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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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# Simplifying rational expressions: common monomial factors

Sal simplifies the rational expressions (14x²+7x)/(14x) and (17z³+17z²)/(34z³-51z²) by taking common factors and canceling them.

## Want to join the conversation?

- at6:56, why doesn't z not equal -1(16 votes)
- Good question. A rational function is allowed to be 0. 0/b is defined as long as b isn't 0. 0/b is just 0 because 0=0b for any real number b. When you plug -1 in for z you just get 0/1 which is 0.(12 votes)

- This video and all explanations for simplifying rational expressions are in Algebra II. Why appears the skill for it in Algebra basics first?(6 votes)
- Algebra Basics is a subset of the easier algebra skills, while Algebra 2 is the full set of skills that are
**required**for Algebra 2. So, if you drew a Venn diagram, Algebra Basics and Algebra 2 would overlap a little. If you learned the Algebra Basics playlist first, then you will see some of those skills in every math course after that. If you did a good job of learning the skills in Algebra Basics then it will be super easy to master those skills when you run into them in the higher courses.(13 votes)

- I don't understand how you find the values that would make the expression undefined. Can somebody please explain?(5 votes)
- It's not as hard as it might appear.

Rational expressions are fractions. Fractions become undefined if the denominator = 0 because we can't divide by 0. The rational expressions have variables in the denominator. Thus, depending upon the value of the variable(s) in the denominator, the denominator might or might not = 0. You need to examine the denominator (before simplifying) to find the value(s) of the variable that can cause the denominator to become = 0.

Look at a couple examples:

2x/5: This fraction has no variable in the denominator and the denominator is not currently 0. So, no value needs to be excluded. This fraction can't become undefined.

2/(5x): A variable is in the denominator. So, we need to consider what values can make 5x become 0. We find this by solving: 5x=0 and we get x=0. So, we need to exclude x=0.

Now look at the 2nd example in the video.

[17z^2(z+1)] / [17z^2(2z-3)]

Notice, I'm using the factored version of the fraction**before**its reduced. Again, we need to look at only the denominator. There are 3 factors in the denominator that contain the variable: 17z, z and (2z-3). If any one of the factors becomes 0, the fraction is undefined. So we look at each factor.

Solve 17z = 0. It becomes 0 when z=0

Solve z = 0. This is already solved and creates the same result as the 1st factor.

Solve 2x-3 = 0. It creates z=3/2.

Put the results together... the values that will make the denominator = 0 are: z=0 and z=3/2. So, we exclude these values.

Hope this helps.(9 votes)

- Y squared + 4
*_________*Im super lost on this one. Can you even simplify this?

Y + 2(3 votes)- Could you please check if it is Y^2 + 4 or Y^2 - 4? (Y^2 - 4)/(Y+2) can be simplified as the numerator is a difference of two squares.

You'd be able to solve it like this:

(Y^2+4)/(Y+2)

((Y+2)(Y-2))/(Y+2)

Y-2(5 votes)

- I have a quick question to clarify what is meant by "undefined":

I've just done a mastery test (related to this video), where I was asked to simplify (14x^2 + 7x)/(14x), so I took out 7x as a factor and was left with (2x+1)/(2) which can be rewritten as x+1/2. I then had to say what value of x makes the expression undefined. So, I set the denominator of the original expression to zero: 14x = 0, so it's undefined for x = 0. Is this still true for the simplified expression (x+1/2)? Or is the expression only undefined at x=0 for the original expression?

It doesn't really make sense to me that the original expression and the simplified expression can both be undefined for x=0, so is an expression "undefined" only if they're written in a certain way?

Could I take any expression and rewrite it to have x on the denominator in such a way that it would cause it to be undefined? Hopefully this question makes sense!? Thanks!(4 votes)- Your simplified expression must be equivalent to the original. Since the original expression had a restriction of being undefined when x=0, the simplified expression needs to carry the same restriction.(2 votes)

- I'm a little confused. At0:38when Sal says that 0 is undefined, what does that mean exactly? Is 0 an undefined answer? Also is there a reason to write that x does not equal zero for every problem?(3 votes)
- What we are avoiding is division by zero.

Since in the example the denominator is 14x, if x=0 then (14)(0)=0 and you have division by zero - not allowed!

Now let's say the denominator was 14 - x. In this case, then, if x was equal to 14 you would have 14-14=0. In this case you would need to say that x does not equal 14 in order to avoid division by zero.(4 votes)

- Also in my school lessons, My "Instructor", Never mentions factoring a numerator number out of the denominator, This may be why I'm so confused, And when your doing factoring, why must you multiply by one if you already know that you are factoring out 17, is it because it is another 17. I think I just answered my own question, but clarification would be amazing! Thank you!(3 votes)
- You did figure it out. When factoring 17 from 17, you are basically dividing 17/17 = 1.

If you aren't sure if your factors are right, it is really easy to check. You multiply your factors and see if they create the original polynomial. If they don't, you know you have a math error.(3 votes)

- I'm not following. Where are you getting the Zs and 1s from?(2 votes)
- If you are talking about3:09the z's are variables in the equation like X is a variable in the previous problem.(2 votes)

- Why can't you divide a number by 0 and just get 0? Why is it undefined?(1 vote)
- Here's Sal's explanation: https://www.khanacademy.org/math/algebra/introduction-to-algebra/division-by-zero/v/why-dividing-by-zero-is-undefined

Here's mine:

If you divide 15 / 3, you find the number * 3 that will = 15. This give us an answer of 5.

Now, apply the same steps to 15 / 0. What number * 0 will = 15? There is no such number!

Any number that you multiply with 0 will = 0. You can never create 15. This is why it is undefined. The problem has no answer.(4 votes)

- But If Z does equal to zero in the solution, then it would be simplified to -1/3(2 votes)

## Video transcript

- [Voiceover] So I have a
rational expression here and my goal is to simplify
it, but while I simplify it, I wanna make the simplified expression be algebraically equivalent. So, if there are certain x values that would make this thing undefined, that I have to restrict
my simplified expression by those x values. So, 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 zero, 14 times zero is zero, it's going to be undefined. And so, we could say this is, we could say x does not,
cannot be equal to zero. For any other x, we can
evaluate this expression. Now let's actually try to simplify it. So, when you look at the
numerator and the denominator, every term we see is divisible by x and every term is divisible by seven. So it looks like we can factor
seven x out of the numerator and seven x out of the denominator. So the numerator, we can rewrite as seven x times, we factor seven x out of 14 x squared, we're gonna be left with two x and then you factor
seven x out of seven x, you're gonna be left with a one. And one way to think about it is we did the distributive
property in reverse, if we were to do it again, seven x times two x is 14 x squared, seven x times one is seven x. All right. And now let's factor seven x out of the denominator. So, 14 x could be rewritten as seven x times two, times two, and remember, I wanna keep
this algebraically equivalent, so I wanna keep the constraint that x cannot be equal to zero. And so we divide the numerator and the denominator by seven x
or one way to think about it, you can divide seven x by
seven x and just get one and we are left with two x plus one over two. Now this was the original expression, x could take on any value, but if we want it to be
algebraically equivalent to our original expression, it has to have the same constraints on. So we're going to have to say x, x does not equal zero, and 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. And 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 inputs. 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 equal zero. This one would have been
defined for x equal zero, this one would not have, and so they wouldn't have
been algebraically equivalent. This makes them algebraically equivalent. And of course you can write
this in different ways. You can divide each of these
terms by two, if you like. So you could divide two x by two and get x and then divide one by two and get 1/2. Once again, we would wanna keep the 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. See if we could simplify it. But as we simplify it, we
really wanna be conscientious of restricting the Zs here so that we get an algebraically
equivalent expression. So, let's think about
where this is undefined. We can think about where it's undefined by factoring the denominator here. Actually, let me just... So, this is going to be equal to, actually, let me just... Let me just do the first
step where I could say, "Well, what's a common factor "in the numerator and the denominator?" Every term here is divided by z squared and every term is also divided by 17, so it looks like 17 z
squared can be factored out. So, 17 z squared can be
factored out of the numerator and then we would be left with, we factor out of 17 z squared
out of 17 z to the third and we're gonna be left with just a z. 17 z squared, you factor
17 z squared out of that and you're gonna be left with just a one. And once again, you can
distribute the 17 z squared, multiply it times z, you
get 17 z to the third. 17 z squared times one is 17 z squared. All right. All of that
is going to be over, we wanna factor out of 17 z squared out of the denominator, 17 z squared times, and so 34 z to the third
divided by 17 z squared, 34 divided by 17 is two, and z to the third
divided by z squared is z and then we have minus
51 divided 17 is three and z squared divided by
z squared is just one. So we'll just leave it like that. And so over here, it
becomes a little bit clear of, well one, how we're gonna simplify it. We're just gonna divided 17
z squared by 17 z squared, but let's be careful to
restrict the domain here. So, we can tell if z is equal to zero, then this, 17 z squared
is gonna be equal to zero and we make the denominator equal to zero and we could see that even looking here. So we could say z cannot be equal to zero. Now what else? Z cannot be equal to whatever makes this two z
minus three equals zero. So let's think about what makes two z minus three equal to zero. Two z minus three is equal to zero. You can add three to both sides and you would get two z is equal to three. Divide both sides by two and you would get z is equal to 3/2. So z cannot be equal to zero and z cannot be equal to 3/2. So that's how we're gonna
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, we're gonna be left with z plus one over two z minus three and we wanna keep that constraint. z cannot be equal to zero. And actually, this second
constraint is redundant because we still have the
two z minus three 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 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 is clearly not defined for z equals 3/2. So, there you go. And so I'm gonna ask you for what values does this
expression not defined, well then you would also include that. Actually, let me just write it. It doesn't hurt to be redundant. z does not equal 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 equals zero, but if we want it to be
algebraically equivalent to this one and that one, it has to be constrained in the same way.