.
Let's try one more of those.
Let's say we have 8m to the 12th, n to the 7th, [ 8m¹² · n⁷ ]
and we're multiplying it by 9m, n squared. [(8m¹² n⁷)(9mn²)]
Notice that on this m in this second monomial,
there's no exponent showing.
A lot of people when they multiply the m's together
will forget that there's an invisible 1 here.
If you have to at the beginning write that 1 in there [ m¹ ]
to remind yourself to add the 1 to the 12,
by all means, please do that on your paper in your work
so that you can see what's happening.
Eventually, you'll wean yourself off of that, though.
[ (8m¹²n⁷)(9m¹n²) ]
So, let's go ahead and multiply our numbers.
8 times 9 gives me 72. [ 8 · 9 = 72 ]
And then our m's,
because we like to keep things in alphabetical order.
So m to the 12th times m is m to the 13. [ m¹²m¹ = m¹³ ]
And then n to the 7th times n squared, [ n⁷ · n² ]
we add those exponents to get n to the 9th. [ n⁷ · n² = n⁹]
[ 72m¹³·n⁹ ]
Now, let's make our multiplying of monomials
just a little bit harder by adding a little bit in there.
Let's do 5x squared y to the 5th [ (5x² · y⁵) ]
times 3x cubed, y, squared, [ (3x³y)² ]
written like that. [ (5x² · y⁵)(3x³y)² ]
Again, notice that this second item here
has an exponent on the outside of your parentheses.
So we need to apply that to everything inside first.
So we're going to start over here. And 3 squared is 9 [ 3² = 9 ]
x cubed squared, [ (x³)² ]
we multiply the exponents to get x to the 6th. [(x³)² = x⁶]
And then notice that our y here doesn't have an exponent on it.
Remember that has that invisible 1 here. [ x¹ ]
So, when we multiply our 1 times our 2,
we get y to the 2nd. [ y² ]
Now, we're going to carry down our first item,
which is 5x squared, y to the 5th, [ 5x²·y⁵ ]
and multiply it by what we came up with
after applying our exponent. [ (5x²y⁵)(9x⁶y²) ]
So now, 5 times 9 is 45.
x squared times x to the 6th is x to the 8th. [ x² ·x⁶ = x⁸ ]
And that's because we added our exponents
because we're just multiplying this time. So 2 + 6 = 8.
And then y to the 5th times y to the 2nd, [ y⁵ · y² ]
we add the 5 and the 2 to get y to the 7th. [ y⁵ · y² = y⁷]
And this here [ 45x⁸y⁷ ] is your final answer.
So that's how you deal with multiplying monomials.
Now we're going to take a look at what happens
when you multiply a monomial by a polynomial.
And you've actually done things like this before.
You did them in a much simpler form,
because you might have had something like
5 times 2x plus 3 in your previous class. [ 5(2x + 3) ]
When you did this, you basically just distributed the 5
and you got a 10x plus a 15. [ 10x + 15 ]
And because there wasn't as much in this,
you could actually do it in your head.
But there is an interim step right here
that you didn't write down that might be helpful
in these problems you're going to be doing now.
And that interim step was to just show your multiplication
of the 5 times the 2x [ 5 · 2x ] and the 5 times the 3 [ 5 · 3 ]
When you write that down, it looks like this. [ 5(2x) + 5 (3) = ]
And then, after you have changed this original problem here
into this, you can actually go back and multiply to get this.
Now, in this problem, you didn't really need to do that;
however, your new problems are going to look more like this one.
You might have something like 6, oops, let's re-write that,
6p squared, q cubed, [ 6p²q³ ]
times 5p minus 7q plus 3. [ 6p²q³(5p - 7q + 3) ]
There's a lot more going on in this particular problem,
so that interim step that we didn't write
on our previous problems might be helpful.
So we're actually going to just distribute
and write the multiplication down and come back to multiply.
So, 6p squared, q cubed [ 6p²q³ ]
times our first term of 5p. [ (6p²q³)( 5p) ]
minus, and then 6p squared, q cubed times 7q. [- 6p²q³( 7q )]
And then plus 6p squared, q cubed, times 3. [ + 6p²q³(3) ]
[ 6p²q³(5p) - 6p²q³(7q) + 6p²q³(3) ]
Now, we go back and do each of these
as separate little monomials times monomials.
So for this first one, we get 30 and then combining our p's,
notice that this p here has a 1 for an exponent. [ p¹ ]
So if we do 2 plus 1,
there will be p cubed here [ p² + p¹ = p³ ]
If you take a look at your q's,
we have q cubed here [ q³ ] and there's no cubes here.
That's like a 3 plus a zero,
which is going to give us a q cubed. [ q³ + 0 = q³ ]
So this first term turned into 30p cubed, q cubed. [ 30p³q³ ]
Minus, now let's look at the next one.
6 times 7 gives you 42, [ 6 · 7 = 42 ]
and then notice that we have a p squared times no p's over here
which will just be our p squared. [ p² + 0 = p² ]
And then we have our q cubed
times a q with no exponent, [q³q ]
which means that that exponent is really a 1. [ q³q¹ ]
So 3 plus 1 gives us q to the 4th. [q³q¹ = q⁴ ]
And now on the last one, we have 6 times 3 is 18. [ 6 · 3 = 18 ]
And since there were no p's and q's in this last particular term,
we just carry down our p squared and our q cubed. [ p²q³ ]
And notice that because this original polynomial
in the parentheses that we were multiplying by
was in descending order,
because we want the variables in alphabetical order
when we don't have more than one
of your one that comes first in the alphabet.
So this is a p which comes before q so it's first.
And all we have is one term with a p and one term with a q.
So, since this was in descending order,
your final answer will automatically be in descending order.
So that's how you do a monomial times a polynomial.
Let's expand this a little.
We're going to do a binomial times a binomial,
which means we're going to have something with two terms
like 2x minus 3 times something else.[ (2x - 3)( + ) ]
with 2 terms like 4x plus 5. [ (2x - 3)(4x + 5) ]
Now, when you do this particular problem,
you can use this same kind of distribution type thing
that we did in the last problem.
Your goal is to make sure
that you multiply everything in this first parentheses
by everything in the last parentheses.
So I could just draw lines and distribute the 2x
to everything over here,
and the negative 3 to everything over here.
And once there's a line from everything up here
drawn to everything over here, I know I've done everything.
We're going to do it that way,
and then I'm going to show you an acronym that many people use
for multiplying binomials times binomials.
Keep in mind that the acronym only works
for binomial times binomial.
If you're doing a binomial times a trinomial,
it no longer applies.
So let's start with the 2x.
I'm going to do 2x times my 4x [ ( 2x )( 4x ) ]
which ends up giving me 8x squared [ ( 2x )( 4x ) = 8x² ]
because we do a 1 plus a 1 on our exponents.
And then, 2x times 5 is 10x. [ ( 2x )( 5 ) = 10x ]
And then I'm done with my 2x.
So now I'm going to start with my negative 3.
And we're using a negative 3, [ -3 ]
because the minus in front of the 3 goes with that.
So negative 3 times 4x is a negative 12x. [ ( -3 )( 4x ) = -12x ]
And negative 3 times the positive 5 [ ( -3)( 5 ) ]
is a negative 15. [ ( -3)( 5 ) = -15 ]
One thing that you might notice about this particular problem
is we ended up with some like terms here.
A lot of times with a binomial times a binomial,
you will end up with some like terms in order to combine.
And you do want to combine those
in order to simplify it into your final answer.
So, 10x and negative 12x is a negative 2x, [ 10x - 12x = 2x ]
which will be our middle term
for the final answer for this problem. [ 8x² - 2x - 15 ]
And this will be your simplified form. [ 8x² - 2x - 15 ]
Now this method that we just saw here will work,
whether you have a binomial times a binomial,
or a binomial times a trinomial,
or something with, that's a trinomial times a trinomial.
You can always draw lines
from everything in the first parentheses
to everything in the second parentheses until they are all done.
I would suggest, however, to draw your first curved line,
multiply the two things, write down your answer,
and then draw your next one,
so that you don't forget to actually do one of them.
Now, let's look at the acronym.
The acronym that we have for remembering binomial times binomial
is F - O - I - L, FOIL.
And many of you might have heard of that before.
The F stands for multiplying the First things in the parentheses.
The First things in the parentheses are the 2x and the 4x,
which when you multiplied gave you that 8x squared. [ 8x² ]
So this was the F out of our acronym.
The O stands for Outer.
And the Outer things in our parentheses are the 2x and the 5,
[ 2x · 5 = 10x ] which we multiplied to get our 10x.
So the 10x was our O in FOIL.
The I here stands for Inner.
Those are the Inner things in the parentheses
which is the negative 3 and the 4x,
which are multiplied to get our negative 12x.
So that was the I for FOIL.
And then, finally, our L stands for Last,
which is the last things in the parentheses
which happens to be our negative 3 and our 5,
which gave us our negative 15. [ -3 · 5 = -15 ]
So this is your L in FOIL.
That little acronym will a lot of times help people
make sure that they multiply everything they need to multiply
when you're doing a binomial times a binomial.
[ First Outside Inner Last ] Let's do one more of these,
and then we'll move on to a binomial times a trinomial,
which will be the largest things that you'll need to multiply.
So, for instance, let's say that we have
3x minus 5y times 4x plus 6y. [ (3x - 5y)(4x + 6y) ]
First, let's multiply our 3x times our 4x
to get 12x squared. [ 3x · 4x = 12x² ]
Then 3x times 6y, which gives us an 18xy. [ 3x · 6y = 18xy ]
Now we're done with our 3x,
so let's move on to our negative 5y. [ -5y ]
Negative 5y times 4x, [ -5y · 4x ]
we'll put our letters in alphabetical order,
so that's going to be a negative 20xy. [ -5y · 4x = -20xy]
Notice that that is a like term with the 8xy.
Now, our negative 5y times our 6y [ -5y · 6y ]
is a negative 30y squared. [ -5y · 6y = -30² ]
[ 12x² + 8xy - 20xy - 30y² ] Let's combine our like terms
here in the middle [ 8xy - 20xy ]
to get a negative 12xy. [ 8xy - 20xy = -12xy ]
And carry down our other items for the final answer
of 12x squared minus 12xy
minus 30y squared. [ 12x² - 12xy - 30y² ]
Now, finally, let's do a binomial times a trinomial.
Let's say that we have 3x plus 2
and we're multiplying it by 4x squared minus 5x plus 3.
[ (3x + 2)(4x² - 5x + 3) ]
The drawing line method that I showed you
for binomial times binomial
will still work for this particular problem.
Let's do it that way,
and then I'm going to show you another way that you might like.
So we're first going to multiply the 3x
times everything in the second parentheses.
So, 3x times 4x squared, remember this x has a 1,
so 1 plus 2 gives us 12x cubed. [ 3x¹ · 4x² = 12x³ ]
And then 3x times a negative 5x [ 3x · -5x ]
is negative 15x squared, [ 3x · -5x = -15x²]
because 1 plus 1 gives us our x squared. [ 3x¹ · -5x¹ = -15x²]
Then, 3x times 3 is 9x. [ 3x · 3 = 9x ]
I'm done with that 3x, so I'm going move on to my 2.
Now, when I do my 2, I want you to watch
where I strategically put these answers.
I'm going to do 2 times 4x squared, [ 2 · 4x² ]
and I'm going to get an 8x squared.
I'm going to write that right here,
because it's right underneath its like term,
which is going to help me when I combine my like terms,
because there's more like terms
when you're multiplying these kinds of items.
So now, let's go do our 2 times our negative 5x. [ 2 · -5x ]
That gives me a negative 10x, [ 2 · -5x = -10x ]
which I'm going to put right here
so it's strategically underneath its like term.
And finally, 2 times 3 is 6. [ 2 · 3 = 6 ]
And notice that I can now draw a line
and just add straight down in order to get my like terms.
So I have plus 6.
9 and negative 10 gives me a negative 1. [ 9 + -10 = -1 ]
And since my variable is x,
that's really a negative 1x [ -1x ]
or I can write it as a negative x [ -x ]
Negative 15x squared and 8x squared [ -15x² + 8x² ]
is a negative 7x squared. [ -15x² + 8x² = -7x² ]
And just carry down my 12x cubed. [ 12x³ ]
And then here is your final,
simplified answer. [ 12x³ - 7x² - x + 6 ]
Now, that's one way to do this particular problem.
I'm going to show you another way
that is very similar to your old fashioned multiplication
where you might have done a 2 digit number times a 3 digit number
Now, I'm going to erase this
to give us a little bit of room for that.
But I'm also going to do a problem
to remind us of our old fashioned multiplication.
You might have had something like say, 312 times 24. [ 312 × 24 ]
When you were doing a problem like this,
you would start with your 4 and you would do 4 times 2 is 8.
[ 4 × 2 = 8 ] And you'd write the 8 down.
And then you would do 4 times 1, which is 4, [ 4 × 1 = 4 ]
and you'd write the 4 down.
And then you'd do 4 times 3, which is 12, [ 4 × 3 = 12 ]
and you'd write the 12 down.
Then, because you're going to move on to the 2
and the 2 is in the tens place, you would put a zero right here.
And then you would continue by doing 2 times 2 which is 4
[ 2 × 2 = 4 ] and write a 4 right here.
And then you would do 2 times 1 which is 2 [ 2 × 1 = 2 ]
and you'd write a 2 here.
And then you would do 2 times 3 which is 6, [ 2 × 3 = 6 ]
and write your 6 here.
Then you'd draw a line and you would add straight down
to get 8, 8, 4, and 7.
And your answer would be 4,488.
That same process can be done with a problem like this.
Here, notice you have a three digit number
on top of the two digit number.
So what we're going to do is re-write this here
so that we have the 3 term item over the 2 term item.
And it's going to look like
4x squared minus 5x plus 3 here [ 4x² - 5x + 3 ]
with 3x plus 2 here. [ 3x + 2 ]
And we're going to do the same process.
We're going to start with our 2 here,
and do 2 times everything up here.
So 2 times 3 gives me 6. [ 2 × 3 = 6 ]
And it's positive, so I put a plus in front of it.
And then 2 times negative 5x is negative 10x. [ 2 × -5x = -10x ]
And since it's negative,
we're going to have a minus in front of it.
2 times 4x squared is 8x squared. [ 2 × 4x² = 8x² ]
I'm now done with the 2, so I'm going to drop down
and put a plus 0 here and move on to this 3x.
3x times 3 is 9x [ 3x × 3 = 9x ]
and it's positive, so I'll put a plus.
3x times negative 5x is negative 15x squared [3x × -5x = -15x²]
And then 3x times 4x squared is 12x cubed. [ 3x × 4x = 12x³ ]
Now, if you draw a line, you can add straight down
to get a positive 6 here [ 6 ] a negative 1x here [ -1x ]
a negative 7x squared [ -7x² ] and a positive 12x cubed [ 12x³ ]
Notice that this is the same answer we got when we drew our lines.
So this method here is called the vertical method.
You can use the vertical method or draw your lines,
whichever method you would like.
They both work perfectly fine.
The key is to make sure you show all your work
and be very careful that you check everything
so that you don't have any simple mistakes
such as not adding your exponents
or adding or multiplying incorrectly.
That is everything that you need to know
for multiplying polynomials for this particular chapter.
You are ready to do any of the problems
for multiplying polynomials in your textbook,
and when you are ready, go ahead and view the next recording,
which is on dividing polynomials. �