Simplifiying Rational ExpressionsDate: 11/29/2004 at 11:16:29 From: marielys Subject: how do I solve rational expressions How do I solve something like this: x + 2 x -------- - ------------- x^2 - 36 x^2 + 9x + 18 We've been working on factoring, and I was doing fine until this section came along. Date: 11/29/2004 at 14:28:08 From: Doctor Ian Subject: Re: how do I solve rational expressions Mi Marielys, Let's look at an example using simpler expressions. Suppose I have 3 2 -- + -- = ? 2x 3y and I want to simplify this. I need a common denominator, right? To get a common denominator, I can multiply each fraction by ()/(), where () is the denominator of the other fraction: 3(3y) 2(2x) 3(3y) + 2(2x) 9y + 4x ------ + ------ = ------------- to give an answer of ------- 2x(3y) 3y(2x) (2x)(3y) 6xy Does this make sense? Your problem is the same, but with more complicated expressions. So in your first problem, you could do this: (x+2)(x^2+9x+18) (x)(x^2-36) ------------------- - ------------------- (x^2-36)(x^2+9x+18) (x^2+9x+18)(x^2-36) Now, before you have a heart attack at the prospect of having to simplify this monster, do you see why it's really the same concept you've always used to add fractions, finding a common denominator and renaming each fraction so it has that denominator? Once you have that down, the rest is just a matter of dealing with increased complexity. Okay, one way to deal with increased complexity is to write things in terms of factors (which is why you are doing this after learning factoring), rather than sums. For example, note that x^2 - 36 is a difference of squares, which can be factored thus: x^2 - 36 = x^2 - 6^2 = (x + 6)(x - 6) So we can rewrite our expression as (x+2)(x^2+9x+18) (x)(x+6)(x-6) --------------------- - --------------------- (x+6)(x-6)(x^2+9x+18) (x^2+9x+18)(x+6)(x-6) It's probably not a coincidence that x^2 + 9x + 18 can be factored too: x^2 + 9x + 18 = (x + 3)(x + 6) So we can make this substitution, too: (x+2)(x+3)(x+6) (x)(x+6)(x-6) -------------------- - -------------------- (x+6)(x-6)(x+3)(x+6) (x+3)(x+6)(x+6)(x-6) There's no more factoring, so we can go ahead and do the addition: (x+2)(x+3)(x+6) - (x)(x+6)(x-6) ------------------------------- (x+6)(x-6)(x+3)(x+6) Now, it's tempting to just go ahead and expand all the factors out again, and do the addition; but one big benefit of having done the factoring is that we can use the distributive property to get some cancellations going: (x+2)(x+3)(x+6) - (x)(x+6)(x-6) ------------------------------- (x+6)(x-6)(x+3)(x+6) (x+6)[(x+2)(x+3) - (x)(x-6)] = ---------------------------- (x+6)(x-6)(x+3)(x+6) (x+2)(x+3) - (x)(x-6) = --------------------- (x-6)(x+3)(x+6) That may not seem like a big help, but it means that if we _do_ want to expand the numerator, we can deal with a quadratic expression instead of a cubic. I don't know about you, but I see that as a win. Okay, expanding, we get (x^2+5x+6) - (x^2-6x) = --------------------- (x-6)(x+3)(x+6) x^2 + 5x + 6 - x^2 + 6x = ----------------------- (x-6)(x+3)(x+6) 5x + 6 + 6x = --------------- (x-6)(x+3)(x+6) 11x + 6 = --------------- (x-6)(x+3)(x+6) Now, that was kind of a mess, but not too bad. What I sort of expected to happen was to end up with something factorable in the numerator, e.g., (x-6)*(...) ------------- (x-6)(x+3)(x+6) so that another of the binomials in the denominator would cancel. Now, note that I could have factored even sooner, and that would have been helpful: (x+2) (x) -------- - ----------- (x^2-36) (x^2+9x+18) (x+2) (x) = ---------- - ---------- (x+6)(x-6) (x+6)(x+3) Now, instead of just taking the 'obvious common denominator', I can find the _least_ common denominator: (x+2)(x+3) (x)(x-6) = --------------- - --------------- (x+6)(x-6)(x+3) (x+6)(x+3)(x-6) (x+2)(x+3) - (x)(x-6) = --------------------- (x+6)(x-6)(x+3) which gets us to the same place before, but with a little less pain. Anyway, the basic idea is always the same: Factor as soon as possible, find a common denominator, do the addition, expand the numerator, factor the numerator again if possible, and look for opportunities to cancel binomial (or other) factors. As I noted earlier, it's really the same sort of thing you've already been doing forever. The hardest thing is to avoid being intimidated. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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