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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).

## Want to join the conversation?

• 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)?
• The order does not matter (8x+7)(3x+1) is the same as (3x+1)(8x+7) based upon the commutative property of multiplication.
Hope this helps.
• Why doesn't he multiply out the (8x+7)(3x+1).
• I think he may not have multiplied them out because it is extra work, and is unnecessary work, as the answer he gave is considered acceptable by most teachers.
Also, it is easier to simplify the fraction, or see if you can simplify it, when the fraction is not multiplied out.
• Why doesn't he FOIL the denominator at , at the end?
• 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.
• 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?
• 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.
• Will the answer still be correct if we multiply the denominator instead of it being factored in the final answer?
• 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.
• how do you do 2x+3/x-4 - x-5/x+2 as a
rational expressions
• 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.
• why is (8x)(-6x^3)= -48x^4? shouldn't it be to the third power?
• 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.
• 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
• I still don't understand how he could know it's totally simplified without factor the numerator.
Thanks!
• 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.