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### Course: Integrated math 3 > Unit 13

Lesson 1: Cancelling common factors- Reducing rational expressions to lowest terms
- Intro to rational expressions
- Reducing rational expressions to lowest terms
- Simplifying rational expressions: common monomial factors
- Reduce rational expressions to lowest terms: Error analysis
- Simplifying rational expressions: common binomial factors
- Simplifying rational expressions: opposite common binomial factors
- Simplifying rational expressions (advanced)
- Reduce rational expressions to lowest terms
- Simplifying rational expressions: grouping
- Simplifying rational expressions: higher degree terms
- Simplifying rational expressions: two variables
- Simplify rational expressions (advanced)

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# Reducing rational expressions to lowest terms

Learn how to simplify rational expressions, which are fractions with variable expressions in the numerator and denominator. This video also shows how to factor the expressions and cancel out common factors, just like with rational numbers. Don't forget to exclude values of x that would make the denominator zero, since those would make the expression undefined. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- do you always have to add the condition? never heard of it from my teacher.(121 votes)
- Sometimes in an Algebra 1 course/text/curriculum, teachers will just teach the simplifying piece, and leave the restrictions for Algebra 2. This is because this is one of the most challenging type of problems in Algebra 1. Brains often melt solving rationals because many students can barely factor and simplify, let alone consider restrictions on the denominator. Even Sal makes mistakes in his examples, he forgets restrictions.(60 votes)

- At7:15, Sal said that x cannot equal -1. However, x also cannot be equal to 2, based off of the answer he got. Why wouldn't you put x cannot equal negative 2 in the final answer, as that would also make the expression undefined?(15 votes)
- Now that the expression has been simplified to (x+5)(x-2), it is obvious that x cannot be equal to 2, but the fact that x could also not be equal to -1 (otherwise division by 0 resulted) in the original non simplified expression has been lost, so we add the condition as a reminder.

You see, we can set x=-1 into (x+5)(x-2) without a problem, but we could not set x=-1 in the original expression - and this simplified expression is based on that.(14 votes)

- Near8:36couldn't you just factor out a 3 and get 3(x^2+x-6) and then factor the rest of it using the quadratic formula or other methods?(11 votes)
- That would work if I'm not mistaken, so if you're comfortable with that, go ahead! I personally prefer the method Sal uses.(5 votes)

- why is it that all math problems have numbers in them?(9 votes)
- I guess an answer would be that numbers can represent an almost unlimited amount of things. I wouldn't be surprised if there were some math problems without numbers, but it is a very essential part of the subject and so it shows up in lots of places. Something to think about would be, if not numbers, what else would be in them?(10 votes)

- in the final example, is it not necessary to also say that x ≠ 2, (ie. x ≠ 2, -1 ) because the fact that x - 2 is still in the denominator means that it's still present? i.e. would it only be necessary to specify x ≠ 2, -1 if somewhere later in the calculation x - 2 was somehow canceled out?(7 votes)
- Good question! The reason is that x ≠ 2 is still clear from our given expression. However, x ≠ -1 isn't (as that information is lost when we cancel the terms). That's why we mention the -1 and not the 2.

However, for the same function, if we were to define the domain, both 2 and -1 would have to be excluded.(10 votes)

- how do you get to the practice problems for this(4 votes)
- Here is a link: http://www.khanacademy.org/math/algebra/rational-expressions/simplifying-rational-alg/e/simplifying_rational_expressions_1

There are several more that you can get to from there.(5 votes)

- at5:00, i still do not get how he got x cannot equal -1 over 3(3 votes)
- First thing to understand this is, you should go through this quick video here:

https://www.khanacademy.org/math/trigonometry/functions_and_graphs/undefined_indeterminate/v/undefined-and-indeterminate

Now, before you cancels out (3x+1) from both, numerator and denominator and remove its trace, you should always leave a note that (3x + 1) cannot be zero and solving it will lead to.... x cannot be -1 over 3 (that is "-1/3").(5 votes)

- I've come across problems in my homework where I don't know what the condition should be. Is it the x value that makes what you cancel equal to zero, that makes the original expression's denominator equal to zero, or that makes the new expression's denominator equal to zero?(4 votes)
- In problems like those in the video you are expected to explicitly state the condition for the x value that makes
**what you cancel**equal to zero.

However, since you're cancelling (and so there is such a factor in the*original*denominator), it implies the same thing**makes the original expression's denominator equal to zero**, or maybe—is ONE OF THE THINGS that**make the original expression's denominator equal to zero**.

If you think about it, the value that**makes the new expression's denominator equal to zero**is the reason for saying "one of the things..." in the previous sentence. Because THE OTHER THING is what**makes the new expression's denominator equal to zero**. It's the restriction (or several restrictions) shared by the old and the new expressions, therefore you don't have to explicitly state it in a task like this.(4 votes)

- I understand that we have to add the restrictions or conditions for the equation, but what if you forget it? Would it affect your graph or something?(4 votes)
- Any restrictions on a rational expression are a consequence of the expression not being defined at one or more values of the variable. At such values,

the graph of the corresponding rational expression function can have a vertical asymptote or a removable discontinuity . Both affect the graph.(3 votes)

- Ramey

a minute ago

Posted a minute ago. Direct link to Ramey's post “Can someone please walk me through this STEP BY ST...”

Can someone please walk me through this STEP BY STEP, regardless of how ridiculously minor the step is? I have tried and failed so many times it's not even funny. I have looked at countless videos, and asked several people for help. I know it is a simple problem, but I am really struggling. Any help would be greatly appreciated.

Two airplanes start at the same time from airports 500 km apart. Each one flies with an airspeed of 200 km/hr directly towards the other airport. But one reaches its airport half an hour before the other plane reaches the other airport. How fast is the wind blowing?

If you can include what laws you use along the way, it would really benefit my understanding. But at the least, i just want to see how to do it. *I AM NOT AFTER THE ANSWER. I AM AFTER THE PROCESS.* I really just want to understand and I am desperate. Thank you in advance!!(3 votes)- Because you say that you don't want to know the answer, I'm guessing that you want the process to lead you to the answer. I would like to say, though, that this is a simple problem, but I get why you are struggling with it because it requires a lot of work to get to the answer.
*Anyways*, if both planes, which we will call the first one Plane A and the second one Plane B, are at opposite airports, and the airports are 500 km apart, that means that if the airplanes are going 200 km/h, then the flight of the planes should be about 2.5 hours, give or take.

Now, one of the airplanes is going a bit faster and the other is going a bit slower because the wind is blowing with plane A and against plane B.

Because Plane A arrives 30 minutes earlier than Plane B, Plane A took 15 minutes less time to get to the opposite airport, and Plane B took 15 minutes more time to get to the opposite airport.

Now, let's say that the airplanes took off at4:00PM. Remember, the flight should have been around 2 hrs 30 min. Because of the wind factor, Plane A landed at6:15PM, and Plane B landed at6:45; a 30 minute difference.

Alright. Now that we know that, we can solve for the wind. If the planes were going 200 km/h, and one arrived 15 minutes early and one arrived 15 minutes late, we will use simple subtraction to find the wind speed:

200(km/h the planes traveled at) - 15(the amount of time the flights were changed) = 175.

Therefore, the wind was going 175 km/h.

If you are confused with a step, all you need to do is ask me.

Hope that helps!(2 votes)

## Video transcript

When we first started learning
about fractions or rational numbers, we learned about the
idea of putting things in lowest terms. So if we saw
something like 3, 6, we knew that 3 and 6 share
a common factor. We know that the numerator,
well, 3 is just 3, but that 6 could be written as 2 times 3. And since they share a common
factor, the 3 in this case, we could divide the numerator by 3
and the denominator by 3, or we could say that this
is just 3/3, and they would cancel out. And in lowest terms, this
fraction would be 1/2. Or just to kind of hit the point
home, if we had 8/24, once again, we know that this is
the same thing as 8 over 3 times 8, or this is the
same thing as 1 over 3 times 8 over 8. The 8's cancel out and we get
this in lowest terms as 1/3. The same exact idea applies
to rational expressions. These are rational numbers. Rational expressions are
essentially the same thing, but instead of the numerator
being an actual number and the denominator be an actual number,
they're expressions involving variables. So let me show you what
I'm talking about. Let's say that I had 9x plus
3 over 12x plus 4. Now, this numerator up here,
we can factor it. We can factor out a 3. This is equal to 3
times 3x plus 1. That's what our numerator
is equal to. And our denominator, we
can factor out a 4. This is the same thing
as 4 times 3x. 12 divided by 4 is 3. 12x over 4 is 3x. Plus 4 divided by 4 is 1. So here, just like there, the
numerator and the denominator have a common factor. In this case, it's 3x plus 1. In this case, it's a variable
expression. It's not an actual number,
but we can do the exact same thing. They cancel out. So if we were to write this
rational expression in lowest terms, we could say that
this is equal to 3/4. Let's do another one. Let's say that we had
x squared-- let me see a good one. So let's say we had x squared
minus 9 over 5x plus 15. So what is this going
to be equal to? So the numerator
we can factor. It's a difference of squares. We have x plus 3 times
x minus 3. And in the denominator we
can factor a 5 out. This is 5 times x plus 3. So once again, a common factor
in the numerator and in the denonminator, we can
cancel them out. But we touched on this a
couple of videos ago. We have to be very careful. We can cancel them out. We can say that this is going to
be equal to x minus 3 over 5, but we have to exclude the
values of x that would have made this denominator equal to
0, that would have made the entire expression undefined. So we could write this as being
equal to x minus 3 over 5, but x cannot be equal
to negative 3. Negative 3 would make this zero
or would make this whole thing zero. So this and this whole
thing are equivalent. This is not equivalent to this
right here, because this is defined that x is equal to
negative 3, while this isn't defined that x is equal
to negative 3. So to make them the same, I
also have to add the extra condition that x cannot
equal negative 3. So likewise, over here, if this
was a function, let's say we wrote y is equal to 9x plus
3 over 12x plus 4 and we wanted to graph it, when we
simplify it, the temptation is oh, well, we factored out a 3x
plus 1 in the numerator and the denonminator. They cancel out. The temptation is to say, well,
this is the same graph as y is equal to the constant
3/4, which is just a horizontal line at y
is equal to 3/4. But we have to add
one condition. We have to eliminate-- we have
to exclude the x-values that would have made this thing right
here equal to zero, and that would have been zero if
x is equal to negative 1/3. If x is equal to negative 1/3,
this or this denominator would be equal to zero. So even over here, we'd have
to say x cannot be equal to negative 1/3. That condition is what really
makes that equal to that, that x cannot be equal
to negative 1/3. Let's do a couple
more of these. And I'll do these in pink. Let's say that I had x squared
plus 6x plus 8 over x squared plus 4x. Or actually, even better, let
me do this a little bit. x squared plus 6x plus 5 over
x squared minus x minus 2. So once again, we want to factor
the numerator and the denonminator, just like we did
with traditional numbers when we first learned about fractions
and lowest terms. So if we factor the numerator,
what two numbers when I multiply them equal 5 and
I add them equal 6? Well, the numbers that pop
in my head are 5 and 1. So the numerator is x plus
5 times x minus 1. And then our denonminator,
two numbers. Multiply negative 2,
add a negative 1. Negative 2 and positive
1 pop out of my head. So this is a positive
1, right? x plus 5 times x plus 1, right? 1 times 5 is 5. 5x plus 1x is 6x. So here we have a positive
1 and a negative 2. So x minus 2 times x plus 1. So we have a common factor
in the numerator and the denonminator. These cancel out. So you could say that this
is equal to x plus 5 over x minus 2. But for them to really
be equal, we have to add the condition. We have to add the condition
that x cannot be equal to negative 1 because
if x is equal to negative 1, this is undefined. We have to add that condition
because this by itself is defined at x is equal
to negative 1. You could put negative
1 here and you're going to get a number. But this is not defined at x is
equal to negative 1, so we have to add this condition
for this to truly be equal to that. Let's do a harder one here. Let's say we have 3x squared
plus 3x minus 18, all of that over 2x squared plus
5x minus 3. So it's always a little bit more
painful to factor things that have a non-one coefficient
out here, but we've learned how to do that. We can do it by grouping, and
this is a good practice for our grouping, so let's do it. So remember, let's factor 3x
squared plus 3x minus 18. So you need to think
of two numbers. This is just a review
of grouping. You need to think of two numbers
that when we multiply them are equal to 3 times
negative 18, or it's equal to negative 54, right? That's 3 times negative 18. And when we add them, a plus
b, needs to be equal to 3x because we're going to split up
the 3x into an ax and a bx. Or even better, not
3x, equal to 3. So what two numbers
could there be? Let's see, our times tables. Let's see, they are
three apart. One's going to have to be
positive and one's negative. 9 times 6 is 54. If we make the 9 positive
and we make the b negative 6, it works. 9 minus 6 is 3. 9 times negative 6
is negative 54. So we can rewrite
this up here. We can rewrite this as 3x
squared, and I'm going to say plus 9x minus 6x minus 18. Notice, all I did here
is I split this 3x into a 9x minus 6x. The only difference between
this expression and this expression is that I split the
3x into a 9x minus 6x. You add these two together,
you get 3x. The way I wrote it right here
you can actually ignore the parentheses. And the whole reason why I did
that is so I can now group it. And normally, I decide which
term goes with which based on what's positive or negative or
which has common factors. They both have a common
factor with 3. Actually, it probably wouldn't
matter in this situation, but I like the 9 on this side
because they're both positive. So let's factor out
a 3x out of this expression on the left. If we factor out a 3x out of
this expression this becomes 3x times x plus 3. And then on this expression, if
we factor out a negative 6, we get negative 6
times x plus 3. And now this is very clear our
grouping was successful. This is the same thing as-- we
can kind of undistribute this as 3x minus 6 times x plus 3. If we were to multiply this
times each of these terms, you get that right there. So the top term, we can rewrite
it as 3x minus 6-- let me do it in the same color. So we can rewrite it as 3x
minus 6 times x plus 3. That's this term right here. I don't want to make it look
like a negative sign. That's that term right there. Now let's factor this bottom
part over here. Scroll to the left
a little bit. So if I want to factor 2x
squared plus 5x plus 3, I need to think of two numbers that
when I take their product, I get 2 times 3, which is equal
to 6, and they need to add up to be 5. And the two obvious numbers
here are 2 and 3. I can rewrite this up here as
2x squared plus 2x plus 3x plus 3, just like that. And then if I put parentheses
over here, and I decided to group the 2 with the 2 because
they have a common factor of 2, and I grouped the 3 with
the 3 because they have a common factor of 3. This right here is 2 and a 3. So here we can factor
out a 2x. If you factor out a 2x, you get
2x times x plus 1 plus-- you factor out a 3 here--
plus 3 times x plus 1. And our grouping
was successful. This is clearly-- let me switch
colors-- this is the same thing as 2x plus
3 times x plus 1. So here we've been able
to factor it as well. We were able to factor out
the denominator as well. Actually, I just realized
that I made a mistake. I wrote here minus 3. I wrote a plus 3 over here. Let me backtrack this. That would have been
a horrible mistake. I would have had to
redo the video. Let me clear all of this, all
of this business over here. Let me clear that. This is 2x squared
plus 5x minus 3. So once again, a times b needs
to be equal to negative 3 times 2, which is negative 6. And a plus b needs
to be equal to 5. So in this situation, it looks
like if we went with 6 and negative 1, that seems to
be a better situation. 6 minus 1 is 5. 6 times negative 1
is negative 6. So that would have been
a horrible mistake. So we can rewrite this up here
as 2x squared, and I'll group the 6 with the 2x squared
because they share a common factor. So plus 6x minus x, this is the
same thing as 5x minus 3. I just had to find the numbers
to split this 5x into. But 6x minus x is 5x. And if I put some parentheses
here, I can factor out of 2x out of this first term. I get 2x times x plus 3. And here I can factor out
a negative 1, so minus 1 times x plus 3. And then our grouping
was successful. We get 2-- let me do this in a
different color-- we get 2x minus 1 times x plus 3. So our denominator here
is equal to 2x minus 1 times x plus 3. And once again, we have a common
factor in our numerator and our denonminator,
the x plus 3. But we have to add the condition
that x cannot be equal to negative 3, because
that would make this whole thing equal to zero. Or not equal to zero, it would
make us divide by zero, which is undefined. So we have to say that x cannot
be equal to negative 3. So this expression up here is
the same thing as 3x minus 6 over 2x minus 1, granted that we
also imposed the condition that x does not equal
negative 3. Hopefully, you found
that interesting.