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## Algebra II (2018 edition)

### Unit 6: Lesson 2

Multiplying & dividing rational expressions- Multiplying & dividing rational expressions: monomials
- Multiplying rational expressions
- Dividing rational expressions
- Multiply & divide rational expressions: Error analysis
- Multiplying rational expressions
- Dividing rational expressions
- Multiply & divide rational expressions
- Multiplying rational expressions: multiple variables
- Dividing rational expressions: unknown expression
- Multiply & divide rational expressions (advanced)

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# Dividing rational expressions

CCSS.Math: ,

Sal divides and simplifies (2p+6)/(p+5) ➗ (10)/(4p+20). Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- I am re-asking JPRandall's question since it is unanswered after after 7 months and I also want to understand this... can someone at least give a hint as to where to find a proof for this?

his question:

"Over the last few videos you say the domain of the simplified expression must be the same as the original expression, to make it mathematically equivalent. Although it makes sense to me, could you perhaps give a more rigorous reason, is it to do with the definition of a function for example?"(46 votes)- The definition of a function involves several things. The main thing is how the number inputted (the argument) is associated with the number outputted (the image). But part of the definition of a function is the domain. Very often the domain is only implied but part of the definition of a function is its domain. If you are changing the function to a form where the domain isn't implied that you have to state it explicitly. There's nothing to prove, it's just a definition.

You can always state the domain, and it's important to do so when you're afraid it's not implied. For example, If you made a function involving money, you might have to say explicitly that the domain is all numbers such that the number of cents is an integer, so you don't have fractions of a cent.(34 votes)

- Hey Sal when you get the answer at3:59, it is acceptable to write the answer as 4p+12/5 or do we have to leave in the factor state?(15 votes)
- I think you can reduce it to 4p+12/5 like you said. He just probably said to himself that it was simplified enough to end the video there.(7 votes)

- How come you don't change the signs of the variables and numbers when you flip them on the second fraction? And do you make Khan Academy for iPhone or iPod, because that would be great.

and comment on this to tell me because you would be the best person ever!(6 votes)- When you flip the variable, what you are really doing is using its inverse.

When you multiply a number by its inverse, the result is always 1.

For instance 1/2 * 2/1 = 1 so 2 is the inverse of 1/2.

And -1/2 * -2/1 = 1, so -1/2 is the inverse of -2.

The inverse of a negative number is always negative, becasue it takes two negative numbers multiplied togather to get a postive 1.

And the inverse of a positive number is always positive,

So when you flip a fraction to use its inverse, you would leave the sign the same.

I hope that helps it click for you.(9 votes)

- If you divide by a fraction, must always the numerator and denomiator be non-zero, since a / (b/c) = a * c/b?(3 votes)
- If a÷b/c then b≠0 c≠0

If b=0 then a÷(0/c)=a/0 this is undefined.

If c=0 then a÷(b/0), b/0 is undefined.

Therefore if a÷b/c then b≠0 c≠0

a÷b/c=a(c/b) for b≠0 and c≠0

You must always add that restriction to make it true.

However when you learn about limit of functions, you'll find that the polynomial of the denominator equals to zero. We then are allowed to rationalize it by algebraically manipulate it because we aren't finding the value that makes the denominator 0, but the value that approaches it. And also it's why we restrict the domain where the denominator is 0.(4 votes)

- What is the domain and why is it helpful?(3 votes)
- A domain is what a variable can or cannot equal. For example, in 3a/a+6 = a-5, a can't equal -6 because then a+6 would equal and dividing by 0 is undefined. That is the domain, and since a ≠ -6, -6 is outside a's domain.(5 votes)

- I would like to get more practices on division of rational numbers. If anyone knows where I can get them pls reply(4 votes)
- At0:40why would you need to put 4p+20=0 if you are going to take the reciprocal? Wouldn't the 4p+20=0 be on top so you wouldn't have to worry about it being a zero?(4 votes)
- The rules you have to keep in mind are the following:

When dividing two rational expressions, the quotient is undefined...

- for any value that makes either of the original rational expressions undefined,

- and for any value that makes the divisor equal to zero.

4p+20 is the denominator of the divisor (second original rational expression). 4p+20=0 makes the divisor undefined. Therefore 4p+20 can not be zero.

Check examples in the following article for more detailed explanations:

https://www.khanacademy.org/math/algebra2/rational-expressions-equations-and-functions/multiplying-and-dividing-rational-expressions/a/dividing-rational-expressions(2 votes)

- At4:02when Sal reached the answer, 4(p+3) over 5, providing p does not equal -5, he did not further simplify the equation. Can you not simplify it into 4p+12 over 5, providing that p does not equal -5?(3 votes)
- In more advanced math, you often have to take your result and do other steps. That is usually easier if the result is in factored form. Sometimes in Calculus you will want to multiply the result all the way out. Practice will help you choose...and also little hints like the instructions you receive with the assignment and choices you are given as possible answers.(4 votes)

- why didn't he do 2p + 6 can not = 0

2p does not equal -6/2 =-3 so p can not equal -3 as well? :/(3 votes)- Because 2p + 6 is the numerator. Numerators are allowed to be 0. 0/16, 0/43213, 0/1 can all be calculated -> 0

The problem occurs when we have a 0 in the denominator.(3 votes)

- Over the last few videos you say the domain of the simplified expression must be the same as the original expression, to make it mathematically equivalent. Although it makes sense to me, could you perhaps give a more rigorous reason, is it to do with the definition of a function for example?(3 votes)
- is that like rates by nearest up hundredth(1 vote)

## Video transcript

Divide and express as a
simplified rational. State the domain. We start off with
this expression. We actually have one rational
expression divided by another rational expression. And like we've seen multiple
times before, these rational expressions aren't defined when
their denominators are equal to 0. So p plus 5 cannot be equal to
0, or if we subtract both sides of this-- we can't call
it an equation, but we could call it a not-equation-- by
negative 5, if we subtract negative 5 from both sides, you
get p cannot be equal to-- these cancel out-- negative 5. That's what that tells us. Over here, we could do the same
exercise 4p plus 20 also cannot be equal to 0. If it was, this expression
would be undefined. Subtract 20 from both sides. 4p cannot be equal
to negative 20. Divide both sides by 4. p cannot be equal
to negative 5. So in both situations, p being
equal to negative 5 would make either of these rational
expressions undefined. So the domain here is the set
of all reals such-- or p is equal to the set of all reals
such that p does not equal negative 5, or essentially all
numbers except for negative 5, all real numbers. We've stated the domain, so now
let's actually simplify this expression. When you divide by a fraction or
a rational expression, it's the same thing as multiplying
by the inverse. Let me just rewrite this
thing over here. 2p plus 6 over p plus 5 divided
by 10 over 4p plus 20 is the same thing as multiplying
by the reciprocal here, multiplying by
4p plus 20 over 10. I changed the division into a
multiplication and I flipped this guy right here. Now, this is going to be equal
to 2p plus 6 times 4p plus 20 in the numerator. I won't skip too many steps. Let me just write that. 2p plus 6 times 4p plus 20 in
the numerator and then p plus 5 times 10 in the denominator. Now, in order to see if we can
simplify this, we need to completely factor all of the
terms in the numerator and the denominator. In the numerator, 2p plus 6, we
can factor out a 2, so the 2p plus 6 we can rewrite
it as 2 times p plus 3. Then the 4p plus 20, we
can rewrite that. We can factor out a 4 as--
so 4 times p plus 5. Then we have our p plus 5 down
there in the denominator. We have this p plus 5. We can just write it down
in the denominator. Even 10, we can factor that
further into its prime components or into its
prime factorization. We can factor 10
into 2 times 5. That's the same thing as 10. Let's see what we
can simplify. Of course, this whole time, we
have to add the caveat that p cannot equal negative 5. We have to add this restriction
on the domain in order for it to be the same
expression as the one we started off with. Now, what can we cancel out? We have a 2 divided by a 2. Those cancel out. We have a p plus 5 divided
by a p plus 5. We know that p plus 5 isn't
going to be equal to 0 because of this constraint, so we
can cancel those out. What are we left with? In the numerator, we have 4
times p plus 3, and in the denominator, all we have is that
green 5, and we're done! We could right this as 4/5 times
p plus 3, or just the way we did it right there. But we don't want to forget
that we have to add the constraint p cannot be equal
to negative 5, so that this thing is mathematically
equivalent to this thing right here.