Factoring by Grouping
Date: 05/27/98 at 18:26:44 From: Ashley Bullard Subject: Algebra 1(factoring by grouping) How do you factor 2r^2 + rt - 6t to the second power? So far I have multiplied the ends of the equation (2r^2 and -6t^2). What do I do now?
Date: 06/01/98 at 13:12:15 From: Doctor Peterson Subject: Re: Algebra 1 (factoring by grouping) Hi, Ashley. This is about the hardest type of problem you'll see for this sort of factoring, because both squared terms have coefficients (that is, multipliers, which in your problem are 2 and -6) that are not one. Let's go through a problem very much like yours (so you can try out what you learn on your own), and then if you need another example you can look on the Web at: http://mathforum.org/dr.math/problems/thorne11.14.95.html for a problem involving bigger and uglier numbers. Here's my problem: Factor 3r^2 - 7rt - 6t^2 We want to get this in the form: (ar + bt)(cr + dt) which multiplies out to: ac*r^2 + (ad + bc)*rt + bd*t^2 You don't have to memorize that. The important thing is that the coefficient of r^2 will be the product of a and c, and the coefficient of t^2 will be the product of b and d. So the first thing to do is to factor the 3 and the 6, and list (at least in your mind) all the products that can produce them: a c ------- 3 = 1 * 3 or 3 * 1 b d ------- -6 = -1 * 6 or -2 * 3 or -3 * 2 or -6 * 1 or 1 * -6 or 2 * -3 or 3 * -2 or 6 * -1 (Maybe this is what you meant when you said you "multiplied the ends." There's no reason to multiply them, but factoring them is the first step.) I don't usually make tables like this, and I don't think about a, b, c, and d; I just write down some parentheses and try filling them in: (r t)( r t) Now we just have to search for a set of a, b, c, and d for which (ad + bc) is -7. The way I do this is to try writing the factored expression with some choice of values and see how it looks. In this case, it doesn't really matter whether I take a and c to be 1 and 3 or 3 and 1, because that would just switch the order of the factors I end up with, so the only real decision is what to try for b and d. I usually start with something more or less balanced (2 * 3 rather than 1 * 6) just to see what will happen: Goal: 3r^2 - 7rt - 6t^2 3 6 / \ / \ / \/ \ / /\ \ (1r 2t)(3r 3t) \ \ / / \ 6 / \ +/- / -- 3 -- === 7 ? The r terms give 3r^2 and the t terms give 6t^2. (We know we'll need a + and a -, but we'll leave that for later.) Try multiplying the two rt pairs: 1r * 3t = 3rt, and 2t * 3r gives 6rt. Subtracting in either order won't give us -7, so this won't work. Let's try something else. Maybe switch the factors of 6: 3 6 / \ / \ / \/ \ / /\ \ (1r 3t)(3r 2t) \ \ / / \ 9 / \ +/- / -- 2 -- === 7 ? Now the rt terms are 1r * 2t = 2rt and 3t * 3r = 9rt. We can combine these to make -7: 2rt - 9rt. So we put a minus sign where it will multiply the 9rt term: (1r - 3t)(3r + 2t) = 3r^2 - 7rt - 6t^2 So we've factored it. As you can see, there's a lot of trial and error. If you don't feel like you know where you're going all the time, don't worry - we're all just feeling around when we do this. When you get enough experience, you get used to the process so it gets a little more orderly. -Doctor Peterson, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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