- [Voiceover] So up here, we are multiplying
two rational expressions.
And here, we're dividing one rational expression
by another one.
Now what I encourage you do is pause this video
and think about what these become when you multiply them.
I don't know, maybe you simplify it a little bit,
and I also want you to think about what constraints
do you have to put on the x values
in order for your resulting expression
to be algebraically equivalent to your original expression.
So let's work it out together just so you realize
what I'm talking about.
So this is going to be, in our numerator,
we are going to get six x to the third power times two,
and our denominator, we're going to have five
times three x.
And we can see both the numerator and the denominator
are divisible by x, so let's divide the denominator by x.
We get one there.
Let's divide x to the third by x.
We get x squared.
And we can also see the both the numerator and denominator
are divisible by three, so divide six by three,
you get two.
Divide three by three, you get one.
And we are left with two x squared times two,
which is going to be four x squared
over five times one times one over five.
And we can also write that as 4/5 x squared.
Now someone just presented you on the street
with the expression 4/5 x squared and say,
for what x is this defined?
I could put any x here, x could be zero
because zero squared is zero times 4/5
is just going to be zero, so it does seems to be defined
for zero, and that is true.
But if someone says, how would I have to constrain this
in order for it to be algebraically equivalent
to this first expression?
Well then, you'd have to say, well, this first expression
is not defined for all x.
For example, if x were equal to zero, then you would be
dividing by zero right over here,
which would make this undefined.
So you could explicitly call it out, x can not be
equal to zero.
And so if you want this to be algebraically equivalent,
you would have to make that same condition,
x cannot be equal to zero.
Another way to think about it, if you had a function
defined this way, if you said, if you said f of x
is equal to six x to the third over five
times two over, times two over three x;
and if someone said, well what is f of zero,
you would say f of zero is undefined.
Undefined.
Why is that?
Because you put x equals zero there, you're going to get
two divided by zero and it's undefined.
But if you said, okay, well, can I simplify this
a little bit to get the exact same function?
Well, we're saying you can say f of x is equal to 4/5
times x squared.
But if you just left it at that, you would get f of zero
is equal to zero.
So now it would be defined at zero, but then this would
make it a different function.
These are two different functions the way they're written
right over here.
Instead, to make them, to make it clear
that this is equivalent to that one, you would have to say
x cannot be equal to zero.
Now these functions are equivalent because now,
if u said f of zero, you'd say all right,
x cannot be equal to zero, you know?
This would be the case if x is anything other than zero
and it's not defined for zero, and so you would say
f or zero is undefined.
So now, these two functions are equivalent,
or these two expressions are algebraically equivalent.
So thinking about that, let's tackle this division
situation here.
So immediately, when you look at this, you say, woah,
what are constrains here?
Well, x cannot be equal to zero because if x was a zero,
this second, this five x to the fourth over four
would be zero and you'd be dividing, you'd be dividing
by zero.
So we can explicitly call out that x cannot be equal
to zero.
And so if x cannot be equal to zero
in the original expression, if the result,
whatever we get for the resulting expression,
in order for it to be algebraically equivalent,
we have to give this same constraint.
So let's multiply this, or let's do the division.
So this is going to be the same thing as two x
to the fourth power over seven times the reciprocal,
times...
The reciprocal of this is going to be four over five
x to the fourth, which is going to be equal to
in the numerator, we're going to have eight x to the fourth.
So we're going to have eight x to the fourth,
four times two x to the fourth,
over seven times five x to the fourth is 35
x to the fourth.
And now, there's something.
We can do a little bit of simplification here,
both the numerator and the denominator are divisible
by x to the fourth, so let's divide by x to the fourth
and we get eight over 35.
So once again, you just look at eight 30,
Well, this is going to be defined for any x.
X isn't even involved in the expression.
But if we want this to be algebraically equivalent
to this first expression, then we have to make
the same constraint, x does not, cannot be equal to zero.
And to see, you know, this even seems a little bit
more nonsensical to say x cannot be equal to zero
for an expression that does not even involve x.
But one way to think about it is imagine a function
that was defined as g of x
is equal to, is equal to all of this business here.
Well, g of zero would be undefined.
But if you said g of x is equal to 8/35,
well now, g of zero would be defined as 8/35,
which would make it a different function.
So to make them algebraically equivalent,
you could say g of x is equal to
8/35,
as long as x does not equal zero.
And you could say it's undefined if you want.
Undefined for x equals zero.
Or you don't even have to include that second row,
and that will literally just make it undefined.
But now, this expression,
this algebraic expression, is equivalent
to our original one even though we had simplified it.