فاکتورینگ - چهارگوشه

سرفصل: بخش ریاضی / سرفصل: جبر، معادلات و نابرابری ها / درس 7

فاکتورینگ - چهارگوشه

توضیح مختصر

  • زمان مطالعه 7 دقیقه
  • سطح متوسط

دانلود اپلیکیشن «زوم»

این درس را می‌توانید به بهترین شکل و با امکانات عالی در اپلیکیشن «زوم» بخوانید

دانلود اپلیکیشن «زوم»

فایل ویدیویی

متن انگلیسی درس

Factoring - Quadratics

The most challenging topic in factoring is knowing how to factor a quadratic trinomial into the product of two linear binomials. So, for example, here’s a quadratic trinomial, three terms, highest power is x squared. And we want to factor this into two linear binomials. Each one of course has just a highest power of x.

That is the factored form. For example, on the test, we might have to factor something like x squared plus 4x minus 21. Now, I’ll point out, this is probably not gonna be an entire test problem. Yes, there may be an entire test problem that says, here’s a quadratic, factor it. But more likely, factoring the quadratic is a skill that you’re going to need in solving a much larger question.

But this is definitely a skill you will need. Let’s think about the details of the FOIL process. Understanding factoring really comes down to understanding the mechanics of foiling. Suppose we start with two generic linear binomials, x plus p and x plus q. So we’ll FOIL these out. Of course, product of the first is x-squared.

Product of the outer, qx. Product of the inner, px. And product of the last, pq. Now notice pq, that’s two numbers. That’s the constant term, x squared is the quadratic term, and the two middle terms are linear terms and they are like terms, so we can combine the like terms and we get this.

Now this is very interesting. Imagine we looked at our quadratic trinomial, and we see this. So, really that linear coefficient is p plus q, and that constant term is p times q. So, in other words, the linear coefficient of the quadratic is the sum of the two unknowns we are looking for.

And the constant term of the quadratic is the product of the two unknowns. Very interesting. Let’s apply this in a problem. Suppose we have to factor this. So, in other words, we’re looking for two numbers whose sum is 8, and whose product is 15.

Well, there aren’t many numbers we can multiply together to get 15. We could do 1 times 15, and that doesn’t have the right sum, or we can do 3 times 5. 3 and 5 are clearly the numbers and that means we can factor this into x plus 3 and x plus 5. That is the factored form.

Notice here, we have a statement that is true for every number on the number line. That’s why we can set an equal sign between these two expressions. We’ll factor this one. We need something, we need a pair, they have a sum of 14 and product of 24. Little bit trickier, because there are many factor pairs that multiply to 24. We could do 4 times 6, that’s not the right sum.

We could do 3 times 8, that’s not the right sum. We could do 2 times 12. That’s the right sum. And so the numbers that we need are 2 and 12, and this is the factor form, x + 2 and x + 12. This one’s a little trickier.

We have a negative sign. So here we’re looking for a sum of positive 4 and a product of negative 21. Let’s think about this. Since the product is negative, one of the two numbers must be positive, and one is negative. Since the sum is positive, it means we’re adding these two numbers and getting a positive.

That means the, the larger number, the number with the larger absolute value, must be positive. So, we’ll think about this. First of all think about factors of 21. Certainly 3 and 7 multiply to positive 21 so let’s make the larger of those two a positive number and the smaller of those two a negative number.

If we have positive 7 and negative 3. Those multiply to negative 21 and they add to positive 4. So those are the two numbers that we’re looking for, positive seven and negative three. And so that means that we can factor this into x plus 7 times x minus 3. And for all of these, incidentally, as you’re practicing factoring, I would suggest once you think you have the factored form, a very good way to practice is FOIL those two on the right and make sure you wind up back at the form on the left.

In other words, if you can factor one way and then FOIL back to what you’ve got, you really understand this very well. Here’s another one. Here we need a sum of negative 2 and a product of negative 35. So, we know that one is negative and one is positive of the two numbers we’re looking for because they have a negative product, and the one with the bigger absolute value is negative.

So what can we multiply to get 35 first of all. Well, one obvious choice is 5 times 7. So let’s make the one with the bigger absolute value negative so we’ll have positive 5 and negative 7. They multiply to negative 35 and they add to negative 2. So these are the numbers that we need and we can factor it in to x minus 7 and x plus 5.

Now this one’s interesting. Here we have a positive product but a negative sum. Well how does that happen? Well, this means that both the numbers are negative. Because in other words, you have a negative times negative, that will give you a positive, but if you have negative plus negative, that will give you a negative.

So both numbers must be negative. So we need two numbers that multiply to 48, and really one way to think about it is they multiply to 48 and have a sum of positive 16, and then we’ll just make them both negative. Well. Certainly 6 times 8, 6 times 8 has a sum of, 6 and 8 have a sum of 14, that doesn’t work.

We could do 4 times 12 4 times 12 has a sum of 16. So negative 4 times negative 12 would be positive 48, but negative 4 plus negative 12 would have a sum of negative 16. These are the two numbers that we need, so we can factor this into x minus 12 and x minus 4. Incidentally, notice once again with multiplication, it doesn’t matter the order that we write things in.

I could have written this as x minus 4 times x minus 12, same thing. It doesn’t matter at all because multiplication is commutative. We can write the factors in any order. Here are some practice problems, pause the video, and factor all of these. And here are the answers.

Notice that this method only works if the coefficient of x squared, the quadratic coefficient, is one. If the quadratic coefficient is something other than one, chances are very good that one of the other factoring methods can be used. Say, the difference of two squares or greatest common factor. It may mean that we have to factor out a greatest common factor.

And then we get an ordinary quadratic so we’ll have to talk about combined factoring techniques where something like that comes into play. And that’s what we’ll talk about in our next video. In this lesson, we learned the method for factoring a basic quadratic into into a product of two linear binomials. The linear coefficient of the quadratic is the sum of the two unknown numbers.

And the constant term of the quadratic is the product of the two unknown numbers.

مشارکت کنندگان در این صفحه

تا کنون فردی در بازسازی این صفحه مشارکت نداشته است.

🖊 شما نیز می‌توانید برای مشارکت در ترجمه‌ی این صفحه یا اصلاح متن انگلیسی، به این لینک مراجعه بفرمایید.