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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.

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  • duskpin ultimate style avatar for user Lezzlly Real
    why did the negative one only go to a part of the numerator instead of being distributed onto the entire one?
    (24 votes)
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    • boggle purple style avatar for user lily J
      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)
  • piceratops ultimate style avatar for user Firedrake969
    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)
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    • mr pants teal style avatar for user Wrath Of Academy
      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)
  • marcimus pink style avatar for user kaplan.sophie4
    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)
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  • aqualine ultimate style avatar for user Ainul Jeffery
    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)
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    • purple pi pink style avatar for user ZeroFK
      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)
  • male robot donald style avatar for user Austin Alexander
    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)
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    • piceratops ultimate style avatar for user Just Keith
      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)
  • blobby green style avatar for user Michael Lasko
    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)
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  • blobby green style avatar for user drabbersquash
    What was the second -1 for?
    (3 votes)
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  • starky tree style avatar for user Izzie M
    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)
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    • cacteye blue style avatar for user Jerry Nilsson
      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)
  • leafers tree style avatar for user Isabelle
    What would you do if you’re question was (x3y3)3 times xy2?
    (2 votes)
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    • boggle purple style avatar for user lily J
      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)
  • blobby green style avatar for user KikiRuiz628
    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)
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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.