Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Division of polynomials. When we have a monomial in the denominator, what we're going to do is we're going to take every term in the numerator and just simplify it. SO16Y to the 4th Z ^2 / -4 Y to the fourth Z -8 Y to the 6th Z to the fourth over -4 Y to the 4th Z + 12 Y to the 8th Z ^3 / -4 Y to the fourth Z. So when we simplify this, we get -4 Z negative divided by negative makes it a positive two y ^2 Z ^3 and a -3 Y to the 4th Z ^2. Now sometimes it's the same type of problem, but it looks a little different. So here we're going to have 16Y cubed over two I ^2 -, 9 Y squared over two y ^2 -, 8 Y over 2 I squared. When we simplify this, we get 8Y -9 halves -4 / y as a final answer. It gets a little more interesting when it's not a monomial that we're dividing by, but a polynomial of some sort. So in this next one, we're going to do long division using polynomials. So we're going to have a cubed. I'm going to put in zero a ^2 as a placeholder -10 A+ 24 and we're going to divide by a + 4. So my thought process needs to be this A here in front times, what gives me a cubed and the answer is going to be a ^2. And I'm going to keep all my squares lined up so my A squares are on top of my A squares. So a ^2 * A + 4. If I write that off here to the side, I can then do the distributive property, realize that that's a ^3 + 4 A squared in division. The next thing we do is we subtract. So a ^3 -, a ^3 is 0 zero, a ^2 -, 4 A squared would give me -4 A squared. And then I'm going to bring down the next term. Now we're going to do the same thought process, this A up here times what gives me -4 A squared, and the answer is going to be negative 4A. So once again, I'm going to keep all my A's in a column to make it nice and orderly. And I'm going to multiply that by A + 4. So we're going to get negative four a ^2 -, 16 A. When I subtract those -4 A squares are going to cancel negative ten a -, a negative minus a negative remembers plus. So that turns into 6A. We're going to bring down the next term. So a times what gives me 6A? Well, a * 6. So we're going to have 6 * a + 4, which is 6A plus 24. And when we subtract, we get a remainder of 0. If this had been anything other than 0, we would have put plus whatever the number is over what we were dividing by. Because it's zero, I don't need that 0 / a + 4, but I wanted to show it. So in this next example, we're going to have a little longer problem. Ten y ^3 plus six y ^2 -, 9 Y plus 10, all divided by five y - 2. The same exact thought process. This 5 Y times what gives me 10 Y cubed and the answer is going to be two y ^2 and I'm going to keep all my Y squareds lined up. So two y ^2 * 5 Y -2 ten y ^3 -, 4 Y squared. When I subtract, the 10 Y cubes will cancel, we're going to get 10 Y squared, and we're going to bring down the next term SO5Y times what gives me 10 Y squared, and the answer is going to be two Y, once again keeping all my YS in a column. So two y * 5 Y -2 ten y ^2 - 4 Y. When I subtract here, the 10 Y squares cancel and we end up with -5 Y +2, so we're going to multiply by a -1. So that's going to give me -5 Y +2. When I subtract here, I get 10 - 2 or 8, so my remainder is a + 8 / 5 Y -2. So this next one is actually going to be just like number six. I think I'll just do #5 and let you practice with six. So 3X to the fourth plus X ^3 - 8 X squared -3 X -3. That whole thing is getting divided by X ^2 -, 3. I'm going to once again put in a placeholder just like we did on problem 3 so that I have 0X's. So when I multiply, I have everything lined up. So X ^2 right here times what gives me 3X to the 4th and the answer is going to be 3 X squared. So I'm going to put the X squareds in a column over the X ^2. So when I take 3X squared and multiply it by X ^2 + 0 X -3, I'm going to get 3X to the fourth plus 0X cubes -9 X squared. The next step is to subtract so the 3X to the fourth cancel we get X ^3 -8 -. A negative 9 is going to give us a +1 and we're going to bring down the next term. So X ^2 times what gives me X ^3, and that would be a single old X. So then when I distribute again, I'm going to get X ^3 plus zero X ^2 -, 3 X Subtracting the X cubes cancel, I get an X ^2 and the -3 X is cancel. But I could put in a 0X as a placeholder if I want, bringing down the next term X ^2 times. What gives me X ^2? The answer is 1. So we'd have 1 * X ^2 + 0 X -3. And when we subtract here, we can see that we actually are not going to get a remainder. So our answer would be three X ^2 + X + 1. Number six is going to be the exact same kind of problem. I'm going to let you go ahead and work it, and then you also have some on the next page. Actually, the thing that emphasized on number six is it's not given in the right order. To start, we want F of X / g of X, but we need to do it in descending order. So 2X to the 5th -3 X to the fourth minus two X ^3 + 8 X squared. Doesn't look like there are any plane axis. So I'm going to put 0X and then a -, 5 and then that's going to get divided by X ^2 + 0 X -1. Go ahead and finish that one up and then do the next page for the triads. Thank you and have a wonderful day.