Hi, guys,
welcome to Algebra 1.
Today's topic
is going to focus
on adding and subtracting
polynomials.
All that you know about
simplifying expressions
is really going to
take you far on this topic.
You ready to get started?
Let's go.
Let me take you back a bit
to the distributive property.
You're going to apply it
when you're working
with polynomials,
when you add
and subtract them,
you apply this a bit.

(female describer)
2 times 4X plus one.

(instructor)
Just to review, you remember
the distributive property.
With that 2 on the left,
I can distribute it throughout
what's in my parentheses.
Somebody probably
showed you these little arrows
to show you
what you were doing.
So I can say,
I need to multiply 2 times 4X,
which is 8X,
and I need to multiply
2 times positive 1,
which is 2.
You could say, "2 times
that quantity, 4X plus 1."
After I apply
the distributive property,
I could write that
as 8X plus 2.
Right?
Let's apply that property
a few more times,
get it in our system.
This expression: 7 plus 3
times the quantity X minus 2.
For this one,
I'm going to work down a bit.
That might help this make
a little more sense to you.
Let's move that over slightly.
Okay.
So the 7 is just going along
for the ride for a bit.
Let me erase that dot so you
don't think it's subtraction.
7 is just going along
for the ride for a second,
so 7 plus...
I do need to apply my
distributive property here,
for the second bit.
So 3 times X, that's 3X.
And 3 times negative 2--
because that's
how I interpret that,
3 times negative 2--
that's negative 6.
So you'd write it
as a minus 6.
Then you look back
at your problem
and you can combine
some like terms here.
You do have your 3X,
which has nothing else
like it in that expression,
so there's nothing
to combine that with.
So it's now
just coming on down here.
I have two constant terms
that I can combine,
to write it more simply.
So 7 minus 6, okay--
so 7 minus 6, that's 1.
So that means I can
write that as a plus 1.
So you apply
the distributive property
and you simplify that original
expression to 3X plus 1.
Okay? Let's keep going.
This one,
let's do the same thing.
Let's move that over
a little bit.
So I have 5 minus
that quantity (3X plus 4).
That 5 is just going
to go along for the ride
for a minute.
When you see something
like this, I hope you remember,
you have this minus and then
this 3X plus 4 in parentheses.
Treat that as if there is a 1
right in front of it.
What's happening is you're
distributing a negative 1
throughout what's in
your parentheses.
Okay?
Negative 1 times 3X,
that's a negative 3X,
and then negative 1
times 4,
that's a negative 4.
Now I'm at the point where
I have like terms to combine.
This negative 3X
doesn't have anything like it.
It's the only term in
this expression with an X,
so I know that this
is negative 3X.
That's just going
to bring on down.
I do have this 5 minus 4.
So 5 minus 4,
I know that's 1,
and it's a positive 1.
So I can write that
as plus 1.
That original expression,
it simplified
to negative 3X plus 1.
Jogged your memory a bit
with the distributive property?
Let's get
into the polynomials.
There are three steps
to follow
when you're adding
and subtracting polynomials.
You first want
to rewrite the problem,
then you want to combine
like terms,
then you want to make sure
your final answer
is in standard form.
Remember, standard form,
that's when you have
your exponents
where they go
in descending order.
So the leading term
has the largest exponent
and then you just go
in descending order from there.
Your final answer always
needs to be in that form.
So while we work
with these problems,
keep these three steps
in mind.
We'll rewrite it,
combine like terms,
then we'll make sure
we're in standard form.
Let's look at the first one.
Deep breath,
because I know it looks massive.
There's a lot going on,
but it's really going to break
right on down to something
that looks a lot more simple.
Our first polynomial--
it's actually a trinomial,
it's got three terms.
So 7X cubed plus 6X squared
plus 2x,
and we're going to add that
to 3X cubed minus 4X squared
plus 7X minus 5.
We're going to start
by rewriting the problem.
When you're adding polynomials,
rewriting the problem
is essentially just
getting rid
of the parentheses.
When you're adding,
the rewriting
is just break those terms
out of the parentheses.
I have 7X cubed
plus 6X squared plus 2X,
plus 3X cubed
minus 4X squared
plus 7X minus 5.
I just broke that problem
out of the parentheses.
Now what I'm going to do,
just so this doesn't
get too crowded here,
is I'm going
to get rid of that problem.
Now we've broken it
out of the parentheses,
so let's just look at it
by itself.
I think it's going to help us
look at this a little better.
Let's click off of it
so I can write.
Here we go.
Now we're at step two,
where we need to combine
like terms.
How I find out
which terms are alike
is the terms have to have
the same variable
raised to the same power.
The coefficient doesn't
need to be the same,
just the same variable
raised to the same power.
Those are our like terms.
And since we know
that we need our answer
to be in standard form,
let's start out
with the largest exponent,
so that by the end,
we've got it written
in the right form.
Scan your terms
and you notice
that you have a 7X cubed
and you have a 3X cubed.
Let's take
those two terms first.
I've got the 7X cubed
and I've got this 3X cubed.
You notice how I took
the sign out front,
because it tells me what--
the sign of that term.
I'm going to come off and do
some scratch work over here.
So 7X cubed plus 3X cubed.
Now here, we're just adding
like terms.
We're not multiplying
anything together,
so we're not going to change
the value of our exponents
because we're not finding
any products here;
we're combining like terms.
So 7X cubed plus 3X cubed,
that's just 10X cubed.
Essentially, all you're doing
is adding your coefficients
and just carrying along
that variable to that power.
Okay?
I know we've got the first set
handled, that 10X cubed,
so that's out of the way.
Go in descending order,
so now let's look
for exponents of two.
And I notice
I have this 6X squared
and I have
this negative 4X squared.
All right?
Let's combine
those like terms.
6X squared--
and I'm combining it
with a negative 4X squared,
so I'm just going
to treat that like subtraction.
So 6X squared
minus 4X squared.
Remember, you're not messing
with your exponents at all,
you're really just focused
on the coefficients.
So 6X squared
minus 4X squared--
well, 6 minus 4,
that's just 2.
So this is 2X squared.
Okay?
Now you've handled this.
So you've handled
the exponent of 3,
you've handled
the exponent of 2,
so let's look
for exponents of 1,
which you really don't see.
I'm going to start
squaring things now,
not as exponents,
but just putting them in boxes
to help me focus.
I have a 2X squared
and I have a positive 7--
I'm sorry, I have a 2X
and I have a positive 7X,
so I'm going to combine
those like terms together.
So 2X plus 7X.
So, essentially, I'm just
combining like terms here.
So 2 plus 7, that's 9,
so this is 9X.
All right?
I'll combine those:
2X plus 7X, that's 9X.
That's taken care of.
All that's left is that
lonely little negative 5.
That's our constant term;
I'm just going
to write it down here.
Now we're at the point
where we can write our answer
and just make sure
it's in standard form.
I'm going to bring all these
individual answers together.
I'm going to write this as...
Let's see the next thing to
erase, make some room.
Let's get rid of our problem
because now
we have the answer.
And I'll write it
right up here.
Let's get that
out of the way.
When we added
those polynomials together,
it all really simplifies
to 10X cubed...
plus 2X squared...
plus 9X
minus 5.
And that is our answer.
That is what we get
when we add those two
polynomials together.
And you can double check
and we do have it
in standard form.
Our exponents are going
from largest
and then in descending order.
Okay? You ready to try
another one?
Let's do it.
All right.
"10A cubed minus 3A squared
plus 4,"
and we're going to add that
to "negative 6A cubed
plus A squared plus 1."
Let's bring that up
a little bit.
You remember, step one
is rewrite the problem.
And when we're adding,
essentially that's just
breaking the terms
out of the parentheses.
So 10A cubed...
minus 3A squared plus 4.
So now look here.
You see we have plus
this negative 6A cubed?
We don't like
to have the signs
bump into each other
like that.
Instead of writing
"plus negative 6A cubed,"
I'm just going to write
"minus 6A cubed."
Didn't change
the meaning at all
because adding a negative
means the same thing
as subtraction.
Remember
that from Pre-Algebra.
I can just rewrite
plus negative 6A cubed
as just minus 6A cubed.
Then just break the other terms
out of parentheses:
A squared plus 1.
Okay?
Now we're at the point
where we can combine
like terms,
so I'm going to give us
a little more room up here.
Let's get rid of our problem
and let's bring this up.
We can get some space to work.
Now look for your like terms
you can combine.
Essentially, you're looking
for the same variable
raised to the same power,
to identify your like terms.
I see I have this 10A cubed
and I have
this negative 6A cubed,
so I can combine this one
and this one.
I'll come off
and do a little scratch work.
So 10A cubed minus 6A cubed.
You're just subtracting your
coefficients here.
So 10A cubed minus 6A cubed,
that's 4A cubed,
because 10 minus 6 is 4,
right?
All right,
you handled that.
Looking for your exponents
in descending order,
so you scan your terms
and you see
you have this
negative 3A squared
and this positive A squared,
so those are your next two
you want to combine.
So negative 3A squared
plus A squared--
you remember, when you don't
see a coefficient,
there's an invisible 1.
So negative 3A squared
plus 1A squared,
that's just
negative 2A squared.
So you've handled
those square terms.
Now I see I have
two constants left.
I have this positive 4
and I have this positive 1.
So 4 plus 1,
that's 5.
And you've combined
all the terms you can combine,
so you're ready
to write your answer.
I have some space,
so I'll write it here.
We're just bringing all those
individual answers together,
since we kind of worked out
our problem in chunks.
So 4A cubed
minus 2A squared
plus 5,
because that 5 down there
is positive.
And you are all done because
you have it in standard form.
All right?
Starting to feel
a little better about it?
Let's try another one.
It usually takes three
to feel like,
"Okay, I think I got it,
I think I'm ready to try one."
Here we have 8B to the 4th minus
5B squared plus B,
and we're adding that
to 3B squared plus 5B.
You remember step one?
Rewrite the problem.
And when we're adding,
that's essentially just
breaking it out of parentheses.
So 8B to the 4th
minus 5B squared
plus B.
We're going to add that
to 3B squared plus 5B.
All right.
Now we're at step two,
where we can
combine like terms.
This is where I get us
some more space to work.
Let's get that
out of the way
and bring that up a bit.
Now let's get to work.
I scan the problem
and I know
I want my answer
in standard form,
so I'm always going to look
for the largest
exponent first,
to start identifying
where my like terms are.
I see this first term
has an exponent of 4.
It's the largest one,
but I notice
there are no other terms
that have an exponent of 4.
So there's nothing
to combine that term with;
it just keeps going along
for the ride.
8B to the 4th just continued
on being 8B to the 4th.
It didn't combine
with anything.
Keep it in mind
that we're trying
to go in descending order
with our exponents.
We look for powers of 3.
We don't see
any powers of 3.
We look for powers of 2,
and I do see a couple terms
that have a power of 2.
I see this negative 5B squared
and I see
this positive 3B squared.
I can combine those.
Negative 5B squared
plus 3B squared.
Remember, essentially we're
focusing on the coefficients,
so negative 5 plus 3,
that's negative 2.
So this is
negative 2B squared.
All right.
We've handled that.
And I see
all we're left with here
is two terms
that both have the variable B,
and those are our last pair
of like terms to combine.
We have this b plus 5B.
So B plus 5B.
Remember, when you don't
see a coefficient,
there's an invisible 1.
So 1B plus 5B,
that is 6B.
Look back at the polynomial,
there's nothing else
to combine.
Now we're at the part
where we can write our answer.
I'll write it up at the top
and get that out of our way.
Then we can
get the answer up here.
All righty.
Let's bring our individual
answers all together,
because we solved
our problem in chunks.
So 8B to the 4th
minus 2B squared
plus 6B.
It's in standard form
because the exponents
are going in descending order.
All done.
All right?
You ready to try one?
Okay, how about two?
Go ahead and press pause,
take a few minutes.
Take your time and work through
these problems.
Take it in chunks,
just like we did.
It'll be easier
to manage that way.
When you're ready to check
your answers against mine,
press play.

(female describer)
Simplify.
Number 1: Open parentheses,
11Y squared plus 3Y minus 2,
close parentheses,
plus, open parentheses,
4Y squared plus Y plus 5.
Number 2: Open parentheses,
negative 5X cubed plus 7X
minus 1, close parentheses,
plus, open parentheses,
8X cubed plus 3X squared
plus 2X plus 2.
All right,
you ready to check?
Let's check these.
That first one:
11Y squared plus 3Y minus 2,
plus 4Y squared
plus Y plus 5.
The answer there,
you should've got 15Y squared
plus 4Y minus 3.
On the second one,
we had negative 5X cubed
plus 7X minus 1,
and we were adding that
to 8X cubed
plus 3X squared
plus 2X plus 2.
And the answer,
you should've got
3X cubed plus 3X squared
plus 9X plus 1.
If you want to see
how I did that,
keep on with me.
Remember step one,
when you're adding,
essentially you're just
breaking those terms
out of the parentheses,
is the first step.
So 11Y squared
plus 3Y minus 2,
plus 4Y squared
plus Y plus 5.
Now that I've got it
broken out of parentheses,
give us some room to work.
I can get rid of that
and move this up a bit.
All right.
Now you're at the point where
you need to scan your terms
because you've got to find
the like terms,
you've got to combine.
We know we want our answer
to be in standard form,
look for the terms that have
the largest exponent first.
I see my largest exponent
in this polynomial is 2,
and the terms
that have that exponent
are 11Y squared, and then
this positive 4Y squared,
so I do some scratch work.
11Y squared plus 4Y squared,
that's 15Y squared.
All right?
So we've handled this.
Now I see I have 3Y plus 1Y.
Right?
So 3Y plus 1Y,
because there's an invisible 1
that I don't see.
3Y plus 1Y, that's 4Y.
So I've handled
those terms.
And I have
two constant terms left.
I have this negative 2
and this positive 5.
So negative 2 plus 5,
that's a positive 3.
Those are my last two terms
that I needed to combine.
Now you can write the answer.
Since we worked
the problem in chunks,
we'll take all our chunks
and put them together.
So 15Y squared
plus 4Y plus 3.
That's how I got
that first one.
Need to see the next one?
Okay.
Going to bring that up
a bit.
Here we go.
Negative 5X cubed
plus 7X minus 1,
plus 8X cubed plus 3X squared
plus 2X plus 2.
So remember for adding,
your first step
is to break it
out of parentheses.
So negative 5X cubed
plus 7X
minus 1,
plus 8X cubed
plus 3X squared
plus 2X plus 2.
So I've broken it
out of parentheses,
and just to help us focus,
I'm going to get rid of that
original problem
and just zone in on this one.
Now I'm looking for
my like terms to combine,
and because I know I want
my answer in standard form,
I'll start with
my largest exponents first.
I see that I have
negative 5X cubed
and this positive 8X cubed.
So negative 5X cubed
plus 8X cubed,
that's 3X cubed.
So you've handled those terms.
Now I'm going
in descending order,
so I'm looking
for X squared;
there's only one little term
that has an X squared.
So it just comes along
for the ride,
there's nothing
to combine it with.
So I know 3X squared
is just going to be around.
Okay?
Now I see I have 7X
plus 2X.
Right? So 7X--
oop, funky-looking X there--
plus 2X...
that equals 9X.
All right?
So I handled that.
And then I had
two constant terms left.
I have this negative 1
combined with this positive 2.
So negative 1 plus 2,
that's a positive 1.
Now we have all
our individual chunks
and we just need to bring it
together for our answer.
I think I left enough space
to write it here.
So 3X cubed
plus 3X squared...
plus 9X plus 1.
And that's how we got
that answer.
All right? Okay.
So you've handled addition;
now let's go to subtraction.
Think back to one
of those early problems
that we did in this lesson,
where we handled this problem
with
the distributive property.
So to jog your memory,
remember what we did
on this one.

(describer)
5 minus, open parentheses,
3X plus 4,
close parentheses.

(instructor)
We said we have
a five out front;
it just stayed along
for the ride.
And we had to subtract
that quantity
in the parentheses,
so we said it's minus this.
There's a 1 here
I don't see,
so essentially
I'm distributing
this negative 1 throughout
this quantity in here.
So negative 1 times 3X,
that's minus 3X.
Negative 1 times 4,
that was a negative 4.
We said, "Okay, now we need
to combine like terms."
The negative 3X
didn't have anything alike,
so it just came along
for the ride--
that term right there,
we've handled it.
Then we had this 5 minus 4,
and 5 minus 4 is 1.
So plus 1.
So remember how we did this.
The part that I want you
to focus in on this,
is that it was
a subtraction type problem.
It was 5 minus that 3X plus 4,
and I want you
to pay attention
to what we had to do with
distributing that negative 1.
We use that same process
when we subtract polynomials.
Look at this one.
We have 8X squared plus 5A,
minus 4A squared
plus 2A minus 6.
Let's give ourselves
a little room;
I'll bring that up a bit.
You remember when
we're dealing with polynomials
and we're adding
and subtracting them,
the first step
is to rewrite the problem.
Now, when we're adding,
rewriting is essentially
just breaking those terms
out of the parentheses,
but when we're subtracting,
it's not just that.
We have to distribute this
negative 1,
that's kind of invisible,
that we're kind of not seeing.
That's the extra step
that's involved
when you're subtracting
polynomials.
I look at that first term,
or that first binomial--
that I'm going to break out
of the parentheses;
nothing's going on
with that one.
As far as this trinomial
is concerned,
this second batch,
there is a minus 1 here
that I don't see;
it's kind of invisible.
So I need to distribute
this negative 1
to each of these terms
before I can go any further
with this problem.
So negative 1
times 4A squared,
that is--
let's show the work,
just so it makes sense.
So negative 1
times 4A squared,
that's negative 4A squared.
That first term
is minus 4A squared.
Now we need to multiply that
negative 1 times the 2A,
so negative 1 times 2A--
that's negative 2A,
so minus 2A.
Now I need to distribute that
negative 1 to that negative 6,
so negative 1
times negative 6,
that's a positive 6,
so plus 6.
Now that I've rewritten
the problem--
and because
it was subtraction,
it involved me distributing
that negative 1 throughout--
now I can continue on
to simplify this expression.
Now it's pretty much the same
as when we were adding.
I think I left myself
enough room to work here.
I see I have this
8A squared minus 4A squared,
all righty.
So let me--
I know what I can do.
I will get rid of this bit
and show that work
right there.
So 8A squared
minus 4A squared,
that is 4A squared.
Okay.
So I've handled this.
Now I have 5A--
and it's positive--
minus 2A, right?
So 5A minus 2A,
that's 3A.
You've handled those terms.
All you have left now
is that constant term, that 6.
It just goes along
for the ride; you've got a 6.
Now we're at the point,
because we solved
our problem in chunks,
we'll bring
our chunks together
and then we'll have
our answer.
I would have 4A squared...
plus 3A plus 6.
And you are all done.
Subtraction isn't too much
unlike addition;
you just have that extra step
where you have to distribute
that negative 1
throughout that second
set of terms.
Let's try another one.
Okay, 7Y to the 4th
plus 3Y squared minus 1,
minus 6Y to the 4th
plus 4Y squared
minus 2Y plus 1.
Step one:
Rewrite the problem.
And because
we are subtracting,
remember to use
the distributive property
for the second set.
For the first set, just break it
out of parentheses.
So 7Y to the 4th
plus 3Y squared minus 1.
All right.
Now that I'm at my second set,
remember there is
an invisible 1 here.
There is a negative 1,
because you're subtracting,
that you have to distribute
throughout those four terms.
You have negative 1
times 6Y to the 4th.
I'll come over here,
so negative 1
times 6Y to the 4th,
that's negative 6Y
to the 4th.
So it's minus 6Y to the 4th.
And I have negative 1
times 4Y squared,
so negative 1
times 4Y squared,
that's negative 4Y squared,
so minus 4Y squared.
And I can see things are about
to run into each other here,
so let's just bring
that down a little bit.
Okay.
Then I need to distribute
this negative 1
to that negative 2Y.
So negative 1
times negative 2Y,
that's a positive 2Y.
All right, so plus 2Y.
Then I have this negative 1
times positive 1, all right?
So negative 1 times positive 1,
that's negative 1.
I have a minus 1 right here,
all right?
Now we're at the point where
we can combine like terms.
I'm going to get that last
bit of work out of our way
because we're going to need
some extra space
to get a little
scratch work in.
All right,
so let's see.
I want to go in
descending order of exponents,
so which exponent
is my largest in this case?
I'm scanning the terms;
it's 4.
I have a 7Y to the 4th
and I have a negative 6Y
to the 4th.
So 7Y to the 4th
minus 6Y to the 4th...
that is 1Y to the 4th.
And I don't
have to write that 1.
So I've handled that.
I don't have any X cubed--
I'm sorry, Y cubed terms,
but I do have Y squared terms.
I see I have a 3Y squared
right here,
and I have
a negative 4Y squared there,
so set's combine those:
3Y squared
minus 4Y squared.
So 3Y squared
minus 4Y squared,
that's negative 1Y squared.
All right?
So now I've handled those.
And I see I do have a 2Y term,
so that just is going along
for the ride.
I can squeeze it in
right here.
2Y is just carrying along.
And I do have a couple
of constant terms here.
I see I have this negative 1
and another negative 1.
Let's get some things
out of our way here.
This problem got large.
Let's get this
out of the way.
And let's do a little erasing
and let's bring
some stuff up a bit.
Let's bring that up
and let's bring the work up.
Now we've got some more room
for scratch work.
All righty, so we were doing
negative 1 minus 1.
So negative 1 minus 1,
that's negative 2.
Now we have combined
all the terms that we can,
so we're at the point where
we can write our final answer.
Y to the 4th
minus Y squared
plus 2Y minus 2.
It's in standard form;
we are all done.
All right?
I think you've got
one more in you.
Let's try another one
before you try them on your own.
Negative 5X cubed
minus X squared plus 3X,
minus negative 4X cubed
plus 4X squared.
All right.
Let's bring that up a bit.
And let's get to work.
So that first polynomial,
essentially I'm breaking it
out of parentheses.
There's no multiplication
going on out here
that's telling me
I need to do anything else.
Negative 5X cubed
minus X squared
plus 3X, okay.
Now I'm subtracting,
so it's like I'm distributing
a negative 1
throughout these terms.
So I need negative 1
times negative 4X cubed--
negative 1 times
negative 4X cubed.
That's a positive 4X cubed,
all righty.
So plus 4X cubed.
Then I have this negative 1
times 4X squared.
So negative 1
times 4X squared,
that is negative 4X squared,
all righty?
So minus 4X squared.
We've gotten everything
out of the parentheses,
we did our distribution,
so now we're at the point
pf combining like terms.
Let's get this old work
out of the way,
make room for some
new scratch work.
I know I want to go in
descending order of exponents.
My largest exponent is 3
in this problem.
So I have a negative 5X cubed
plus a 4X cubed.
So negative 5X cubed
plus 4X cubed,
that's negative 1X cubed,
right?
Okay. So I've handled that.
And I noticed I have some
X squared terms.
So negative X squared
minus 4X squared.
So negative X squared minus
4X squared,
that's a negative 5X squared.
You've handled
the X squared terms,
then all that's left
is that 3X.
There's nothing
to combine with it,
so it's just going along
for the ride.
Now we've handled
everything we can
as far as
combining like terms,
so we're at the point
of writing our final answer.
We would have
negative X cubed
minus 5X squared plus 3X.
Got it in standard form,
so we are all done.
Okay?
All right,
why don't you try a couple?
Go ahead and press pause
and take a few minutes,
then really take your time
and work your way
through these problems.
To compare your answers
with me, press play.

(describer)
Simplify.
Number 1: Open parentheses,
negative 6C to the 5th
plus 5C cubed plus C,
close parentheses,
minus, open parentheses,
4C to the 5th
plus C cubed minus 8C.
Number 2:
Open parentheses, 5X cubed
plus 6X squared minus 1,
close parentheses,
minus open parentheses,
negative 2X cubed
plus X squared plus 2.
Ready to check?
Let's go.
For that first one,
negative 6C to the 5th
plus 5C cubed plus C,
minus 4C to the 5th
plus C cubed minus 8C.
That one was negative 10C
to the 5th
plus 4C cubed plus 9C.
That should've been your answer
for the first one.
5X cubed plus 6X squared
minus 1,
minus negative 2X cubed
plus X squared plus 2.
That one,
7X cubed plus 5X squared
minus 3.
If you want to see
how I did those...
For the first trinomial,
essentially we just need
to get it
out of the parentheses
because there's
no multiplication
of anything here
that I need to handle.
So I would have
negative 6C to the 5th
plus 5C cubed
plus C, okay?
And because
I'm subtracting,
there's a negative 1 out here
that I need to distribute
throughout those terms.
I need to do negative 1
times 4C to the 5th,
negative 1
times C to the 3rd,
negative 1 times negative 8C.
I'll come off to the side
and do that scratch work.
So negative 1
times 4C to the 5th,
that's negative 4C
to the 5th, right?
So minus 4C to the 5th.
Then I have negative 1
times C to the 3rd.
So negative 1
times C to the 3rd.
That's negative C to the 3rd,
right?
So minus C to the 3rd.
Then I have negative 1
times negative 8C.
So negative 1
times negative 8C,
and that's a positive 8C.
That means I have
a plus 8C.
If you notice,
when you're subtracting,
essentially those terms
right there,
where I'm distributing
that negative 1 throughout,
their signs
just become the opposite
of whatever they were
to start with.
If you notice that pattern,
that when you distribute
the negative 1,
it changes all the signs
in that second group,
you can jump to that step;
that's fine.
Algebra is all about patterns
that lead to shortcuts.
If you continue
and do that distribution
and really show the steps,
that's fine too,
because you'll get
really comfortable
with what we're doing here,
with this process.
Keep that in mind
and let's keep going.
We're at the point
of combining like terms,
so let me get us
more room here to work.
Let's get that out of the way.
Let's erase this right here,
let's get rid of that.
Let's move this up.
Now we've got some room
to work here.
So I know I want my answer
in standard form,
so scan your terms--
you're looking for
the largest exponent first.
And I see in this problem,
my largest exponent's a 5.
I'm going to combine
negative 6C to the 5th
with negative 4C to the 5th,
right?
So negative 6C to the 5th
minus 4C to the 5th,
that's negative 10C
to the 5th.
Right?
Okay,
so I've handled those.
I don't have
any C to the 4th terms,
but I do have
some C to the 3rd terms.
So 5C to the 3rd,
and I see I have
this minus C to the 3rd.
So 5C cubed minus C cubed,
that's 4C cubed.
Right? Okay.
So we handled
the C cubed terms.
I see I have two terms left
and they're alike--
they're both C.
Or they both have
a variable for C.
I have this plus C
and I have this plus 8C--
they're both positive--
so C plus 8C, that's 9C.
All right?
And I've handled all the terms
in that polynomial.
Now I can write my answer
in standard form.
All righty, so I'd have
negative 10C to the 5th
plus 4C to the 3rd plus 9C.
And you are all done
with that one.
Okay?
Let's look at how
I did that next one.
All right.
For this one,
let's move it up a bit.
Okay.
What you want to do first,
check out
that first polynomial.
There's no multiplication
or anything going on out front,
so just break it
out of parentheses.
5X cubed
plus 6X squared
minus 1.
Now I'm subtracting,
so remember you're
distributing this negative 1
throughout these terms.
I need to multiply
each of those terms
by negative 1.
I'll come off to the side
and do a little scratch work.
For the first term, negative 1
times negative 2X cubed.
So negative 1
times negative 2X cubed,
that's a positive 2X cubed.
Okay, so plus 2X cubed.
Now I have this negative 1
times X squared.
So negative 1 times X squared,
that's a negative X squared.
All righty.
So minus X squared
for that bit.
Then I have a negative
1 times this positive 2.
So negative 1 times 2,
and that's negative 2.
So, come up here,
I've got my minus 2.
Then we said
you might have noticed
it just basically changes all
the signs in that second bit
when you distribute
that negative 1.
Maybe you took that shortcut;
maybe you did all the work.
Either way is fine.
Now we've reached the point
of combining like terms,
so let's give ourselves
a little more room
to work up here.
Let's get rid of that,
and let's get rid of this,
and get that out of our way...
and move this up.
Okay, let's start combining.
I know I want my answer
in standard form,
so I'm going to look
for the largest exponent first.
I see I have 5X cubed,
2X cubed,
so I can combine those.
So 5X cubed
plus 2X cubed,
that's 7X cubed.
You've handled
those two terms.
Now you're scanning again.
You see you have
some X squared terms.
I have 6X squared
and minus X squared.
So 6X squared
minus X squared,
that's 5X squared.
You've handled
the X squared terms.
Then you look again
and you have two constant terms
left to combine.
So I have this negative 1
and that negative 2.
Negative 1 minus 2,
that's negative 3.
Now we can put each of
our individual chunks together
and write our final answer
in standard form.
So 7X cubed
plus 5X squared
minus 3,
and you are all done.
All right?
Okay, well,
I hope you really got a grip
on how to add
and subtract polynomials,
and I hope you saw
how the distributive property
really helps you with those
subtraction problems.
See you soon
for more Algebra 1.
Bye, guys.