Hey, everyone. So in math and algebra, so far, we've seen how to multiply numbers and polynomials. I could take, like, 2 times 3, and I know that multiplies to 6. And I can take this expression and use the distributive property and know that it multiplies to \(x^2 + 3x\). Well, in some problems now, they're actually going to give you the right side of the equation, and they're going to ask you for the left side. They're asking you for what things you have to multiply to get here. And that's a process called factoring, and we're going to take a look at how to do that for polynomials. And what I'm going to show you here is that, basically, whereas multiplication was taking simpler terms and multiplying into more complicated expressions, now we're going to do the opposite. Factoring is the opposite of multiplication. We're going to take a complicated expression and break it down into its simpler factors. So I'm going to show you how to basically factor something like 6 into just 2 times 3 or this expression into \(x \times (x + 3)\). Alright? So it turns out there are actually four ways to factor polynomials, and we'll be discussing each one of them in great detail in the next couple of videos. We're going to take a look at the first ones. I'm going to show you how this works. The first thing you should do is you should always look for the greatest common factors inside of your expressions. This has an acronym, the greatest common factor. It's called the GCF, and so I'm going to show you how to factor out this GCF using this example over here. So we're going to take a look at this example of \(2x^2+6\) and, basically, what the greatest common factor is, is it's the largest expression that evenly divides out of each of the terms in the polynomial. I have to figure out the largest thing that divides out of everything in that polynomial. Alright? And here's how to do this. Here's a step-by-step process. The first thing I like to do is write what's called a factor tree for each one of the terms. In factor trees, you may have seen them before. It's basically just a way to break down larger things into things that multiply. So, for example, 12, there are two numbers that multiply to 12. I have 2 times 6. Right? You also could have used 4 and 3 and actually would have been perfectly fine. It would have worked out the same way. Because what happens is if I do 2 and 6, I can't break down 2 anymore, but I can break down 6. 6 actually just breaks down into 2 times 3. So in other words, the two 2s and the 3s, these are all factors of 12. I can also do the same thing for things with variables. In other words, the \(6x^2\) breaks down to \(6 \times x \times x\), but then I can also just keep breaking down the 6 into 2 times 3, and I've already seen that. So in other words, the 2, the 3, and the \(x\) and the \(x\), those are all just factors of \(6x^2\). We're going to be doing this exact same thing, but now for each one of these terms in this expression. I'm going to do some color coding over here. I'm going to do the \(2x^2\), and I'm going to do the 6 over here. So what does the \(2x^2\) break down into? It just breaks down into \(2 \times x \times x\). Can I break down anything else? No, because 2 can't be broken down any further. What about the 6? I've seen that the 6 can break down into 2 times 3. So once I've done all the factor training for each one of the terms, I'm going to put a big parenthesis around them and include the sign that it was in the original expression. And now what I'm going to do is I want to figure out the largest thing that is evenly divisible out of each one of the terms or the largest thing that pops up in between both of the terms. So if I take a look at this expression, what are the common items that appear in these terms? Well, I see a 2 that pops up in the left term, and it also pops up in the right term. Are the \(x\)'s common between both terms? No, because they only appear on the left side. What about the 3? Is that common? Well, no. Because that only appears on the right term. The one thing that pops up in both is the 2, so that is the greatest common factor. That leads us to now the second step. What do you do with this 2? Well, basically, what we're going to do is we're, basically, just going to move it and extract it outside of the parentheses and kind of just remove it from this whole entire expression, and then we're going to leave everything insideạt we had from step 1. So here's how this works. I take the 2 and I pull it to the outside of parentheses. And then what was left over? And what was left over was the two factors of \(x\) and then one factor of 3. So, in other words, what I've done here is I've done 2, and then I have \(x^2 + 3\). That was what was left over on the inside, and this was my greatest common factor. One easy way to check this, just in case you ever worry that you didn't do it correctly, is if you do the distributive property, you should basically just get back to your original expression. That's basically what the GCF is. It's like the opposite of the distributive property. Alright? That's all there is to it. Let's take a look at a couple more examples here. So I have \(7x^2\) and \(5x\). So here what happens is I'm going to write down the factor tree again. So I have \(7x^2\) and \(5x\). This just becomes \(7 \times x \times x\), and this just becomes \(5 times x\). Now I keep a minus sign over here, and then I just put a big parenthesis. Now can I break down the 7 or 5 any further? Well, actually, I can't because 7, the only two things that multiply 7 are 7 and 1. Same thing with 5. Only things that multiply are 5 and 1. So these are all prime factors. So what are the common items between the two? Well, it's not 7 and 5, but I do see one power of \(x\) that's between both of them. Why is it at 2? Well, it's basically just because there's only 2 in the left term, but there's only one power of \(x\) in the right term. So the thing that's the largest thing that's common between both of them is just one power of \(x\). So now what do I do with this \(x\)? I pick it up, and I basically just move it to the outside of the expression over here. So I write an \(x\), and then I have a parenthesis. And then what was left over in my expression? I have \(7x\) and then a \(5\), and then you always have to include the sign that was in here. So this just becomes \(7x - 5\). If you distribute this, you should get back to your original expression. Now let's take a look at the last one over here. So this, \(8x^2\), \(8x^3 + 16x\). Alright. So let's do the factorization of this, and let's do the factorization of this. Now one of the things you actually can notice here is that 8 and 16 are multiples of each other. We didn't have to worry about those two in the first two examples, but 8 is just a multiple of 16. So one of the shortcuts that you can use is instead of having to factor completely, you can basically just write this as the product of you can do this \(8 \times x \times x \times x\), and then this over here, the 16, can be written as \(8 \times 2\). So notice how if I do it this way, if I notice that they're multiples of each other, I'll end up with the same thing in both of the factor trees. So now what happens is I have \(8 \times 2 \times 1\) power of \(x\) over here. Now one thing that you also wanna sort of consider or be just, you know, watch out for is that I had a negative sign over here in the beginning. And, basically, the way I like to account for that is I like to put a little negative one times 8, so I don't forget that there's a negative sign there. Alright? Now there's a plus sign over here. What are the common items between the two terms? I see an 8 here and an 8 over here, and I see 1 power of \(x\) over here and 1 power of \(x\) over here. There's nothing else that's common, so that means that my greatest common factor is just \(8x\). So what do I do? I take the \(8x\) and I pull it to the outside, and move all this stuff to the outside, and remove it from the parentheses. And this just becomes \(8x\). This becomes \(8x\)( and then what was left over on the inside? I see a negative one, 2 powers of \(x\), and then I see a 2 over here. So, basically, this becomes \(-x^2 + 2\). You distribute, you should get back to your original expression. Alright? So, these are the answers. That's how to factor using the greatest common factor. Let's go ahead and take a look at some more practice.
Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
0. Review of Algebra
Factoring Polynomials
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