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# Multiplying rational expressions

Sal multiplies and simplifies (a²-4)/(a²-1) X (a+1)/(a+2). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• is there a simpler way to solve this?
(35 votes)
• You could skip the rearranging he does at . He apparently did that so that the cancellation step would be easier to see, but that step certainly wasn't necessary.
(38 votes)
• in you can see that in the numerator that there is a (a - 2), woudnt you have to write a cannot be equal to 2 aswell? does the "not equal part" only occur for the denominator?
(4 votes)
• If you put 0 divided by any number in a calculator, you will get an answer. if you but any number divided by 0 you get an error message. I think this is because theoretically, if you have 10 pieces of candy, you could choose divide it into 0 groups (meaning you didn't sort it into groups) whereas you can't do any dividing with nothing!
(2 votes)
• why doesnt he simplify it further? my answer was 2/1 because a-2/a-1, the a's will cancel out to 1? therefore it will be -2/-1 = 2/1
(4 votes)
• The response from "coaster1235" is correct. The "a"s do not cancel. When you are reducing fractions, you must following the rules for creating equivalent fractions. To create equivalent fractions when reducing fractions, you divide out (or remove common factors). Factors are things that are multiplied. In the example, the "a"s are terms (they are being subtracted/added with the constants). For example: 4/8 = (4x1)/(4x2). You can reduce the fraction to 1/2 by dividing out the common factor of 4. If you had 3a/5a, since the "a"s are factors (being multiplied with the numbers), they can be cancelled/divided out to create 3/5. Factors can be numbers, variables; or in more complicated cases polynomials (which is what is typically found in rational expressions).
(4 votes)
• (n^3+3n)/(n^2-9)÷(n^2+5n-14)/(n^2+4n-21) how would I solve this and find the restrictions??
(2 votes)
• I'm having trouble with this idea that "a" is still undefined for -2,1,-1 at the end of simplifying. The point of simplifying is that we can change an expression without affecting anything important. The changes are all superficial. So how do we deal with the fact that different values of "a" can be undefined for different forms of an expression?

- at the end of simplifying, have we created a new expression? Do the rules of math need to be amended so that we're not allowed to do this?

- if it's possible for a value of "a" to go from undefined to defined just through rearranging an expression, could I rearrange it differently so that "a" is undefined at, say, 6? or 3? or -234 or whatever? How do I know that I've found all the undefined values for all the possible configurations of the expression?

- this is crazy. Can anyone make this make sense?
(2 votes)
• The only way an expression is undefined is when a factor (or factors) in the denominator equals zero.
Once we note those values as being excluded, only then we are allowed to do the division that will help to simplify the expression.
As part of the new simplified fraction and so that the simplified expression is algebraically the same as the original expression., we need to note the values of x for those factors that were completely divided out of both numerator and denominator Otherwise no one would know that those values made the original expression undefined.
(3 votes)
• so , 10x*2/x*2-y*2. Cannot be simplified? Why?
(2 votes)
• If you mean these to have exponents, you need to use the ^ (shift-6 key). All of your asterisks (*) means multiplication. So, your expression is: 20x / (2x-2y)
This can be reduced if you factor the denominator: 20x / [2(x-y)]
-- Cancel out a common factor of 2 and you get: 10x / (x-y)

If you mean the asterisks to be exponents: 10x^2 / {x^2-y^2)
This expression can not be reduced. When you reduce fractions, you can only cancel factors (items being multiplied). The x^2 in the denominator is a term (it is being subtracted with y^2).
We can't cancel terms.
If you first factor the denominator: 10x^2 / [(x+y)(x-y)]
This doesn't help either as you can see there is no common factor to cancel.

Hope this helps.
(3 votes)
• At . Why don't we say that A can't be equal to 2 since in the numerator their is (a-2) if a was 2 than (a-2) would be 0 making the numerator 0. Why would that not make the equation undefined? Thanks:)
(2 votes)
• Making the numerator equal to zero does not make the expression undefined.
It is only when the denominator equals zero that the expression is undefined --- because DIVISION (not multiplication) by zero is never allowed.
(2 votes)
• At Sal said factor but didn't he mean simplify?
(2 votes)
• Hi Knight of Stone / Ninjormon (Mormon ninja),

Sal did say factor, but meant simplify. They have added the correction to the video in the box which shows up at .

Hope that helps!
- JK
(2 votes)
• I wonder, how come 'a' can be a positive 2?
It seems to me that if 'a' cannot be 0, then 'a' cannot be 2 as well.
The final answer in the video is that 'a' cannot be -2, -1 or 1.
Unless I misunderstood how it works, this is missing positive 2.
Can you further explain why isn't that so?
Thanks!!
(2 votes)
• Let's consider the initial expression. Why can't a be equal to -2, -1 or 1? Here's why:

When a=-2 , then the second fraction becomes (-2+1)/(-2+2) = -1/0.

Mathematicians say that when you divide something by zero, you get something that just doesn't make sense. At all. Our task is to find and exclude all such numbers that turn our expression into a sense-less thing. I will leave some food for thought for you here, as to why it doesn't make sense.

Cool, so we now know that you don't want to divide anything by zero! Let's find all such values of a that, when plugged into our expression, turn denominators of either fraction into zero.

When a=-1, the first fraction has a zero in the denominator
When a=1, that same fraction again has zero in the denominator.

Case when a=-2 I covered above - that turns the second fraction into a bizarre no-idea-what thing.

O-kay. Did we cover all senseless cases? Well yeah, we did! There's nothing special about a=2, contrary to what you think. In fact, when a=2 no denominator turns into 0. Numerator of the first fraction, on the other hand, does turn into 0, which renders our whole expression equal to zero.
(2 votes)
• Why is part 2 before part 1?
(2 votes)

## Video transcript

Multiply and express as a simplified rational. State the domain. Let's multiply it, and then before we simplify it, let's look at the domain. This is equal to, if we just multiplied the numerators, a squared minus 4 times a plus 1, all of that over-- multiply the denominators-- a squared minus 1 times a plus 2. Now, the a squared minus 4 and the a squared minus 1 might look familiar to us. These are the difference in squares, a special type of binomial that you could immediately, or hopefully maybe immediately recognize. It takes the form a squared minus b squared, difference of squares, and it's always going to be equal to a plus b times a minus b. We can factor this a squared minus 4 and we can also factor this a squared minus 1, and that'll help us actually simplify the expression or simplify the rational. This top part, we can factor the a squared minus 4 as a plus 2-- 2 squared is 4-- times a minus 2, and then that times a plus 1. Then in the denominator, we can factor a squared minus 1-- let me do that in a different color. a squared minus 1 we can factor as a plus 1 times a minus 1. If you ever want to say, hey, why does this work? Just multiply it out and you'll see that when you multiply these two things, you get that thing right there. Then in the denominator, we also have an a plus 2. We've multiplied it, we've factored out the numerator, we factored out the denominator. Let's rearrange it a little bit. So this numerator, let's put the a plus 2's first in both the numerator and the denominator. So this, we could get a plus 2 in the numerator, and then in the denominator, we also have an a plus 2. In the numerator, we took care of our a plus 2's. That's the only one that's common, so in the numerator, we also have an a minus 2. Actually, we have an a plus 1-- let's write that there, too. We have an a plus 1 in the numerator. We also have an a plus 1 in denominator. In the numerator, we have an a minus 2, and in the denominator, we have an a minus 1. So all I did is I rearranged the numerator and the denominator, so if there was something that was of a similar-- if the same expression was in both, I just wrote them on top of each other, essentially. Now, before we simplify, this is a good time to think about the domain or think about the a values that aren't in the domain, the a values that would invalidate or make this expression undefined. Like we've seen before, the a values that would do that are the ones that would make the denominator equal 0. So the a values that would make that equal to 0 is a is equal to negative 2. You could solve for i. You could say a plus 2 is equal to 0, or a is equal to negative 2. a plus 1 is equal to 0. Subtract 1 from both sides. a is equal to negative 1. Or a minus 1 is equal to 0. Add one to both sides, and you get a is equal to 1. For this expression right here, you have to add the constraint that a cannot equal negative 2, negative 1, or 1, that a can be any real number except for these. We're essentially stating our domain. We're stating the domain is all possible a's except for these things right here, so we'd have to add that little caveat right there. Now that we've done that, we can factor it. We have an a plus 2 over an a plus 2. We know that a is not going to be equal to negative 2, so that's always going to be defined. When you divide something by itself, that is going to just be 1. The same thing with the a plus 1 over the a plus 1. That's going to be 1. All you're going to be left with is an a minus 2 over a minus 1. So the simplified rational is a minus 2 over a minus 1 with the constraint that a cannot equal negative 2, negative 1, or 1. You're probably saying, Sal, what's wrong with it equaling, for example, negative 1 here? Negative 1 minus 1, it's only going to be a negative 2 here. It's going to be defined. But in order for this expression to really be the same as this expression up here, it has to have the same constraints. It has to have the same domain. It cannot be defined at negative 1 if this guy also is not defined at negative 1. And so these constraints essentially ensure that we're dealing with the same expression, not one that's just close.