Simplifying rational expressions: common monomial factors | High School Math | Khan Academy

So, I have a rational expression here, and what my goal is, is to simplify it. But while I simplify it, I want to make the simplified expression be algebraically equivalent. So if there are certain x values that would make this thing undefined, then I have to restrict my simplified expression by those x values.

You could pause this video and take a go at it, and then we'll do it together. All right, so let's just think real quick: what x values would make this expression undefined? Well, it's undefined if we try to divide by zero, so if ( x ) is 0, ( 14 * 0 = 0 ), it's going to be undefined. So we could say, we could say this: ( x ) cannot be equal to zero. For any other ( x ), we can evaluate this expression.

Now let's actually try to simplify it. When you look at the numerator and the denominator, every term we see is divisible by ( x ), and every term is divisible by 7. So it looks like we can factor ( 7x ) out of the numerator and ( 7x ) out of the denominator.

So the numerator we can rewrite as ( 7x ) times... if you factor ( 7x ) out of ( 14x^2 ), you're going to be left with ( 2x ), and then you factor ( 7x ) out of ( 7x ), you're going to be left with a 1. One way to think about it is we did the distributive property in reverse. If you do it again, ( 7x * 2x ) is ( 14x^2 ), ( 7x * 1 ) is ( 7x ).

All right, and now let's factor ( 7x ) out of the denominator. So ( 14x ) could be rewritten as ( 7x * 2 * 2 ), and remember I want to keep this algebraically equivalent. So I want to keep the constraint that ( x ) cannot be equal to zero.

So we divide the numerator and the denominator by ( 7x ). One way to think about it, you can divide ( 7x ) by ( 7x ) and just get 1, and we are left with ( \frac{2x + 1}{2} ). Now, if this was the original expression, ( x ) could take on any value, but if we wanted to be algebraically equivalent to our original expression, it has to have the same constraints on it.

So we're going to have to say ( x ) does not equal 0. This is a real, it's a very subtle but really important thing. For example, if you defined a function by this right over here, the domain of the function could not include zero.

So if you simplified how you defined that function to this, if you want that function to be the same, it needs to have the same domain. It has to be defined for the same input, and so that's why we're putting the exact same constraints for them to be equivalent. If you got rid of this constraint, these two would be equivalent everywhere except for ( x = 0 ). This one would have been defined for ( x = 0 ); this one would not have, and so they wouldn't be algebraically equivalent.

This makes them algebraically equivalent, and of course, you could write this in different ways. You could divide each of these terms by 2 if you like, so you could divide ( 2x ) by 2 and get ( x ), and then divide 1 by 2 and get ( \frac{1}{2} ). But once again we would want to keep ( x ) cannot be equal to zero.

Let's do another one of these.

So it's a slightly hairier expression, but let's do the same drill. Let's see if we could simplify it, but as we simplify it, we really want to be conscientious of restricting the ( z )s here so that we get an algebraically equivalent expression.

So let's think about where this is undefined. We could think about where it's undefined by factoring the denominator here. Let me just... so this is going to be equal to... actually let me just do the first step where I could see, well what's a common factor in the numerator and the denominator? Every term here is divided by ( z^2 ), and every term is also divided by 17, so it looks like ( 17z^2 ) can be factored out.

So ( 17z^2 ) can be factored out of the numerator, and then we would be left with... you factor out ( 17z^2 ) out of ( 17z^3 ), and you're going to be left with just ( z ). For ( 17z^2 ), you factor ( 17z^2 ) out of that, you're going to be left with just a 1.

Once again, you can distribute the ( 17z^2 ), multiply it times ( z ), you get ( 17z^3 ), ( 17z^2 * 1 ) is ( 17z^2 ). All right, all of that is going to be over... we want to factor out ( 17z^2 ) out of the denominator: ( 17z^2 ) times... and so ( 34z^3 ) divided by ( 17z^2 ) is ( 34 / 17 = 2 ) and ( z^3 / z^2 = z ), and then we have ( -51 / 17 = 3 ), and ( z^2 / z^2 = 1 ), so we'll just leave it like that.

And so over here, it becomes a little bit clearer of, well, how we're going to simplify it. We're just going to divide ( 17z^2 ) by ( 17z^2 ), but let's be careful to restrict the domain here. So we can tell if ( z ) is equal to zero, then ( 17z^2 ) is going to be equal to zero; it would make the denominator equal to zero, and we could see that even looking here.

So we could say ( z ) cannot be equal to zero. And what else? ( z ) cannot be equal to whatever makes ( 2z - 3 = 0 ). So let's think about what makes ( 2z - 3 ) equal to zero. You can add 3 to both sides and you would get ( 2z = 3 ). Divide both sides by two, and you would get ( z = \frac{3}{2} ).

So ( z ) cannot be equal to zero, and ( z ) cannot be equal to ( \frac{3}{2} ). So that's how we're going to restrict our domain, but now let's simplify.

So if we simplify it, these two cancel out, and we are going to be left with ( \frac{z + 1}{2z - 3} ) and we want to keep that constraint: ( z ) cannot be equal to zero. Actually, this second constraint is redundant because we still have the ( 2z - 3 ) here. If someone were to just look at this expression, like, well, the denominator can't be equal to zero, and so ( z ) cannot be equal to ( \frac{3}{2} ).

So this is still... if we just left it the way it is, we don't even have to rewrite this; that would be redundant since looking at this expression, it's clearly not defined for ( z = \frac{3}{2} ). So there you go. If someone asks you for what values is this expression not defined, well then, you could... you would also include that. Actually, let me just write it; it doesn't hurt to be redundant: ( z ) does not equal ( \frac{3}{2} ).

But this constraint right here is really important because it's not obvious by looking at this expression. This expression, by itself, would be defined for ( z = 0 ), but if we wanted to be algebraically equivalent to this one and that one, it has to be constrained in the same way.