Factoring by GroupingDate: 11/24/97 at 20:36:09 From: Tammy Subject: Factoring by grouping This is Algebra II. The directions say to factor the expression by grouping. Here is the problem: xy + ay - bx - ab Here is what I did so far. I put xy+ay together making it y(x+a) I put bx-ab together making it b(x-a) now my problem is y(x+a)-b(x-a). I know that somehow, by making the first (x+a) negative, I can use that as the first part. It would be something like (a-x)-(y-b) but I am not sure. How do I get (x+a) and (x-a) to be the same and how do I do this problem? Some other problems that I am having trouble on are: 2xy - ay - 6bx + 3ab x^2 + 2x - 3x^2 - 6 ANY help that you could give me would be greatly appreciated. Tammy Date: 12/01/97 at 12:16:01 From: Doctor Mark Subject: Re: Factoring by grouping Hi Tammy, First of all, when you tried to factor by grouping the first two terms, and the second two terms, it actually did work, though you made a small mistake that made it look like it didn't. You wrote xy + ay - bx - ab = [xy + ay] - [bx - ab] But that's not quite right: isn't - [bx - ab] = -bx + ab? (You want it to equal - bx - ab!). What you should have gotten was: xy + ay - bx - ab = [xy + ay] - [bx + ab] Don't feel *too* bad, though; this is the kind of "stupid" mistake that smart people make *all the time*; you just have to be extra careful around minus signs, since they can drive you *nuts* if you're not careful. I've probably done about a million problems like this, and I still make minus sign mistakes (though not that often anymore). So let's talk about factoring by grouping... Factoring by grouping when the polynomial has four terms is really easy, though books don't make it look like it is. Here's how to do it. Let's look at your example: xy + ay - bx - ab. We can group the first term (xy) with either of the three remaining terms (with the ay, the - bx, or the - ab). Let's try each of those possibilities: (1) xy + ay - bx - ab = [xy + ay] - [bx + ab] (grouping the first with the second, and remembering that - bx - ab = - [bx +ab]). (2) xy + ay - bx - ab = [xy - bx] + [ay - ab] (grouping the first with the third) (3) xy + ay - bx - ab = [xy - ab] + [ay - bx] (grouping the first with the fourth) Notice that when we group the first term with one of the other terms, the remaining two terms automatically get grouped together. Now, for each of these three groupings, we look for a common factor of each of the pairs we grouped together: For 1: [xy + ay]: common factor of y, so take it out: [xy + ay] = y(x + a) [bx + ab]: common factor of b, so take it out: [bx + ab] = b(x + a) Now look at the quantities in the two parentheses. Are they the same? Here, yes, they are both equal to (x + a). That means that we can factor out (x + a) : xy + ay - bx - ab = [xy + ay] - [bx + ab] = y(x + a) - b(x + a) = (y - b)(x + a) and we're done: there's no reason to look at numbers 2 or 3, since we already have the answer, but let's see how it would go anyway. For 2: [xy - bx]: common factor of x, so take it out: [xy - bx] = x(y - b) [ay - ab]: common factor of a, so take it out: [ay - ab] = a(y - b) Now look at the quantities in the two parentheses. Are they the same? Here, yes, they are both equal to (y - b). That means that we can factor out (y - b), obtaining: xy + ay - bx - ab = = [xy -bx] + [ay - ab] = = x(y - b) + a(y - b) = = (x + a)(y - b). Notice that this is the same one we got before, except that the order of the factors is reversed. Now let's try grouping 3: [xy - ab] : no common factors here [ay - bx] : and no common factors here either. What this means is that we can't factor this polynomial by grouping the first term with the fourth term (and hence the second with the third). This doesn't mean that we can't factor the polynomial by grouping (how could it? we already factored it using number 2 or number 3!), it just means that we can't factor the polynomial by grouping in this way! So here's the deal: either a polynomial can be factored by grouping, or it can't. If it can be factored by grouping, then two of the three groupings will "work," and the other one won't. Which ones will work? You can't tell until you try it. If the polynomial cannot be factored by grouping, then none of the three groupings will work. Said another way, try grouping the first term with the second. If that "works," then fine: you're done. If it doesn't "work," try grouping the first term with the third term. If it works, fine; if it doesn't, try grouping the first with the fourth term. If that works, fine; if it doesn't, then since none of the three possible groupings "works," the polynomial cannot be factored by grouping. Here's something to think about: if you've tried two of the groupings, and neither "worked," then you don't have to try the third grouping: it won't work either. Do you see why? Let's try this method on one of the other examples: x^2 + 2x - 3x^2 - 6. Now I don't think you meant to write that third term as - 3x^2; I think you meant to write it as - 3x. If you *did* mean it to be -3x^2, you can still factor it by grouping, but it would be easier to just combine the x^2 and the - 3x^2 as x^2 - 3x^2 = - 2x^2, and then factor the equivalent polynomial: - 2x^2 + 2x - 6 = - 2(x^2 - x + 6). The polynomial in () cannot be factored (at least not "over the integers"), so we're done. So let's try factoring the polynomial x^2 + 2x - 3x - 6, using grouping. Now it would be simpler to just combine the two linear terms and factor as usual, but let's see how it goes using grouping: First, let's try grouping the first term (x^2) with the second term (2x): x^2 + 2x - 3x - 6= [x^2 + 2x] - [3x + 6]. {It is + 6, right Tammy? ;<) } The terms in the first group, [x^2 + 2x], have a common factor of x, so take it out: [x^2 + 2x] = x(x+2). The terms in the second group, [3x + 6], have a common factor of 3, so take it out: [3x + 6] = 3(x + 2). So this way of grouping works, since we found a common factor of (x + 2): (remembering the minus sign!) x^2+2x-3x-6 = [x^2 + 2x] - [3x + 6] = x(x+2) - 3(x + 2) = (x - 3) (x + 2) Since this grouping "worked," you'd have to be a math masochist to try the others, but let's see: Grouping the first with the third: x^2+2x-3x-6 = [x^2 - 3x] + [2x - 6] = x(x - 3) + 2(x - 3) = (x + 2)(x - 3) which is the same answer as before, but with the order of the factors reversed. Grouping the first with the fourth: x^2+2x-3x-6 [x^2 - 6]+[2x - 3x] = [x^2 - 6]+x[2 - 3] = [x^2 - 6]+x[-1] The two groups do not have a common factor, so this grouping doesn't work. Hope this was of help Tammy, and be sure to write back if there was anything I said that was not clear. -Doctor Mark, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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