Adding Rational Expressions Example 2

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

Adding Rational Expressions Example 2

相關課程

0 / 750

Let's add some more rational expressions.
So let's say I have 1 over x squared minus 9.
And to that I'm going to add 2 over x squared
plus 5x, plus 6.
So just like we did when we first learned to add
fractions, or to add rational numbers, we always had to come
up with a common denominator.
So once again, we're going to have to come up with a common
denominator.
And what we really want to do is find the least common
multiple of these two expressions.
And when I say that, let me just remind you.
Let's say we wanted to add 1/4 plus 1/6.
Well, we know that 1/4 is 1 over 2 times 2, and that 1/6
is 1 over 2 times 3.
So to be able to add these two things, we incrementally-- we
both have a common 2.
We both have this common 2.
But in order for this guy to go into whatever our common
denominator is, we're going to have to be
divisible by another 2.
So let me do that.
So our common denominator's going to have to be-- we have
this 2 in each of them.
So that takes care of that first 2.
But then we need the second 2.
We need the second 2 there.
2 times a second 2.
And then we need this 3 here.
We need this 3 here.
We have to have all of the prime factors of both of them.
This has two 2's, so this guy has to have two 2's.
This guy has a 2 and a 3, so this guy has to
have a 2 and a 3.
And so our least common multiple here is 2 times 2,
times 3, which is 12.
And then we can add accordingly.
That is our common denominator.
That's the least common multiple.
You could have just multiplied these two out, but you would
have gotten 24, and that's larger than necessary.
You could have still worked it out, but you would have had
more simplifying left for you to do.
The same idea we can do right here.
The same idea we can do here.
We can factor both of these denominators, so x squared,
that is x plus 3, times x minus 3.
It's just a difference of squares.
And x squared plus 5x, plus 6.
That is what?
That is x plus 2, times x plus 3.
2 times 3 is 6.
2 plus 3 is 5.
So our common denominator's going to be what?
What's going to be our common denominator?
Let me clear up this real estate over here.
What's our common denominator going to be?
Our common denominator-- so this is going to be equal to--
our common denominator's going to be-- our common
denominator's going to have to have an x plus 3 in it.
And that takes care of this x plus 3 as well.
It's going to have an x minus 3 in it.
It's going to have to have an x minus 3 in it.
So that's an x minus 3.
And it's going to have that x plus 2 in it.
Another way you could have thought about is, like, look,
it's going to have to have all of the factors of this guy,
which are x plus 3, times x minus 3. x plus 3
times x minus 3.
And then once you write those down, you say, well, it also
has to have all of the factors for this guy.
Well, we already have an x plus 3 there, so we just have
to only add on the x plus 2.
And that saves us from multiplying both of these
expressions together.
This is the least common denominator.
Now, if this is the least common denominator, what do we
have to change the 1 by?
Well, let me actually rewrite this a little bit.
Well, I'll write it like this.
So what do you have to do to go from-- if I start with 1
over x squared-- let me write it this way-- 1 over x plus 3,
times x minus 3.
This and this is the exact same thing.
If I want an x plus 2 in the denominator, what
do I have to do?
I have to multiply the numerator and the denominator
by x plus 2.
So this expression right here, which is this expression right
here, will become x plus 2 over x plus 3, times x minus
3, times x plus 2.
Those would cancel out and you're just left with a 1.
That's that right there.
So this is the first term.
That is this term.
This term over here, the second term over here, this is
2 over x plus 2, times x plus 3.
That's in its original form.
Now, if we also want an x minus 3 here in the
denominator, we can multiply it by x minus 3
over x minus 3.
Or essentially, we can multiply the
numerator by x minus 3.
So this is going to be equal to 2 times x minus 3 is 2x
minus 6 over x plus 2, times x plus 3, times x minus 3.
So this expression right here is the exact same thing as
this expression right over here.
So if we were to add these two expressions, they have the
same denominator.
I just switched the order that I'm multiplying.
But they're the exact same denominator, which is exactly
the same thing as this.
So this, this first expression, we're going to get
our x plus 2 here.
x plus 2 over all of this would cancel, and you'd get 1
over x plus 3, times x minus 3, over 1 over x
squared minus 9.
And then to that you're going to add plus 2x minus 6.
And why am I adding 2x minus 6?
Because the second term is 2x minus 6 over all of this
business, because we were just multiplying.
This would also be written as 2 times x minus 3, over all of
this. x minus 3's would cancel out and you'd get this again.
So this is what we get when we add the 2, and so this will be
equal to-- let me scroll down a little bit-- this will be
equal to x plus 2x is 3x.
2 minus 6 is negative 4.
All of that over this thing.
We can either keep it multiplied out, or factor it
out, or we could multiply it if you like, but it'll take
some time to multiply it.
So I'll just say x plus 3, times x minus 3,
times x plus 2.
And we are done.

留言:

載入中...

新增一則留言(尚未登入)

adding rational expressions example 2 討論區:

載入中...