Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. General strategy for factoring. We're going to start by looking for the greatest common factor. So we're going to look at the polynomial and see if there's anything that all of the terms have in common. This first example, there's AY squared in common, so that would leave us AY +9. We could also, if we want to, take out a negative y ^2 and that just flips the signs of what's in the parentheses. So a negative y ^2. If we pulled it out, it'd be a negative y -, 9, Both of those are correct. This next one, 1035 and 10 all have A5 in common. P to the 4th P ^3, P ^2, so AP squared. We look for basically the lowest quantity of PS:, Q to the 4th Q ^3 and Q ^2. So when we pull out five P ^2 Q ^2, that leaves me two P ^2 Q ^2 + 7 PQ +2. That can't factor any further because these two twos in the ( 2 * 2 is 4 and there aren't 2 numbers that would multiply to give me 4 but add to give me that middle coefficient of seven. Here we're going to pull out an X and AY, and that's going to leave us X -, y. Remember, we could, if we wanted to pull out a negative and all that does is it changes all of the terms in the parentheses. We could have done that back here also if we had wanted to. So I'm not going to keep emphasizing it, but just keep that in mind in case you need to. Here. I'm going to pull out an A, and that's going to give me negative a ^3 plus two a ^2 -, 13. Or if I wanted a negative A, it'd be a cubed minus two a ^2 + 13. So you're going to try a couple. The next thing is, once we've pulled out anything that's in common, count the number of terms. If there's 2 terms, think about is it the difference of squares or the sum or difference of cubes? This case it's the difference of two squares. SO3X times 3X gives us 9X squared, and 16 is going to be a +4 and a -4 because it needed to be a -, 16. And remember, when we do the outer terms and the inner terms this way, one's +1's negative and they cancel this. Next, pull out the three first look to see if we can continue factoring, and we can. This is the difference of two squares. So we get three times the quantity t -, 3 T +3, and it doesn't matter which order the t + 3 T -3 come, T to the 8th -1 T to the fourth plus one T to the 4th -1. Now this T to the 4th -1 is actually the difference of two squares again. So we're going to keep going. The T to the fourth plus one can't factor because sum of squares doesn't. So this is t ^2 plus one t ^2 -, 1. Guess what, that t ^2 -, 1 down there at the end, it continues to factor one more time. So our final answer is going to look like this. Going to give you a few to try. The difference and sum of cubes. 5X cubed 120 five X ^3 + y ^3. Sorry, that's 5X quantity cubed plus y ^3. So when we plug it into our formula, we get five X + y * 20 five X ^2 -, 5 XY plus y ^2 m ^3 -, 64. Remember that M quantity cubed -4 ^3. So just using our formula m -, 4 M squared plus four m + 16 is a final answer. 50 four X ^3 + 2. We're going to pull out the two. We're going to get 20 seven X ^3 + 1. So 2 times three X + 1 * 9 X squared minus three X + 1. Here's a few for you to try. So now we're going to keep counting. If there's three terms, we're going to look for a trinomial with A1 in front. If we have a one in front, we look at the last number and we think about what 2 numbers multiply to give us a positive 21, but add to give us whatever the middle coefficient is. So 3:00 and 7:00. So we get X + 3 and X + 7 three times. 7 is 21/3 plus 7 is the 10, so -30 this time. So 1 and 32 and 15, three and 10, five and six. The one we want, we need it to be a -1, and it needs to add to give me a -30. So the bigger numbers are all going to be negative so that when I add I get a negative. So we could see it's X + 5 and X -, 6 +5 negative 6 multiplies to give me -30 negative 6 and +5 add to give me -1. This last one, we don't have them in the right order. So we're going to start by factoring out a -1 and putting them in D Oh, a negative X. Let's factor out a negative X and put it in decreasing order. So X ^2 -, X -, 56. So the two numbers that are going to multiply to give me 56 negative 56, 156 to 28, four and 16, seven and eight. So 7 and 8 is going to be our the one we need. We need it to be a negative that it adds to. So it's going to be a positive 7 and a -8 so negative ** +7 X -, 8. Now there are several different variations. We could have had ** plus seven negative X + 8 just by taking this -1 and distributing it. So there's a quite a few different ways that that one might look. You go ahead and try a few. Next we're going to look for a perfect square trinomial. So this first one's a square and the second or the last terms a square. So the question is that when I multiply it out, do I get the middle term? So 8A and 8A gives me 16 a. So that one is a perfect square. Sometimes we need to rearrange the order first. So 20 five y ^2 -, 80 Y plus 64. Remember to put it in descending order 5 Y and five Y -8 negative 8. Does that give us the middle? So -40 negative 40? That one does. So once again, a few for you to try. If the coefficient in front of the squared term is not one, we have to use a different strategy. So we start by factoring on anything in common and we're going to multiply the three and the 8 and get 24. So if I list all the things that multiply to give me 24, and then I'm going to look for the ones that have a sum that gives me -10 So we know that they're both going to be negative because a negative times a negative is a positive. So I need the sum that's going to give me -10. So I'm going to use -4 negative 6/1 method to do this. Is a - 4 / 3 and a - 6 / 3 the -4 negative 6 over the leading coefficient. Reduce those fractions as far as they can, and then if there's still a denominator, pull the denominator out in front of the leading coefficient. So first step, there wasn't anything in common. So 14 * -3 is -42 one and 42 two and 21 three and 14 seven and oops, 6. Actually, let's do six and seven. I need them to add to give me -19 but they had to multiply to give me a negative. So one had to be positive when the other is negative. So I can see it's going to be +2 and -21 So we're going to actually factor out our X ^2, so we get 14 X squared -19 X -3 and then we're going to have X + 2 / 14 and X -, 21 / 14. Reducing those fractions as much as possible. And reducing just really means taking any number that will go into both. Oops, that was a 14. So 21 / 14 is 3 halves. If there's a denominator, we're going to pull it out in front. So seven X + 1 two X -, 3 keep counting. If there's 4 terms, we're going to factor by grouping. So we can factor by grouping in two different ways. We can group the 1st 2 together and the second two together and if that doesn't work then we think about grouping the first 3 together and the last one, or the first one and the last three. I always recommend trying two and two first. Frequently. That will work. So when we look here we get a * a -, 3 left. If we look at the last two and we pull out AY, we get a -, 3. So we went from 4 terms to two terms and these two terms have an A -, 3 in common. If we had gotten here and they didn't have anything in common, that's when I would have gone back and tried three and one and or one and three as my groupings. If I pull an A -, 3 out of both, I get an A + Y left and that's my final solution here. We're going to pull out an X to the 4th to start, and that's going to give me X ^2 -, X minus wait, X ^2 -, X. Oh, we didn't have it written in the correct order to start. Let's pull out an X ^3 and write it in descending order when we do it. X ^3 -, X ^2 + X -, 1. So 4 terms. We're going to factor by grouping and we're going to pull out an X ^2 in these first ones, leaving us an X -, 1. Going to pull out a one because they don't have anything in common other than a one in the second two terms. So now we can see those each have an X -, 1 in common. And if I pull that out, that leaves me X ^2 + 1. So we have a couple that are tricky here. Here we could think of this as something like six a ^2 + 13 a -5 instead of that X -, 7 quantity. And we could do the same skills, so we'd have a -30. So 1 and 32 and 15, three and 10, five and six. We want a +13, but they need to multiply to give me a negative. So we're going to have -2 and +15. So if we thought about having this as a - 2 / 6 * a + 15 / 6 and reduce it a -, 1/3 A plus 5 halves. So we'd have three a -, 1 and two a + 5. But that's really, really, really not the original problem. The original problem had XS, so instead of the A's, let's put in what the A was representing, which is this X -, 7. Now there are lots of ways to do this problem. We could have multiplied it all out and then combine like terms and factored if we wanted to. So 3 X -21 -, 1 would be -22 two X -14 + 9 negative 9. So that's our final solution there. Here, this is an example where two and two groupings not going to work. If I pulled out an X, I'd get X + 2 Y. And down here I can't pull out anything. I might see it's difference of two squares, but now I have two terms and I still don't have anything in common. So we're going to group these three and one. Some of the keys to look at when you're deciding whether you're going to do 3 and one or one and three is do the first 3 combined a factor. And the answer is yes, this is X + y quantity squared. And then it's -9 well, now I have two terms or a binomial and there are difference of two squares. So we're going to have X + y + 3 and X + y - 3. And that's our final solution. Because X + y * X + y is the X + y ^2. When we do the outer terms and the inner terms, they would cancel one's -1's positive, so additive inverses. And when we do the three and -3, that gives us the -9 at the end. So let's do one more. This one, we're going to factor these first 3 together, and it's going to give us m + 2 N quantity squared. Plus these last two have A5 in common, and that's going to give us an m + 2 N Now I have taken 5 terms to two terms. These two terms have an m + 2 N in common. If I pull out an m + 2 N, that leaves me m + 2 N +5 a couple for you to try. We're going to always factor completely, so we're going to start by pulling out a three. Here we have two terms left in the parentheses. That's going to be difference of two squares. But once we factor the difference of two squares, we can see we still have a difference of two squares. So we got to make sure that we go all the way with our factorization. We're going to let you try one. Sometimes polynomials can't factor. When I multiply the one and two, well, the only thing that multiplies to give me two is 1 and 2, and it can't add to give me 5. This is called a prime polynomial. But we can always factor out a common one. If nothing else, they always have a one in common. Here the sum of two squares never factors. So just seeing a square plus a square, we know that that doesn't work. Thank you and have a wonderful day.