Class 4 -- Expressions II I. Quadratics A. Multiplying Binomials B. Factoring When a = 1 C. Factoring When a != 1 D. Differences of Squares II. Algebraic fractions A. Cancelling B. Multiplying and Dividing C. Adding and Subtracting In the last class we introduced the idea of factoring: taking an algebraic expression and breaking it into component parts. Today we'll resume our discussion of factoring, starting with the most important application: factoring quadratics. I. Quadratics A. Multiplying binomials Before we factor, let's consider the reverse operation. Let's use the distributive property to multiply two things in parentheses. (x - 4)(y + 6) We can rewrite this as follows: (x - 4)(y + 6) = x(y + 6) - 4(y + 6) = xy + 6x - 4y - 24 At each step we simply used the properties we've already derived. This leads to a general rule for multiplying things like this (a + b)(c + d) = ac + ad + bc + bd The easy way to remember this is FOIL. FOIL stands for FOIL = first, outside, inside, last We got 4 terms when we multiplied the two binomials. The first one is xy, which is just the product of the first terms of each binomial. So first means multiply the first two terms, in this case x and y. Next we have 6x, which is just the product of the two outside terms, x and 6. After that, we have -4y, which is the product of the inside two terms, -4 and y. Finally, we have -24, which is the product of the two last terms, -4 and 6. So FOIL means that to multiply two binomials, multiply first, then outside, then inside, then last, and add them all together. It's just a useful mnemonic. Practice: multiply (2x - 1)(3x + 6) B. Factoring When a = 1 Expressions that look like the one we just got, ax^2 + bx + c, where a, b, and c are numbers, are called quadratics. We often get expressions of this sort from FOILing. It is often quite useful to be able to go backwards and figure out what two things we had to FOIL together to make a given quadratic. This is called factoring the quadratic. Let's consider an example. In this case, we'll start with the binomials, then get the quadratic, then try to see how to go backwards. (on board) (x + 3)(x - 2) Let's multiply these using FOIL. Someone tell me how to do that. (let class walk through it) (on board) (x + 3)(x - 2) = x^2 - 2x + 3x - 6= x^2 + x - 6 So, suppose we were presented with x^2 + x - 6 and asked to factor it. How would we do that? We know it's going to be something of the form (on board) (_x + _) (_x + _) The challenge is to figure out what numbers to fill into the blanks. The first thing to notice is the first part of foil. The quadratic we have starts with 1x^2, not 2x^3, or 3x^2, or something like that. Thus, we know that the coefficients in front of the x's in our fill-in problem have to give us 1. Since we're dealing with all integers here, and we want to keep it that way, this means there's only one possibility: the fill-in slots in front of x are both 1's. This gives (on board) (x + _) (x + _) Now we need to fill in the last two slots. Now we can look at the "last" part of foil. The last terms have to multiply to give -6, so we need to try pairs of numbers that, when multiplied, give -6. There's no way to do this other than by trial and error. So let's make a list of integers we can multiply to get -6. Someone tell me some. (let class make list) (on board) Possibilities for -6: 1,-6 -1,6 -2,3 2,-3 Note that we have to consider all possible permutations of where the - sign could be. Now we just have to try these possibilities one by one until one of them works. Let's do that: (on board) (x + 1)(x - 6) = x^2 - 6x + x - 6 = x^2 - 5x - 6 -- doesn't work (x - 1)(x + 6) = x^2 + 6x - x - 6 = x^2 + 5x - 6 -- doesn't work (x - 2)(x + 3) = x^2 + 3x - 2x - 6 = x^2 + x - 6 -- this work Once we find the right pair, we're done. We've factored the quadratic: x^2 + x - 6 = (x - 2)(x + 3) Let's try another example, this time without knowing the answer in advance. (on board) Factor x^2 - 8x + 16. How do we do this? (let class walk through it) (on board) (x + _)(x + _) (on board) Possibilities for 16: 1, 16 -1, -16 2, 8 -2, -8 4, 4 -4, -4 (on board) Trying each: (x + 1)(x + 16) = x^2 + 17x + 16 -- doesn't work (x - 1)(x - 16) = x^2 - 17x + 16 -- doesn't work (x + 2)(x + 8) = x^2 + 10x + 16 -- doesn't work (x - 2)(x - 8) = x^2 - 10x + 16 -- doesn't work (x + 4)(x + 4) = x^2 + 8x + 16 -- doesn't work (x - 4)(x - 4) = x^2 - 8x + 16 -- this works (on board) x^2 - 8x + 16 = (x - 4)(x - 4) C. Factoring When a != 1 Things are only slightly more complicated when the thing in front of the x^2 isn't a 1. Let's consider: (on board) Factor 2x^2 + 11x + 12 We can go back to our template for what the answer has to look like: (on board) (_x + _)(_x + _) Looking at the first term, we notice that we have 2x^2, so we're going to have to multiply two things to get 2. We can make a possibilities list here: (on board) Possibilities for 2: 1,2 -1,-2 We make the same list for 12: (on board_) Possibilities for 12: 1,12 -1,-12 2,6 -2,-6 3,4 -3,-4 To check these systematically, unfortunately, we need to consider every possible was of combining the pairs for 2 and for 12. For the pairs (1,2) and (1,12), note that there are two possible ways of combining them. (on board) (1x + 1)(2x + 12) (2x + 1)(x + 12) For the pairs (1,2) and (2,6) we have: (on board) (1x + 2)(2x + 6) (2x + 2)(x + 6) We need to check each possible combination, and two orders for each combination. As you can see, this quickly becomes extremely long. There is a way of shortcutting this process, which we'll learn next week. However, for now there's nothing to do but try to use some intuition to see which pair can work. It helps to be able to do the multiplications in your head, since that's quicker than writing them down. Does anyone have a guess for the combination they'd like to try here? (let class answer) The on that works turns out to be: (on board) (2x + 3)(x + 4) = 2x^2 + 8x + 3x + 12 = 2x^2 + 11x + 12 Really the only way to be able to do these quickly is to practice a lot of them. However, I can point out some tips that may help shorten your search. These are rules of thumb to keep in mind. (on board) 1. If all the numbers in the quadratic are positive, don't bother checking negative possibilities. 2. You get larger numbers in the middle by taking possibilities for first and last that are far apart, and smaller numbers in the middle by taking first and last possibilities that are close to each other. D. Differences of Squares Before we move on, I'd like to point out one special case of factoring ax^2 + bx + c: the case when b, the coefficient of the x term, is zero. Consider (on board) x^2 - 9 How can we factor this, using our standard procedure? (let class answer) (on board) Possibilities for -9: 1,-9 -1,9 -3,3 Using our trick, we note that we're looking for a pair of numbers that add up to the coefficient of x -- which is zero in this case. Looking at this list, we can see that the answer has to be -3, 3, since only they add up to 0. Thus, the answer is (on board) (x-3)(x+3) We can see a pattern that's going to be true whenever we have a difference of squares, something like (on board) x^2 - a^2 The answer is just going to be: (on board) (x - a)(x + a) Thus, for example, what is the factorization of (on board) x^2 - 64 (let class answer) (on board) (x - 8)(x + 8) III. Fractions Let's end by discussing another type of algebraic expression: algebraic fractions. These are just things where there is one algebraic expression divided by another, for example (6x + 3) / (2x + 1). Things that look like this are also called rational expressions. A. Cancelling Factoring 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) B. Multiplying and Dividing 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. C. Adding and Subtracting 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.