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Subtracting rational expressions: unlike denominators

Sal rewrites (-5x)/(8x+7)-(6x³)/(3x+1) as (-48x⁴-42x³-15x²-5x)/(8x+7)(3x+1).

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  • blobby green style avatar for user anvaz1111
    Does it matter the order of the denominator towards the end of the problem (as shown in of the video)? For example, would it matter if I did (3x+1)(8x+7), rather than (8x+7)(3x+1)?
    (7 votes)
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  • cacteye yellow style avatar for user Emma🖤
    Why doesn't he multiply out the (8x+7)(3x+1).
    (8 votes)
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  • blobby green style avatar for user erinikk.
    Why doesn't he FOIL the denominator at , at the end?
    (6 votes)
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    • stelly blue style avatar for user Kim Seidel
      When working with fractions, we keep the denominator in factored form because after it is converted to the common denominator, the addition / subtraction does not change the denominator. We only add/subtract numerators.

      Once the addition/subtraction is complete, we need to try and reduce the fraction. At this point, both the numerator and denominator need to be factored. Thus, if you FOIL it earlier on, you need to un-FOIL (or factor it) at this point. So, FOILing the denominator actually creates extra work.

      Hope this helps.
      (8 votes)
  • male robot johnny style avatar for user John Gendron
    I'm confused as to why (-48x^4-42x^3-15x^2-5x)/(8x+7)(3x+1) is a completely simplified equation. Wouldn't you factor out the numerator into -x(48^3+42^2+15x+5)/(8x+7)(3x+1) and then even go a step further to create -x(6x(8x+7)+5(3x+1))/(8x+7)(3x+1)? I can't tell if you can reduce from there (trying to write it out like this messes me up) but wouldn't that simplify it more? Would that be an acceptable answer?
    (0 votes)
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    • stelly blue style avatar for user Kim Seidel
      Your 1st version is somewhat better than the original because you have factored the numerator and we can clearly see there is no common factor and the fraction is fully reduced.
      -x(48^3+42^2+15x+5)/(8x+7)(3x+1)

      Your second version: x(6x(8x+7)+5(3x+1))/(8x+7)(3x+1) would not be considered simplified. You are correct to try and factor the polynomial further. And your work so far looks correct, but since the 2 terms do not have a common binomial factor, in other words (8x+7) does not equal (3x+1), the polynomial is not factorable. Leaving it in this partially factored form is not done. It should be fully factored or each factor should be in the form of simplified terms. So you should revert back to your 1st version which contains fully simplified factors.
      (13 votes)
  • blobby green style avatar for user russel_semana
    Will the answer still be correct if we multiply the denominator instead of it being factored in the final answer?
    (3 votes)
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    • stelly blue style avatar for user Kim Seidel
      It is not incorrect. But, the convention is to leave rational expressions in their factored form. I believe this is done because then you can quickly look at the fraction and see that it is fully reduced. Sal should have factored the numerator to verify / show that the fraction is fully reduced.
      (4 votes)
  • female robot amelia style avatar for user saketh125
    In my math homework of adding and subtracting rational expressions, I don't get most of the problems. I didn't get what the LCD is for them. How do you take the LCD of these numbers.
    (3 votes)
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  • male robot johnny style avatar for user D'Andre' Stroman
    how do you do 2x+3/x-4 - x-5/x+2 as a
    rational expressions
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      The steps you need to follow are the same as when subtracting 2 numeric fractions. They are just a little more complicated because we're working with polynomials within the fractions.

      1) Find the common denominator: LCD = (x-4)(x+2)
      -- The 2 binomial denominators are not factorable. So, they act like prime numbers. This makes their LCD = (x-4)(x+2)

      2) Convert each fraction to the common denominator. Note: simplify the numerators since we need to add/subtract the numerators once we have a common denominator.
      (2x+3)/(x-4) * (x+2)/(x+2) = [(2x+3)(x+2)] / [(x-4)(x+2)] = [2x^2 + 7x + 6] / [(x-4)(x+2)]
      -(x-5)/(x+2) * (x-4)/(x-4) = - [(x-5)(x-4)] / [(x-4)(x+2)] = - [(x^2 - 9x + 20)] / [(x-4)(x+2)]
      The expression is now: [2x^2 + 7x + 6 - (x^2 - 9x + 20)] / [(x-4)(x+2)]

      3) Distribute the subtraction (minus sign) across the 2nd numerator to subtract the entire fraction.
      [2x^2 + 7x + 6 - (x^2 - 9x + 20)] / [(x-4)(x+2)] = [2x^2 + 7x + 6 - x^2 + 9x - 20)] / [(x-4)(x+2)]

      4) Combine like terms in the numerator
      [2x^2 + 7x + 6 - x^2 + 9x - 20)] / [(x-4)(x+2)] = [x^2 + 16x -14)] / [(x-4)(x+2)]

      5) Try to factor the numerator to reduce the fraction. The numerator is not factorable. So, we're done. The result is: [x^2 + 16x -14)] / [(x-4)(x+2)]

      Hope this helps.
      (3 votes)
  • female robot amelia style avatar for user warmour2005
    why is (8x)(-6x^3)= -48x^4? shouldn't it be to the third power?
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      Remember, exponents are shorthand for repetitive multiplication. So, multiplication / division will change an exponent.

      In this problem: x * x^3 means you are multiplying x * x*x*x
      There are 4 Xs being multiplied together, so the exponent needs to go up to 4.

      Hope this helps.
      (3 votes)
  • leaf green style avatar for user Alex Pyzhianov
    I was just thinking that multiplying both the numerator and the denominator of a standalone fraction by an expression that contains a variable (like 3x+1 or 8x+7) changes it, because it becomes undefined when x=-1/3 or x=-7/8. And there is no way to add a condition that would make it equal the original expression because you can't just say that it is defined at x=-1/3 for example because you still won't be able to evaluate the resulting expression, because what would you do with a zero in the denominator, that makes no sense. That does not mess up the example in the video though, because it is undefined at both of those Xes anyway
    (3 votes)
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  • boggle purple style avatar for user lily J
    I still don't understand how he could know it's totally simplified without factor the numerator.
    Thanks!
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      I agree, he should have factored the numerator. The only thing that could be factored in the numerator is to remove a common factor of "x". This would not help to reduce the fraction. The remaining portion of the polynomial is not factorable. This is likely why Sal stopped where he did. But, it would have helped for him to demonstrate that it is not factorable other than the GCF.
      (2 votes)

Video transcript

- [Voiceover] So right over here, we have one rational expression being subtracted from another rational expression. I encourage you to pause the video and see what this would result in. So actually do the subtraction. Alright, let's do this together. And if we're subtracting two rational expressions, we'd like to have them have the same denominator. And they clearly don't have the same denominator and so we need to find a common denominator. And a common denominator is one that is going to be divisible by either of these and then we can multiply them by an appropriate expression or number so that it becomes the common denominator. So the easiest common denominator I can think of especially because these factors, these two expressions have no factors in common, would just be their product. So this is going to be equal to, so we could just multiply these two. So this is going to be, so let me do, let me do this one right over here, in magenta. So this is going to be equal to the common denominator. If I say, if I want to just multiply those two denominators. For this one, I'll have my 8x plus seven and now I'm going to multiply by 3x plus one. Multiplying it by the other denominator and I had negative 5x in the numerator but if I'm going to multiply the denominator by 3x plus one and I don't want to change the value of the expression, we'll have to multiply the numerator by 3x plus one, as well. Notice, 3x plus one divided by 3x plus one is just one and you'll be left with what we started with. And from that, we are going to subtract all of this. Now there's a couple of ways you can think of this subtraction. I could just write a minus sign right over here and do the same thing that I just did for the first term. Or another way to think about it and actually for this particular case, I like thinking about it better this way, is to just add the negative of this. So if I just multiply negative one times this expression, I'd get negative 6x third over 3x plus one. If I had more terms up here, in the numerator, I would have to be careful to distribute that negative sign, but here, I only have one term. So I just made a negative and so I can say this is going to be plus and, let me do this, in a new color, do this in green. Our common denominator, we already established, is the product of our two denominators so it is going to be 8x plus seven times 3x plus one. Now if we multiply the denominator here was 3x plus one, we're multiplying it by 8x plus seven. So that mean's we have to multiply the numerator by 8x plus seven as well. 8x plus seven times negative 6x to the third power. Notice, 8x plus seven divided by 8x plus seven is one. If you were to do that, you would get back to your original expression right over here, the negative 6x to the third over 3x plus one. And now, we're ready to add. This is all going to be equal to, I'll write the denominator in white. So we have our common denominator, 8x plus seven times 3x plus one. Now, in the magenta, I would want to distribute the negative 5x, so negative 5x times positive 3x is negative 15x squared. And then, negative 5x times one is minus 5x. And then, in the green, I would have, let's see, I'll distribute the negative 6x to the third power, so negative 6x to the third times positive 8x is going to be, negative 48x to the fourth power. And then negative 6x to the third times positive seven is going to be negative 42, negative 42x to the third. And I think I'm done because there's no more, there's, you know, I only have one fourth-degree term, one third-degree term, one second-degree term, one first-degree term, and that's it. There's no more simplification here. Some of you might want to just write it in descending degree order, so you could write it as negative 48x to the fourth minus 42x to the third minus 15x squared minus 5x. All of that over 8x plus seven times 3x plus one. But either way, we are all done. I mean, it looks like up here, yeah, there's no, nothing to factor out. These two are divisible by five. These are divisible by six but even if I were to factor that out, nothing over here, down here, no five or six to factor out. Yeah, so it looks like we are all done.