Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Factoring polynomials. First look for the greatest common factor. So in y ^3 + 9 Y squared they each term there are two terms. Each term has AY squared in common. If I pull out the Y ^2 I get left y + 9. I put that in parentheses and I took 2 terms or a binomial down to one term, which is a monomial, and the next 110 P to the fourth Q to the fourth plus 35 P cubed Q ^3 + 10 P squared Q ^2. There's three terms. That's a trinomial. If we look for what each of those 3 terms have in common, they have a five P ^2 Q ^2. If we factor that out, we get left two P ^2 Q ^2 + 7 PQ +2. Now it's a monomial, or a single term, because it's one thing times another thing. That second thing is all in parentheses, so it all acts together. If you have X ^2 y minus XY squared, those each have an XY in common. We factor it out, and we get the quantity X, -, y left. So we went from a binomial 2 terms to a monomial of one term. Sometimes we want to factor out a negative coefficient so that our leading term in the parentheses is a positive. In this case, we're going to take out a negative A and we get a cubed minus two a ^2 + 13. Next, once we've taken out the greatest common factor, we're going to count how many terms there are. If there are two terms, there are a couple possibilities. 1 is the difference of two squares, so nine X ^2 - 16 two terms binomial will factor into the monomial of three X + 4 * 3 X -4. Another possibility, three t ^2 - 27. They both had a three in common, so we're going to factor the three out first. Now we end up with difference of squares for t ^2 - 9. So even though we went from a binomial to a monomial, that monomial wasn't completely done because that t ^2 - 9 could still factor. So our final answer should be three t + 3 * t - 3. We could have something even more complex, like T to the 8th -1. That's the difference of squares. It factors into T to the fourth plus 1 * t to the 4th -1. Now the sum of squares never factors, so that T to the fourth plus one won't ever go any further. But T to the 4th -1 is the difference of squares again. So it's going to factor into t ^2 + 1 * t ^2 - 1. Now the t ^2 + 1, because it's a sum of squares, won't factor. But that t ^2 - 1 is the difference of squares, so it does factor and AT plus 1 * t - 1. So when we look at our bottom line, that T to the fourth plus one, t ^2 + 1, T plus one, and t - 1, none of those terms can break down or factor any further. So that's the final answer. We also have sum indifference of cubes when we have two terms and it's a formula. It's a ^3 + b ^3 equal the quantity A+B times the quantity a ^2 minus AB plus b ^2 and also a ^3 -, b ^3, which is the quantity a -, b * a ^2 plus AB plus b ^2. So when we look at this 120 five X ^3 + y ^3, that's really 5X quantity cubed plus y ^3. So the first thing that I think about is drop the cubes SO5X plus Y in the first parenthesis. The second parenthesis, take that 5X and square it, then multiply the two together, changing the sign and the last 1 ^2. Sometimes we think of it as MOP. Match the signs, opposite signs, positive sign. So that will factor into five X + y times the quantity 20, five X ^2 -, 5 XY plus y ^2 m ^3 - 64. We're going to make it into M quantity cubed -4 quantity cubed. The first parenthesis just drop the cube, so you get m -, 4. The second parenthesis 1st 1 ^2, the 2 multiplied together with the opposite sign and the last 1 ^2, so m ^2 + 4 M plus 16. The next example in cubes is the fact that they had a 2IN common, so we're going to factor it out first. Then we get 20 seven X ^3 + 127 X cubed is really the quantity three X ^3 + 1 ^3. Drop the cubes, take the first one and square it. Multiply the two together, the last 1 ^2. What if we had three terms? So we're going to look for trinomials with one in front of the first term. If we have a one in the front, we just really have to look at the last number, the constant and the coefficient on the middle term. We want 2 terms that are going to multiply to give us 21 and add to give us 10. Well, sometimes I have students write out all the things that could multiply to give us 21. So one in 21, three and seven. Obviously three and seven adds to give us the 10. The next example, we want 2 numbers that are going to multiply to give us -30. So when we list all of these, we can see that it's going to be a positive 5 and a -6 because we need it to add to give me a -1. So X -, 6 * X + 5. If it's not in the correct order, start by rearranging it so that it's in descending order. If necessary, take out anything they have in common. In this case, we're going to take out a negative X so that the X ^2 is positive. Now it's a one in front of that X ^2 coefficient. So we want 2 numbers that are going to multiply to give us -56 but to add to give us -1. So if we list all of our possibilities, we're going to get +7 and -8. So negative X * X -, 8 * X + 7. Something else to look for is to look for a perfect square. Perfect square is something of the form where we have a ^2 + 2 AB plus b ^2 and that factors into A+B quantity squared. Or it could be a ^2 -, 2 AB plus b ^2, which factors into a -, b quantity squared. So for this one, the two numbers that multiply to give us 64 and add to give us 16 are 8 and 8. So it's a + 8 quantity squared. This next 125 Y squared would be five y * 5 Y 64 is 8 * 8, and if we did the outer terms, we'd get -40 Y and the inner terms would give us another -40 which is the middle term of that -80. So that factors into the quantity five y -, 8 ^2. If we have a coefficient in the front, the easiest way I can tell you to do it is to multiply the coefficient and the constant we get 24. List everything that gives us a product of that 24. Find the two that give us the middle term in this case -10. So we want -4 and -6. Now if you take a - 4 / 3 and a - 6 / 3, that three came from the leading coefficient. Reduce each of those fractions, and then if there's still a fraction, put the denominator number as the leading coefficient. So three a ^2 -, 10 A+, eight factors into three, a -, 4 and a -, 2. This next example, first factor out the X ^2 and then we're going to multiply 14 * -3. We're going to list everything that multiplies to give us 42. Let's see 3 and 14, 6:00 and 7:00. We want the one that adds to give us that -19 so we want 2 and -21. So if we take and, we say X + 2 over the leading coefficient of 14 and X -, 21 over that leading coefficient of 14. If we reduce those fractions, then if there's anything in the denominator, we're going to put that as the leading coefficient in front of the polynomial. So we get 7X plus one and two X -, 3. If there are 4 terms, we're going to factor by grouping, and we're either going to group the 1st 2 and the last two, or possibly 3 and 1. So in this first example, we're going to factor two and two. The first two had an A in common, leaving a -, 3. The second two had AY in common, leaving a -, 3. So now we're down to two terms here. We started with four, and these two terms each have an A -, 3 in common. So if I factor out the A -, 3 and write it down, that leaves me that a + y, which is one term. If we look at the next one, we have 4 terms. We're going to factor these X ^3 out of all of them and put it in the correct descending order. Then these first two have an X ^2 in common and we're going to have X -, 1 left. The second two don't have anything in common, but there's always an understood 1. So if I factor out a one, I get X -, 1. Now this has an X - 1 and this has an X -, 1. So we're going to be able to factor out an X -, 1. That left us the X ^2 + 1, which is going to go in parentheses together. So finally we're going to watch out for tricky ones, like if we had six times the quantity X -, 7 ^2 + 13 times the quantity X - 7 - 5. Think about that X -, 7 being some new variable, maybe a U. If we had that, we now have six U ^2 + 13 U -5, and we know how to factor that. We could multiply the outside terms 6 and -5 to get -30. We'd multiply. We'd list all the products that multiply to give us -30 and we'd find the one that adds to give us the 13. Well, if it has to add to give us 13 but multiply to give us a negative, we're going to have -2 and +15. So if we had U -, 2 sixths and U + 15 sixths, if we reduce that, we'd get U -, 1/3 and U + 5 halves. If there's a fraction, we're going to put that denominator as a leading coefficient with that variable. But instead of U now we're going to substitute in U was X - 7. And then we're going to distribute and combine like terms. So we'd get three X - 21 - 1, and we'd have two X - 14 + 5 or three X - 22 * 2 X -9. What if we had 4 terms and it's not a grouping problem of two and two? We're going to group the first 3 together and realize that that first 3 is a perfect square trinomial. It factors into X + y quantity squared -9 Well, now that's really of the form a ^2 -, b ^2, which would tell us A+B times a -, b, or that first term X + y + 3 times that first term X + y - 3. We could have one with five terms. And when we have 5 terms, we're going to look and we're going to see that these first 3 is really a perfect square trinomial. Again, we get m + 2 N squared. The second two have a 5 that can factor out, leaving us m + 2 N. Now each of these new terms have an m + 2 N in common. If I take one m + 2 N out of a quantity squared, that leaves me one of them. If I take an m + 2 N out of the second term, that leaves me 5. So our final answer of m + 2 N times m + 2 N +5, we always want to factor completely. So we start by pulling out anything they have in common, in this case a three that's now different to squares. So we're going to factor m ^2 plus one m ^2 -, 1. We're going to remember that sum of squares never factors, but that difference of squares m ^2 - 1 can go further. It can go into m + 1 and m -, 1. So the last thing. In summary, sometimes there are polynomials that can't factor. They're called primes. We can always pull out a one, but other than pulling out A1, we can't factor this one because there aren't 2 numbers that multiply to give us 2 and add to give us 5. Sum of squares can never be factored other than being able to pull out a one, so that one's also a prime. Thank you and have a wonderful day.