Factoring Sucks
Well, to be specific, factoring trinomial expressions with a 2nd-degree coefficient that's not one sucks. Woe be to the algebra teacher who assigns this problem:
Factor: 6x2 + 16x + 10
Not only are there two distinct ways to factor this, it's a lot of work. But let's take a step back for a minute, and look at easy factoring:
Factor: x2 + 8x + 12
This is easy because we can just think about the factors of 12 (1*12, 2*6, 3*4) and immediately see which two factors can be combined to make 8 (in this case, 2 and 6). So there aren't many choices (even if the constant term has a lot of factors, like 12 or 60), and it's easy to reach the solution in our heads:
x2 + 8x + 12 = (x + 2)(x + 6)
Looking back at our original problem, we see that we first have to figure out how we're going to get our 2nd-degree coefficient. There are only two ways: 1*6 and 2*3, so that doesn't seem so bad. We know, then, that our solution will either look like:
(x + ...)(6x + ...)
or
(2x + ...)(3x + ...)
We also have a pretty easy time determining how to get the constant term: it's just the factors of 10 (1*10, 2*5). So we know the other numbers in the parenthesis will either be 1 and 10 or 2 and 5. So far so good.... But let's consider the possible combinations:
(x + 1)(6x + 10)
(x + 2)(6x + 5)
(2x + 1)(3x + 10)
(2x + 2)(3x + 5)
That doesn't seem too bad...but wait! We're not done yet. Unlike our simple factoring above, the ordering is important, so we also have the following possibilities:
(6x + 1)(x + 10)
(6x + 2)(x + 5)
(3x + 1)(2x + 1)
(3x + 2)(2x + 5)
So we've now got a total of 8 possible combinations. The worst part? We have to FOIL them all to figure out if we get that magical 16x term. Well, there is one shortcut we can take: we don't have to do the F part (because we know that we'll always get 6x2 in every case), nor do we have to do the L part (because we know that we'll get 10 in every case). So all we have to do is OI (OI!). So we get:
(x + 1)(6x + 10) O: 10x I: 6x
(x + 2)(6x + 5) O: 5x I: 12x
(2x + 1)(3x + 10) O: 20x I: 3x
(2x + 2)(3x + 5) O: 10x I: 6x
(6x + 1)(x + 10) O: 60x I: x
(6x + 2)(x + 5) O: 30x I: 2x
(3x + 1)(2x + 1) O: 3x I: 2x
(3x + 2)(2x + 5) O: 15x I: 4x
(Forgive the ugly font; it was the easiest way to get things to line up.) Well! Finally we can glance over the OI products and figure out which of them will give us our 16x, namely 10x and 6x. So we finally have two possible factorizations:
(x + 1)(6x + 10)
(2x + 2)(3x + 5)
It's worth noting here that if we set both of these expressions to zero, we'll end up with the same solutions. So, end the end, it doesn't matter which factorization we use.
I just spent an hour trying to figure out a way to make this nonsense easier and came up with...nothing. It's just an ugly, hard slog. So I challenge you, my dear readers, to come up with a method of solving these problems that isn't so painful. You've got mnemonics? Bring 'em on. A fancy table? I'd love to see it. Something ultra-clever? Maybe it's out there....
In the meantime, should you not give your students these problems? Hell no! Of course you should assign them these problems. Maybe not all the time, but don't get the little buggers get away with easy factorizations. Sometimes math is easy and elegant, and sometimes it's a hard dig in a cold ditch, and the sooner students realize that the better. Woe be to the student who makes it to differential equations thinking that there's an easy solution to every problem.
Well, to be specific, factoring trinomial expressions with a 2nd-degree coefficient that's not one sucks. Woe be to the algebra teacher who assigns this problem:
Factor: 6x2 + 16x + 10
Not only are there two distinct ways to factor this, it's a lot of work. But let's take a step back for a minute, and look at easy factoring:
Factor: x2 + 8x + 12
This is easy because we can just think about the factors of 12 (1*12, 2*6, 3*4) and immediately see which two factors can be combined to make 8 (in this case, 2 and 6). So there aren't many choices (even if the constant term has a lot of factors, like 12 or 60), and it's easy to reach the solution in our heads:
x2 + 8x + 12 = (x + 2)(x + 6)
Looking back at our original problem, we see that we first have to figure out how we're going to get our 2nd-degree coefficient. There are only two ways: 1*6 and 2*3, so that doesn't seem so bad. We know, then, that our solution will either look like:
(x + ...)(6x + ...)
or
(2x + ...)(3x + ...)
We also have a pretty easy time determining how to get the constant term: it's just the factors of 10 (1*10, 2*5). So we know the other numbers in the parenthesis will either be 1 and 10 or 2 and 5. So far so good.... But let's consider the possible combinations:
(x + 1)(6x + 10)
(x + 2)(6x + 5)
(2x + 1)(3x + 10)
(2x + 2)(3x + 5)
That doesn't seem too bad...but wait! We're not done yet. Unlike our simple factoring above, the ordering is important, so we also have the following possibilities:
(6x + 1)(x + 10)
(6x + 2)(x + 5)
(3x + 1)(2x + 1)
(3x + 2)(2x + 5)
So we've now got a total of 8 possible combinations. The worst part? We have to FOIL them all to figure out if we get that magical 16x term. Well, there is one shortcut we can take: we don't have to do the F part (because we know that we'll always get 6x2 in every case), nor do we have to do the L part (because we know that we'll get 10 in every case). So all we have to do is OI (OI!). So we get:
(x + 1)(6x + 10) O: 10x I: 6x
(x + 2)(6x + 5) O: 5x I: 12x
(2x + 1)(3x + 10) O: 20x I: 3x
(2x + 2)(3x + 5) O: 10x I: 6x
(6x + 1)(x + 10) O: 60x I: x
(6x + 2)(x + 5) O: 30x I: 2x
(3x + 1)(2x + 1) O: 3x I: 2x
(3x + 2)(2x + 5) O: 15x I: 4x
(Forgive the ugly font; it was the easiest way to get things to line up.) Well! Finally we can glance over the OI products and figure out which of them will give us our 16x, namely 10x and 6x. So we finally have two possible factorizations:
(x + 1)(6x + 10)
(2x + 2)(3x + 5)
It's worth noting here that if we set both of these expressions to zero, we'll end up with the same solutions. So, end the end, it doesn't matter which factorization we use.
I just spent an hour trying to figure out a way to make this nonsense easier and came up with...nothing. It's just an ugly, hard slog. So I challenge you, my dear readers, to come up with a method of solving these problems that isn't so painful. You've got mnemonics? Bring 'em on. A fancy table? I'd love to see it. Something ultra-clever? Maybe it's out there....
In the meantime, should you not give your students these problems? Hell no! Of course you should assign them these problems. Maybe not all the time, but don't get the little buggers get away with easy factorizations. Sometimes math is easy and elegant, and sometimes it's a hard dig in a cold ditch, and the sooner students realize that the better. Woe be to the student who makes it to differential equations thinking that there's an easy solution to every problem.
posted by Ethan Brown at 6:16 PM
1 Comments:
Good point -- especially about finding simple answers in DE! I think it's a good idea to let students struggle a little bit; it makes them ask questions they wouldn't normally think to ask. Thanks!
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