Class 8 I. Rational Expressions A. Definition B. Domains C. Simplifying II. Multiplying and Dividing Rational Expressions III. Adding and Subtracting Rational Expressions I. Rational Expressions A. Definition This week we'll introducing something closely related to polynomials: rational expressions. A rational expression is just one polynomial divided by another. For example, the following are all rational expressions: (x^2 + 2) / (x - 1) 1 / (x - 10) (xy^3 + z^2 - xyz) / (xz + y^2) We encountered things like this a few weeks ago when we introduced the idea of dividing polynomials. On the other hand, these are not rational expressions: sqrt(x) 2^x B. Domains A rational expression involving variables isn't always well defined for all possible values of the variable. Here's an example: 1/(x-1) is a rational expression. I can evaluate this for x = 2, 0, and -5, for example: 1/(2-1) = 1 1/(0-1) = 1/(-1) = -1 1/(-5-1) = 1/(-6) = -1/6 However, I can't evaluate it for x = 0, because the denominator would be zero. We can therefore describe the domain of a rational expression as the set of x values, or values of whatever variable, that I can use without making the denominator 0. Example: What's the domain of x/(x^2 + 2x - 3)? To figure this out, we need to factor the denominator and see what values of x make it zero. We therefore want to solve: x^2 + 2x - 3 = 0 (x - 1)(x + 3) = 0 x = 1, -3 The domain is therefore all real numbers except 1 and -3, because either x=1 or x=-3 will make the denominator 0. C. Simplifying Factoring also gives us a tool for simplifying rational expressions. Just like we can simplify fractions by getting rid of common factors, we can simplify rational expressions. Example: 4/8 = (4 . 1) / (4 . 2) = 1/2 Similarly, consider (x^2 + 5x + 6) / (2x^2 + 5x + 2) To simplify this, we factor both the numerator and the denominator: x^2 + 5x + 6 = (x+2)(x+3) 2x^2 + 5x + 2 = (2x+1)(x+2) Therefore (x^2 + 5x + 6) / (2x^2 + 5x + 2) = [(x+2)(x+3)] / [(2x+1)(x+2)] We can cancel out the common factor, which gives (x^2 + 5x + 6) / (2x^2 + 5x + 2) = [(x+2)(x+3)] / [(2x+1)(x+2)] = (x+3) / (2x+1) Let's try one more example like this: Simplify: (x^3 - 8) / (x^2 - 7x + 10) Again, we have to factor: x^3 - 8 = (x-2)(x^2 + 2x + 4) x^2 - 7x + 10 = (x-2)(x-5) So: (x^3 - 8) / (x^2 - 7x + 10) = [(x-2)(x^2 + 2x + 4)] / [(x-2)(x-4)] = (x^2 + 2x + 4) / (x-4) II. Multiplying and Dividing Rational Expressions Now that we know how to simplify, we are ready to multiply and divide rational expressions. Again, this process is exactly the same as it is for fractions. To multiply, you just multiply the tops and the bottoms. Usually it is easiest if you factor the tops and the bottoms first, because that way you can cancel out common factors before multiplying rather than after. That will save you work. Example: [(9 - n) / (n^3 - 64)] . [(n^2 + 5n - 36) / (n^2 - 81)] Step 1: Factor everything, and rewrite in factored form. 9 - n is already factored n^3 - 64 = (n - 4)(n^2 + 4n + 16) n^2 + 5n - 36 = (n + 9)(n - 4) n^2 - 81 = (n - 9)(n + 9) So the problem becomes: {(9 - n) / [(n-4)(n^2+4n+16)]} . {[(n+9)(n-4)] / [(n-9)(n+9)]} Step 2: Cancel out common factors. For this problem, n-4 and n+9 are obviously common factors. It's a little less obvious that the 9-n and n-9 are common factors as well. That's because 9-n = -1(n-9), so (9-n)/(n-9) = -(n-9)/(n-9) = -1 Thus, after cancelling all the common factors, we get [-1/(n^2+4n+16)] . [1/1] Everything in the second term cancelled out, and we were just left with 1/1! Step 3: Multiply what's left. In this case there's no work left to do, so we have the answer: -1 / (n^2+4n+16) Note that, there is no step 4, write things in lowest terms, because if we factored first and cancelled out everything we could, there is no reduction left to be done. Division works the same way as for fractions: just flip the second term, then multiply. Example: [(3a^2 - 13ab + 4b^2) / (3a^3b - 4a^2 b^2 + ab^3)] / [(7b^2 - 6ab - a^2) / (21 a^3b^2 + 3a^4 b)] Turning this into a multiplication: [(3a^2 - 13ab + 4b^2) / (3a^3b - 4a^2 b^2 + ab^3)] / [(21 a^3b^2 + 3a^4 b) / (7b^2 - 6ab - a^2)] Now we do what we'd do for any other multiplication. Factor first: 3a^2 - 13ab + 4b^2 = (3a - b)(a - 4b) 3a^3b - 4a^2 b^2 + ab^3 = ab (3a^2 - 4ab + b^2) = ab (3a - b) (a - b) 21 a^3b^2 + 3a^4 b = 3a^3b (7b + a) 7b^2 - 6ab - a^2 = (7b + a)(b - a) Write the multiplication in factored form: {[(3a - b)(a - 4b)] / [ab (3a - b) (a - b)]} . {[3a^3 b (7b + a)] / [(7b + a)(b - a)]} Cancel where possible: [(a - 4b) / (a - b)] . [3a^2 / (b - a)] Then just combine the results to get the final answer 3a^2 (a - 4b) / [ (a-b)(b-a) ] That's it. III. Adding and Subtracting Rational Expressions Addition and subtraction of rational expressions is also closely analagous to addition and subtraction of fractions. One has to find a common denominator by factoring before adding or subtracting. Here's a basic example: 4r/(r^2 - 6r + 9) + r/(r^2 - 11r + 24) Step 1: Factor everything. In this problem the numerators are already factored, so we just have to factor the denominators: r^2 - 6r + 9 = (r - 3)(r - 3) r^2 - 11r + 24 = (r - 3)(r - 8) Step 2: Rewrite in factored form. 4r / [(r-3)(r+3)] + 2r / [(r-3)(r-8)] Step 3: Find the least common denominator. The least common denominator is found by just taking the smallest combination of factors that contains all the factor in the denominator of every term. In our example, one denominator is (r-3)(r+3), and the other is (r-3)(r-8). The least common denominator is therefore (r-3)(r+3)(r-8). It contains every factor in either denominator. Step 4: Multiply top and bottom of both terms to convert to common denominator. {4r / [(r-3)(r+3)]}.{(r-8)/(r-8)} + {2r / [(r-3)(r-8)]}.{(r+3)/(r+3)} = [4r^2-32r] / [(r-3)(r+3)(r-8)] + [2r^2+6r] / [(r-3)(r+3)(r-8)] Step 5: Add or subtract the numerators. 6r^2 + 26r / [(r-3)(r+3)(r-8)] Step 6: Factor the numerator and reduce if possible 3r(r+13) / [(r-3)(r+3)(r-8)] In this case no reduction is possible, so we're done. Let's try another example: (1 - 2n) / (n^2 + 6n + 9) - (1 - 5n) / (n^2 - 3n - 18) Step 1: Factor n^2 + 6n + 9 = (n+3)(n+3) n^2 - 3n - 18 = (n+3)(n-6) Step 2: Rewrite in factored form (1-2n) / [(n+3)(n+3)] - (1-5n) / [(n+3)(n-6)] Step 3: Find LCD In this case the LCD is (n+3)(n+3)(n-6). Note that since (n+3) appears twice in one of the denominators, it must appear twice in the LCD. Step 4: Multiply to put in terms of LCD. {(1-2n) / [(n+3)(n+3)]} . {(n-8)/(n-8)} - {(1-5n) / [(n+3)(n-6)]} . {(n+3)/(n+3)} = (-2n^2 + 17n - 8) / [(n+3)(n+3)(n-8)] - (-5n^2 - 14n + 3) / [(n+3)(n+3)(n-8)] Step 5: Subtract. = (3n^2 + 31n - 11) / [(n+3)(n+3)(n-8)] The numerator can't be factored, so we're done.