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Advanced Numerical Factoring
So now we’re gonna take a break from algebra, and go back to just pure numbers. Now remember, we will return to the idea of prime factorizations in the Integer Properties module. Now if you haven’t seen that module, I would highly recommend watching that lesson on prime factorization so you appreciate how important it is to find the prime factorization of a number.
Once we have the prime factorization, we can find all the other factors. We can find the number of factors. We can do any one of a number of things, if we have the prime factorization of a number. Well it turns out, no that we have learned these algebraic factoring techniques, especially the difference of two squares, we can use some of these to factor numbers that otherwise would be very hard to factor.
So for example. Suppose you have to find the prime factorization of 1599. Well hm. That would be hard. We could probably tell from our divisibility tricks, that, since the sum of these is, let’s see, 24, then that’s divisible by three, so the number would be divisible by 3.
But even if we divide this by 3, then we’re gonna get some ugly 3 digit number that’s not gonna be easy to factor. Well it turns out there’s a much easier way, a very elegant way, to get to the prime factorization. Simply notice, 1599 is 1600 minus 1. That’s a difference of squares because of course 1 squared is one, and 40 squared is 1600.
So what we have here is 40 squared minus 1 squared. Well following the difference in two squares pattern, this is 40 plus 1, that’s 40 minus 1, or in other words, 41 times 39. 41 is a prime number but 39 we can factor into 3 times 13, and these are the 3 prime factors. That’s the prime factorization of the number.
So, a very elegant way to get to the prime factorization of the number. Suppose we had to find the prime factorization of 2491. I’m gonna tell you at the outset, this would be a very hard number to factor, even if you had a handheld calculator in your hand, it would be hard to find the prime factorization of this number. Well, what can we do?
Well, notice this one also, is 2500 minus 9. 2500 is 50 squared, and of course 9 is 3 squared. So this is 50 squared minus 3 squared, a difference of squares. We can write this as 50 minus 3, times 50 plus 3. 47 times 53. That is the prime factorization.
Those are two prime numbers right there. And again that would be incredibly hard to find even if you had a handheld calculator at your disposal. But here, we can find it directly and elegantly using the difference of squares. Suppose we have to prime the prime factorization of 9975. Well, hm, looks like it would be divisible by 25, but then we’d get some other ugly number that we had to divide.
Hm. That would be difficult doing it the conventional way. But notice that this 10,000 minus 25. 10,000 is 100 squared. And of course, 25 is 5 squared. So this is 100 squared minus 5 squared.
100 plus 5, 100 minus 5. That gets us down to 105 minus, times 95. And we can factor each one of those. 105 is 5 times 21, and 95 is 5 times 19. 21 we can factor into 3 times 7, and here’s the complete prime factorization. 5 squared times 3 times 7 times 19.
So in an instant we get the prime factorization simply by making use of the difference of two squares formula. If there is a large 4 digit number on the test, and to solve the problem, you realize that you would need to have the prime factorization of the number, it is a very, it is very likely that the test is expecting you to recognize that that large number can be expressed with the difference of two squares.
This technique can also help with factoring many large numbers, and it can help with decimals. Hm, decimals let’s think about this. Consider this decimal, suppose for some reason we needed to find two numbers that we can multiply together to get that decimal. Well, hm, it doesn’t sound like a very fun thing to do.
But, notice, 0.9991 is 1 minus 0.0009. And that’s a difference of squares, because 0.0009 is 0.03 squared. So we can write this as the difference of squares. Use the pattern, 1 plus 0.03 times 1 minus 0.03. This gives us 1.03 times 0.97. And so we have expressed that decimal as a product of two decimals.
Here’s a practice problem. Pause the video, and then we’ll talk about this. Okay. So clearly we’re gonna write both of these as 1 minus something. The top is 1 minus .000049, the bottom is 1 minus .007, double 0, 7.
And notice that that top decimal 0.000049 is 0.007 squared. And so this means we can write the top in the difference of two squares pattern. We get a 1 plus 0.007, and a 1 minus 0.007. That second factor, the subtraction, appears in the denominator so we can cancel it. We’re left with one plus 0.007.
And we get 1.007. And that is, that is the quotient, and we get that very easily using the difference of squares pattern. Here’s another practice problem. Pause the video and then we’ll talk about this. Okay. In that fraction we’re gonna write each one as 1 minus something.
1 minus 0.000144. And then the denominator 1 minus 0.012. Well 0.012 squared is 0.00014, so on the top we have a difference of squares. We’ll cut directly to the difference of squares pattern, we’ll cancel the subtraction and then, now we just have addition and we can do the 1 minus 1.
They cancel and we’re left with 0.012. And so that entire ugly function at the beginning, simplifies to this lovely little decimal. Remember that the difference of two squares pattern can be powerful in finding the prime factorization of large numbers or in simplifying decimals just less than one.
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