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## Integrated math 3

### 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: opposite common binomial factors

Sal simplifies and states the domain of (x^2-36)/(6-x). Created by Sal Khan.

## Want to join the conversation?

- why did the negative one only go to a part of the numerator instead of being distributed onto the entire one?(24 votes)
- Do you mean that in (x+6)(x-6)(-1)(-1), why should -1 only multiply to (x-6) but not (x+6)(x-6)?

If that's what you mean, maybe I know the answer of this question.

Do you know the equation "a×b×c×d = a×(b×c)×d"? That's the equation that uses here. Imaging that (x+6) is "a", (x-6) is "b", the first -1 is "c" and the second -1 is "d".

In multiplication, the numbers could change order and wouldn't change the result. Like "1×2×3×4 = 2×3×1×4".And that's why the -1 only multiply to (x-6), but not the (x+6)(x-6).

But if it's in addition, let's say the if it's (a+b)×c×d, you could say that it's equal to (a×c+b×c)×d. The reason why that's fine to change it like this is because (a+b)×c×d = (a+b)×(c×d) = a×(c×d)+b×(c×d) AND (a×c+b×c)×d = a×c×d+b×c×d.

Hope that helps!(2 votes)

- I was working on the problem set for this, and found that even if the denominator turned out to be something like (y + 5)(y - 6), the answer would only accept y =/= 6. Why?(6 votes)
- I looked at the "Simplifying rational expressions" exercises and I don't see any problems where the denominator has multiple roots factored out. Note that if it asks when the simplification is valid for
`(x+1) / ((x+1)(x+2))`

and you simplify to`1/(x+2)`

, you only need to mention that`x != -1`

. Saying that`x != -2`

is still there in your answer equation, so there's no need to state it again. They only want to know when the "simplification is valid", not when the whole equation is valid.(6 votes)

- Can somebody please tell me what I did wrong in solving this problem?

1. (x^2-36)/(6-x)

2. ((x+6)(x-6))/((-1)(x+6))

3. (x-6)/(-1)

4. (6-x), x cannot = 6(5 votes)- When you factor -1 from (6-x), you get (-6+x) or (x-6). You have (x+6)

Hope this helps.(3 votes)

- I've seen in almost all the the equation that the domain is (All real numbers). Why can the domain not be imaginary number? and if it can be, then what happens?(3 votes)
- You're right, there's no real reason to exclude imaginary and complex numbers. For example there is no reason the x in x^2 + 3x -5 = 0 can't be i or 5+3i.

However, we don't usually involve complex numbers without a reason. In the previous example, using real numbers gives a simple quadratic equation. If we plot it, it's just a parabola. But if you say the domain is all complex numbers, suddenly you need a 3D plot. In fact, you need 2 plots, both 3D: one for the real part and one for the imaginary part. There is no need to add that kind of complexity unless you know the input can be a complex number.

There is also a convention to use x for real variables, and z for complex ones (and n for natural).(6 votes)

- Is simplifying rational expressions the same as simplifying rational equations? I've been told this by a friend but I'm still not sure.(2 votes)
- No, it is not the same, but they are related. A rational equation is a math problem that involves rational expressions. So, you would, as part of solving the rational equation, simplify the rational expressions contained within the equation. But, then you would still need to solve the rational equation after you simplified it.

So, in short, simplifying rational expressions is one of the steps involved in solving a rational equations.(6 votes)

- Wouldn't it be clearer to multiply the numerator and denominator both by -1 (-1/-1 = 1 after all). you would get -1(x+6)(x-6) / -1(6-x). This would turn the problem into (-x-6)(x-6) / (-6+x) and you could cancel from right there.(3 votes)
- so then how would you solve a problem like 4/4y-8(2 votes)
- You can't solve it because it doesn't equal something. "Solve" is when you actually find the values of all variables, which can only happen with "equations". Note that in this video, you're dealing with "expressions". Your thing you wrote (4/4y-8) is an expression. If it was, say, 4/(4y-8) = 2, it would be an equation (note it becomes an "equation" when it "equals" something).

To simplify your expression that you wrote, you would divide both the numerator and the denominator by their common factor (4). So top divided by 4 is 1. Bottom divided by 4 is y-2. Therefore, the simplified form of 4/(4y-8) is 1/(y-2).(2 votes)

- As I was trying to solve the expression I factored the numerator like Sal did (x+6)(x-6) and then factored a -1 out of the denominator to set it equal to -1 (x-6) so the x-6 would cancel out, ending with the expression (x+6)/-1, and simplifying to -x-6, but I noticed instead of factoring a -1 out of the denominator, Sal multiplies the numerator with -1.

I understand that we both came to the same answer, but is Sal's method somehow more efficient or is it a matter of preference?(1 vote)- Basically, what Sal did was to factor out (−1) from the numerator, but in the process he also chose to at least kind of explain what "factoring out" actually means, which is to multiply the expression by 1, then rewrite the 1 as a product of two factors and distribute one of those factors over the expression, thus giving us (𝑥 + 6)(𝑥 − 6) = (−1)(−1)(𝑥 + 6)(𝑥 − 6) = (−1)(𝑥 + 6)(6 − 𝑥).(3 votes)

- What would you do if you’re question was (x3y3)3 times xy2?(2 votes)
- If you mean [(x^3)×(y^3)×3]×(xy)^2, here's the steps to solve it:

[(x^3)×(y^3)×3]×(xy)^2

= 3×(x^3)×(y^3)×(x^2)×(y^2)

= 3×[(x^5)(y^5)]

= 3×[(x×y)^5]

The equations used to solve this problem:

(ab)^c = (a^c)×(b^c)

(a^x)×(a^y) = a^(x+y)

Hope that helps!(1 vote)

- I came across several websites to solve a problem I am on. It is the following:

6m^2√128m^2 n^4 p^7

I don't know how to simplify it completely. Please answer soon --- thank you!(1 vote)- Could you edit it with () to show what portions belong under the square root.

If you meant 6m²√(128m² n⁴ p⁷)

Then that simplifies to 48 m³ n² p³√(2p)(3 votes)

## Video transcript

Simplify the rational
expression and state the domain. Let's see if we can start with
the domain part of the question, if we can start
with stating the domain. Now, the domain is the set of
all of the x values that you can legitimately input into
this if you view this as a function, if you said this is
f of x is equal to that. The domain is a set of all x
values that you could input into this function and
get something that is well-defined. The one x value that would make
this undefined is the x value that would make the
denominator equal 0-- the x value that would make
that equal 0. So when does that happen? Six minus x is equal to 0. Let's add x to both sides. We get 6 is equal to x, so the
domain of this function is equal to the set of all
real numbers except 6. So, x could be all real numbers
except 6, because if x is 6 then you're dividing
by 0, and then this expression is undefined. We've stated the domain, now
let's do the simplifying the rational expression. Let me rewrite it over here. We have x squared minus
36 over 6 minus x. Now, this might jump out at you
immediately, as it's that special type of binomial. It's of the form a squared minus
b squared, and we've seen this multiple times. This is equivalent to a plus
b times a minus b. And in this case, a
is x and b is 6. This top expression right here
can be factored as x plus 6 times x minus 6, all of
that over 6 minus x. Now, at first you might say,
I have a x minus 6 and a 6 minus x. Those aren't quite equal, but
what maybe will jump out at you is that these are the
negatives of each other. Try it out. Let's multiply by negative 1 and
then by negative 1 again. Think of it that way. If I multiply by negative 1
times negative 1, obviously, I'm just multiplying the
numerator by 1, so I'm not in any way changing
the numerator. What happens if we just multiply
the x minus 6 by that first negative 1? What happens to that
x minus 6? Let me rewrite the
whole expression. We have x plus 6, and I'm
going to distribute this negative 1. If I distribute the negative 1,
I have negative 1 times x is negative x. Negative 1 times negative
6 is plus 6. And then I have a negative
1 out here. I have a negative 1 times
negative 1, and all of that is over 6 minus x. Now, negative plus 6. This is the exact same thing
as 6 minus x if you just rearrange the two terms.
Negative x plus 6 is the same thing as 6 plus negative
x, or 6 minus x. Now you could cancel them out. 6 minus x divided by 6 minus x,
and all you're left with is a negative 1-- I'll write it
out front-- times x plus 6. If you want, you can distribute
it and you would get negative x minus 6. That's the simplified
rational expression. In general, you don't have to
go through this exercise, multiplying by a negative
1 and a negative 1. But you should always be able
recognize that if you have a minus b over b minus a that that
is equal to negative 1. Or think of it this way: a
minus b is equal to the negative of b minus a. If you distribute this negative
sign, you get negative b plus a, which
is exactly what this is over here. We're all done.