Hey, guys,
welcome to Algebra I.
Today's lesson focuses
on dividing polynomials.
All you know about factoring
and the laws of exponents
will get you through
this lesson.
You ready? Let's go.
All right.
Let me jog
your memory first.
Do you remember how to simplify
something like this?
If we apply
the laws of exponents
to simplify 20x to the 5th
over 4x squared--
Let me be sure
I've got my pen here.
To handle this one,
the first thing that we do
is handle the coefficients.
We'd go ahead and divide 20
divided by 4,
and say,
"All right, that's 5."
Then x to the 5th
divided by x squared--
the laws of exponents say,
in a situation like this,
a quotient of powers,
subtract the exponents.
So 5 minus 2, that's 3.
The other part
is x to the 3rd.
We did that
earlier in the course.
That helps us learn how
to divide polynomials.
Look at this one.
Deep breath
'cause that's a long one.
We've got 35y to the 4th
plus 20y squared
minus 18y--
that whole quantity--
we're going
to divide it by 5y.
It looks
a little intense,
but it breaks down
to something simple.
We're going to take each of
these terms and divide it by 5y.
I'll rewrite this differently.
It might make you
more comfortable
if you saw it like this:
35y to the 4th
divided by 5y
plus 20y squared
divided by 5y
minus 18y divided by 5y.
That's all that we'll do here.
Essentially, you are
taking that 5y
and dividing each term by it.
Whatever our answer is
is the answer.
Okay?
I need more
room to work.
Let's get rid
of some stuff.
Let's get rid
of example one.
Let's cut that
out of the picture.
Let's move this up,
and then let's move this up
so we can get
more space up here.
Now let's get our pen back,
and we're really to rock.
If we look at this
first chunk right here,
we'll handle coefficients first,
like you already know.
You ignore
the rest of this problem
and just focus your
attention on that first bit.
So 35 divided by 5,
I know that's 7.
Now, y to the 4th
divided by y.
Remember, there's an invisible
1 we don't see, right?
To handle
y to the 4th
divided by y to the 1st,
subtract exponents.
That'd be
y to the 3rd.
That first part just simplifies
to 7y to the 3rd.
Let's look
at the second part.
It's plus,
so put that in there--
20y squared divided by 5y
to the 1st.
Get that in there.
What are we're doing first?
Right, handle
those coefficients--
20 divided by 5, that's 4.
y squared divided
by y to the 1st,
subtract the exponents.
2 minus 1, that's 1.
This is y to the 1st,
but when it's a 1,
we don't typically
write the exponent,
so 4y for the second one.
Let's look at the last one.
We have 18 divided by 5.
That might have made
you uncomfortable
because it's not
dividing evenly,
but we're not afraid
of fractions in Algebra I.
I can't simplify this
to anything simpler
than 18 divided by 5,
so I'll leave it as 18 over 5.
Just leave it
like a fraction.
I've got y to the 1st, right,
divided by y to the 1st--
so 1 minus 1, that's 0.
Think back--
I'm coming off to the side.
Remember, anything
to the 0 power is 1.
So I'm going to be multiplying
18 divided by 5 times 1,
but that doesn't
change anything.
Leave it as is.
Just leave 18 divided by 5,
and you've got it.
This polynomial
is your answer.
When you divide what we
started with divided by 5y,
that's what you get.

(female describer)
7y cubed plus 4y
minus 18 over 5.

(teacher)
That's your answer for that one.
Let's try another one.
Find the quotient.
We have 18x to the 6th
plus 27x to the 5th
minus 45x cubed,
all divided by 9x squared.
This problem
is set up differently,
but you are doing
the same thing.
You're going
to find the quotient.
I break this into chunks,
like we just did.
In math,
if you can chunk things up
and look at them in bits,
it's easier to process
a problem that way.
You can manipulate this the way
you want to get through it.
As long as you follow the rules
for math, do whatever you want.
I get that
out of our way.
Better move that up a bit;
get some room.
I'm going to chunk
this problem up
so it's easier to process.
I'm going to write it as 18x
to the 6th
divided by 9x squared.
Then I have plus...
27x to the 5th
divided by 9x squared.
Then I have minus
45x to the 3rd
divided by 9x squared.
I've got that
all broken up.
I'm going to handle each piece
separately and get through it.
If I look at the first bit,
the coefficients first,
so you have 18 divided by 9--
that's 2.
I bet your mental math
is good now
that you've been working
with these numbers.
Now look at
that variable part--
x to the 6th
divided by x squared.
So what do we do there?
Subtract the exponents.
So, 6 minus 2, that's 4.
So, 2x to the 4th.
I'm done with that first part.
I'm crossing it out.
Look at the middle bit.
Look at our coefficients.
I'm bringing
the plus sign down--
27 divided by 9, that's 3.
And then as far
as our exponents go,
what will we do?
Subtract.
So, 5 minus 2, that's 3.
This part's going
to be x cubed.
Let's look
at the last part.
I brought my minus down.
So, 45 divided by 9,
that's 5.
Okay?
And I handle
my exponents over here.
I need to subtract--
3 minus 2, that's 1.
When it's an exponent of 1,
we typically don't write it.
That crazy-looking problem
that started out
simplifies to this--
2x to the 4th
plus 3x cubed minus 5x.
And we're all done.
Look at the next one.

(female describer)
Example 3:
Find the quotient.
20a to the 6th,
b to the 4th
minus 16a to the 4th,
b squared,
all over 4a cubed, b.
All right,
so this one's different
because we've got
two different variables
in this problem,
but the process
is still the same, okay?
I told you that
you'll never run
into the same type
of problem every time,
so don't memorize
exactly the steps I'm taking,
but memorize the process.
It's the same process
I'm taking every time.
What I do here to have room
to work is get rid of stuff.
I'm going to slide that up.
The first thing
I do every time
is chunking it out,
just breaking it up into chunks.
This one's short, so I'm not
going to have but two chunks.
I'll have 20a to the 6th,
b to the 4th
divided by 4a cubed b.
Got the first bit.
Then I have minus 16a
to the 4th,
b squared
over 4a cubed b, okay?
Got it into chunks.
I simplify
what I've got here.
Same thing--handle
your coefficients first.
You have 20 divided by 4--
mental math--that's 5.
If you don't want to do mental
math, pull out a calculator.
Handling the "a"s,
I have 6 and 3,
so I need to subtract those.
6 minus 3 is 3,
so a to the 3rd.
Then I have a b,
so I know it's an invisible 1.
Subtract these.
So 4 minus 1,
that's also 3,
so b cubed.
That first part just simplified
down to 5a cubed, b cubed,
and you've handled that.
You're done with that.
Bring down the minus sign,
then we handle this part.
Coefficients first--
16 divided by 4, that is 4.
Let's handle these exponents
with the "a"s.
We've got 4 and 3,
I need to subtract.
4 minus 3, that's 1.
So, a to the 1st, but we don't
generally write the 1.
If I handle my "b"s--
there's an invisible 1
to help me subtract.
2 minus 1, that is 1.
So, then I've got just b.
There you go for that one.
That quotient just simplified
to 5a cubed, b cubed minus 4ab.
Let's keep going.
What do you know?
It is your turn.
Press pause,
take a few minutes,
and work through
these problems,
and when you're ready
to compare answers, press play.

(female describer)
Find each quotient.
1. Open parenthesis,
28x to the 7th
minus 49x to the 6th
plus 21x to the 4th,
close parenthesis,
divided by 7x cubed.
2. 30x to the 5th,
y cubed plus 50x to the 4th,
y squared,
all over 10x squared,
y squared.

(teacher)
Let's see how you did.
Let me get my pointer
to move these out of the way.
For the first one,
your answer should have been
4x to the 4th
minus 7x cubed plus 3x.
And for the second one,
3x cubed y plus 5x squared.
Now, if you want to see
how I got those,
I'll show you.
Okay.
I followed the same process
we've been doing.
I started out by rewriting this
into my chunks,
into those three chunks.
At first--I'm going to need
a little more room.
Get this out of the way,
and let's move that up.
Here we go.
Get my chunks going.
28x to the 7th divided
by 7x cubed
minus 49x cubed--
I'm sorry, x to the 6th--
divided by 7x cubed
plus 21x to the 4th
divided by 7x cubed.
Okay?
I've got it in my chunks,
now I simplify.
I handle
my coefficients first--
28 divided by 7, that's 4.
Subtract my exponents here.
So 7 minus 4, that's 4--
I'm sorry, 7 minus 3--
that's 4, all right,
minus 49 divided by 7.
That's 7.
Subtract my exponents,
and I get 3.
All right, got that chunk.
Last bit,
got my plus sign.
21 divided by 7 is 3.
Subtract my exponents,
and I get 1.
We don't typically write
the exponent when it's 1.
There we go.
That is how
I got the first one.
If you want to see
the second one, here we go.
Same process.
Break it up into chunks.
Let's move some things
out of the way.
All right,
let me get my chunks.
I have 30x to the 5th,
y cubed,
and I divide that
by 10x squared, y squared--
I've got the first bit--
plus 50x to the 4th,
y squared,
divided by 10x squared,
y squared.
Let's just break it down:
30 divided by 10 is 3.
Subtract my exponents
with my "x"s.
I'd get x to the 3rd.
Subtract my exponents
with my "y"s,
and I'd get y to the 1st,
which, remember,
we don't generally write.
I've handled that.
Plus 50 divided by 10--
that's 5.
Subtract the exponents,
so 4 minus 2, that's 2.
So, x squared--
2 minus 2, that's 0.
Remember, anything to the 0
power is 1, so essentially,
it would be 5x squared times 1,
which doesn't change anything,
so just disregard it.
That's how we got
the second one--
3x cubed, y plus 5x squared.
All right.
There's a little more to this.
There's also a bit of this
where you have to know
how to factor
to find how to divide
some polynomials.
To jog your memory, let me
bring you back to fractions.
When we simplify fractions,
basically what we're doing
is finding the greatest common
factor and canceling it out.
For example, 12 divided by 15--
I find factors of 12 and 15,
and I want there to be
a greatest common factor
between them.
I'm thinking about 12 and 15
and about the factors.
The pair that's going
to work for me--
I know 4 times 3 is 12,
and I know
that 5 times 3 is 15,
so 3 is the greatest
common factor here.
You can cancel out 3s,
which would leave you
with 4 over 5.
You could say 12/15
simplifies to 4/5
because you factored the 12,
you factored the 15,
and then you canceled out
the greatest common factor.
Keep that process in mind
because we're going to do
the exact same thing
but with polynomials.

(female describer)
Example 4: Find the quotient.
x squared minus 2x minus 3,
all over x minus 3.
We're going to factor
that numerator.
Our denominator already
has one factor,
and we're going to cancel
some things out
to get through this one.
Think about factoring
because we need to factor
x squared minus 2x minus 3.
I come down--let's go here
and do scratch work.
I'm trying to factor this.
So, if you remember,
you're looking
for a pair of factors--
First, factor the 3.
Then you want to find the pair
that the sum is negative 2.
Let's factor 3.
I'm ignoring
the sign to start with.
1 times 3--3 is prime,
so that's a short list--
1 times 3 gives me 3.
Now consider your signs.
You really need
a negative 3
and you need to combine these
to get a negative 2.
When I'm multiplying,
if my answer is negative,
I've multiplied a negative
number times a positive number.
I figure which one
is going to be negative
so that when I combine them,
I also get negative 2.
What if the 1 was negative?
If 1 was negative, negative 1
times 3 gives me negative 3,
but negative 1 plus 3
gives me positive 2,
and I need negative 2.
That's telling me, okay,
the 1 is not
what's negative here.
It'll have to be 3
that's negative,
because 1 times negative 3
will give you negative 3,
and 1 plus negative 3
is negative 2.
This trinomial will factor
to x plus 1, x minus 3.
Now I'm going to go back
to my problem,
rewriting that numerator
by its factors instead.
I replace x squared
minus 2x minus 3
with the factors,
x plus 1, x minus 3.
Let's do a little moving
around here, get some space.
I replace that numerator
with x plus 1, x minus 3.
I'll get that
scratch work out of our way
so things don't get
too cluttered up here.
I've replaced the numerator
and written it by its factors.
My denominator is still
my denominator--
still x minus 3.
Now I step back
and look at it.
Notice a factor
in your numerator
that's also
in your denominator?
The x minus 3.
Like the fractions, where we
crossed out those factors,
we're going to cross out
those common factors.
The answer is just this factor
I'm left with,
this x plus 1.
And you're all done.
That's what it was here.
With these problems,
the part that takes
a little work is just
factoring the numerator.
After you've factored your
numerator, it's smooth sailing.
Cross out the common factors.
Let's try the next one.
Our numerator--
we're finding a quotient.
We've got x squared
plus 7x plus 10.
We divide it by x plus 5.
Remember, start out
by factoring the numerator.
I'm going to scoot that a bit.
I come to the side;
do a little scratch work.
So x squared plus 7x plus 10.
From what I know,
I need to factor 10.
So, for 10--
I know 1 times 10 is 10,
2 times 5 gives me 10.
I look at the trinomial
and consider the signs.
I need a positive 10
for my product.
That means I'm multiplying
two positive numbers together
or two negative
numbers together.
Looking at my trinomial,
I need a sum of positive 7.
I could not add
negative numbers together
and get something positive,
so that means my factors here
are going to be positive.
Let's test it out.
I need to find out which pairs,
when added,
is going to give me positive 7.
I know 1 plus 10 is 11,
so it's not that pair.
2 plus 5 is definitely 7,
so this is the pair
that works for my factors.
I go to the problem
and rewrite that
x squared plus 7x plus 10
by its factors,
x plus 2 and x plus 5.
I write it here
so you understand
before I move over there.
I'm going to replace that
trinomial with its factors.
That'll help me simplify
this quotient.
All righty.
I do need to erase;
I'll run into that up here.
I'll leave that trinomial.
I leave those factors
and copy those.
I write x plus 2, x plus 5,
and then for my denominator,
I still just have
that x plus 5.
I have handled the numerator;
I've got the denominator.
I get this out of our way.
Remember what we did
in the last problem--
this is where you see
what you cancel out,
what's common to the numerator
and the denominator.
In this case, it's x plus 5,
so cancel those out,
and you're left with x plus 2,
and that is your answer.
All right?
Let's keep going.
Let's try another.
Find the quotient
of x squared minus 7x plus 12.
What do I do first?
Right, factor the 12.
I'll come off to the side,
okay?
I know 1 times 12 is 12.
So is 2 times 6,
and so is 3 times 4.
And that's it.
If I consider my signs here--
Let me rewrite the trinomial
so we look at that by itself.
Okay, so if I consider
my signs--
I need to multiply
and get positive 12.
That means I'm multiplying
two positive numbers together
or two negative
numbers together,
but because I need
a sum of negative 7,
that's telling me
it's the two negatives--
that's going to be the case.
Run down your list
and see which pair works.
If these were negative 1
and negative 12,
if I combine those,
I get negative 13.
That's not the pair.
If I had negative 2
and negative 6--
if I combined those,
I'd get negative 8.
So that's not the pair.
If I had negative 3
and negative 4,
I'd get negative 7,
so that's the right pair.
This trinomial
is going to factor
to x minus 3, x minus 4.
Going back to our problem,
'cause we finished
our side work,
we replace that numerator
and write it by its factors.
We write the x minus 3, the x
minus 4 for the numerator.
Let's scoot a few things
around here, get some space.
I scooted that off too far,
I think--there we go.
So, x minus 3, x minus 4.
All right?
And that's typically
when things get crowded,
so let me get rid
of some things.
All right.
Then I throw my denominator
in there, that x minus 4.
All right,
what do we do now?
We get rid of stuff.
We cancel that out, and you're
just left with x minus 3.
And you're all done.
All right, I think
it's about to be your turn.
Take a few minutes;
press pause.
Work through these problems.
When you're ready, press play.

(female describer)
Find each quotient.
1. x squared plus 8x minus 9
all over x minus 1.
2. x squared plus 5x minus 24
all over x plus 8.

(teacher)
All right, you ready?
Let's see how you did.
Let's get the pointer here
to move these.
For number one,
the answer was x plus 9.
That was your quotient.
For number two,
x minus 3.
If you want to see my work,
this is what I did:
the process we've been doing.
Start out by factoring
the numerator.
So I need factors of 9,
and I believe
I do not have my pen.
So I need factors of 9,
so 1 times 9 gives me 9
and so does 3 times 3.
Now I consider my signs.
I need negative 9,
so that means in each pair,
one number's negative,
but I need to be able
to combine those pairs
and get a positive 8.
Is there any way to combine
1 and 9 and get positive 8?
Mm-hmm.
If the 1 were negative.
Negative 1 plus 9
is positive 8.
That means that's
the right pair of factors.
Now I'm over here,
back in my problem.
I replace that trinomial
and write it by its factors:
x minus 1, x plus 9.
I'll get that scratch work
out of the way
so it doesn't confuse you.
You do not need to erase
your scratch work.
I do it to get a clear view
of what's going on.
We've got
the numerator factored.
Let me write the denominator
in here, x minus 1.
You're at the fun part--
cancel what's common:
x minus 1, x minus 1.
And that just left you
with x plus 9.
That's how I got
that first one.
I'll show you
how I got the second one.
Did the same thing.
Factor 24, because I'll ignore
the signs to start with.
24's got a longer list.
I know 1 times 24,
2 times 12,
3 times 8,
and 4 times 6.
That completes the list
for the factors of 24.
We need a pair of factors
that gives us negative 24.
Out of these pairs,
one of the numbers is negative,
but when we combine,
we get a positive 5.
Let's run down the list.
There's no way I could
combine 1 and 24,
whichever was negative,
and get a positive 5,
because if the 1 was negative
and I combined it with 24,
I'd get positive 23.
If the 24 was negative,
I'd get negative 23.
There's no way.
Okay, 2 and 12--
let's consider both cases.
If 2 were negative,
it'd be a positive 10.
If 12 were negative,
it'd be negative 10.
That's not the pair.
I've got a good feeling
about this pair.
If 3 were negative
combined with a positive 8,
I get positive 5;
I would get it.
Negative 3 plus 8
is positive 5.
This is the pair
that I want.
I replace that trinomial
in my numerator
and write it by its factors.
So, x minus 3 and x plus 8,
and then I still have
my denominator, x plus 8.
Okay?
I move my scratch work
so it doesn't get confusing.
Now you start
crossing stuff out.
So, that x plus 8,
that's common--
I can cancel that out.
And I'm left
with x minus 3.
And you're all done.
Okay? All right.
I hope you're feeling good
about dividing polynomials
and how to use
the laws of exponents
in certain problems and
factoring in other problems
to get through
how to divide polynomials.
Hope to see you here soon.
Bye.