Dividing Rational Expressions
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For dividing rational expressions, you will use the same method as you used for dividing numerical fractions: when dividing by a fraction, you flip-n-multiply.
To refresh, let's look at a division of numerical fractions:
- Perform the indicated operation:
To simplify this division, I'll convert it to multiplication by flipping what I'm dividing by; that is, I'll switch from dividing by a fraction to multiplying by that fraction's reciprocal. Then I'll simplify as usual:
In my answer above, can the 2s cancel off from the 20's? No! Of course not! Trying to do so would mean taking the original fraction, looking at it in the form of (20 + 0)/(20 + 7), which is approximately equal to 0.74, and claiming that it somehow equal to (1 + 0)/(1 + 7), which is approximately equal to 0.59, which obviously isn't true; or else just cutting off the 2s entirely, leaving me with 0/7 = 0, which is even worse!
There were no common factors that could be cancelled off, and we can never reach inside a number and cancel off digits. Therefore, the final form above is is as simplified as that fraction gets.
Division works for rational expressions as it does for numerical fractions.
How do you divide rational expressions?
To divide rational expressions, follow these steps:
- Convert the division to multiplicaion by flipping the second rational expression and replacing the "division" symbol with a "multiplication" sign.
- Factor all of the polynomials.
- Cancel off any common factors.
- Multiply out the resulting numerator.
The fourth step above may be omitted, if your instructor doesn't care. And, on occasion, an instructor will want the denominator multiplied out, in addition to the numerator. Make sure you are clear on what your instructor wants here.
What is an example of dividing rational expressions?
- Perform the indicated operation:
![[ (x^2 + 2x - 15) / (x^2 - 4x - 45) ] ÷ [ (x^2 + x - 12) / (x^2 - 5x - 36) ]](rational/mult16.gif){kind=small}
To simplify this, first I'll flip-n-multiply. Then, to simplify the multiplication, I'll factor the numerators and denominators, and then cancel any duplicated factors. My work looks like this:
![[(x^2+2x-15)/(x^2-4x-45)] ÷ [(x^2+x-12)/(x^2-5x-36)] = [(x^2+2x-15)/(x^2-4x-45)] × [(x^2-5x-36)/(x^2+x-12)] = [(x+5)(x-3)(x-9)(x+4)]/[(x-9)(x+5)+4)(x-3)] = 1](rational/mult17.gif)
I cancelled out everything in the denominators. Yes, I cancelled out everything from the numerators, too, but numerators don't create division-by-zero problems; those come from denominators.
Because of the disappeared denominators, I have to take note that the variable cannot equal any of −5, −4, 3, and 9. To make sure that my simplified form is equal to what they'd given me in the first place, I'll need to add these restrictions to my answer.
Then my hand-in answer is:
1, for x ≠ −5, −4, 3, 9
Your instructor may not require the restrictions on the allowable x-values, in which case your answer would be just "1". But don't guess or assume; specifically ask your instructor regarding including these restrictions with your answers.
Of course, the exercises you'll be given won't usually simplify as much as did the above example. The following exercise is much more typical:
- Simplify the following expression:
![[ (x^2 + 3x - 40) / (x^2 + 2x - 35) ] ÷ [ (x^2 + 2x - 48) / (x^2 + 3x - 18) ]](rational/mult19.gif){kind=small}
First, I'll need to flip the second fraction, and convert from division to multiplication. Then I'll factor, and see if anything cancels.
![[(x^2+3x-40)/(x^2+2x-35)] ÷ [(x^2+2x-48)/(x^2+3x-18)] = [(x^2+3x-40)/(x^2+2x-35)] × [(x^2+3x-18)/(x^2+2x-48)] = [(x+8)(x-5)(x+6)(x-3)/(x+7)(x-5)(x+8)(x-6)] = [(x+6)(x-3)]/[(x+7)(x-6)]](rational/mult20.gif)
Can I cancel off the 6's? or any of the x's? No! I can only cancel off entire factors, and none of the remaining factors match; the above is as simplified as this gets!
I do note, however, that I cancelled two factors from denominators, which made two division-by-zero issues magically disappear. But this disappearance changes the domain; as it stands, my simplified expression is not equal to the original expression, because the variable in my simplified form is allowed to be equal to −8 and 5. To make my version truly equal to the original expression they gave me, I'll need to cut those two values out of the domain. So my final answer is:
![[ (x + 6)(x - 3) ] / [ (x + 7)(x - 6) ] = [x^2 + 3x - 18] / [ (x + 7)(x - 6) ] for x not equal to -8 or 5](rational/mult21.png){kind=small}
For reasons which will become clear when adding and subtracting rationals, the numerator is usually multiplied out (that is, the numerator is usually "simplified") to get rid of the parentheses, while the denominator is usually left in factored form.
Make sure you know how to factor quadratics and cubics, because, as you have seen, it is required for many of the problems you'll be doing. Also, make sure you are careful to cancel only factors, not terms. If you can keep these things straight in your head, then you'll probably do fine.
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