Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Division of polynomials. When we have a polynomial multiple terms divided by a monomial, what we're going to do is we're going to take each of the terms in the numerator and divide it by the denominator and then reduce each term. So we're going to take all of these numerator terms and divide them -7 into 21 gives us negative three a ^3 / a. Remember, when the bases are the same, we subtract the exponents so that give us a ^2 -7 into 7 is going to give us -1 A into a ^2 is A -3 / -7 is going to give us a positive and the A's are going to cancel. So we're going to get 3 sevenths -14 / -7 is going to give us a positive 2 and there aren't any A's to reduce, so the A is going to stay in the denominator. If we look at this next one, we're going to have 16Y to the 4th Z ^2 over -4 Y to the 4th 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. If we simplify those terms, 16 / -4 is going to be -4 the Y to the fourth cancel Z ^2 / Z is going to leave US1 Z -4 into -8. It's going to give us 2Y to the 6th over. Y to the fourth is y ^2. Z to the 4th over Z is Z ^3 -4 into 12 is going to be -3 Y to the 4th Z ^2. If we look at the next one, we're going to have the same kind of problem, but right now it doesn't look the same. If we rewrote this, it'd be the first polynomial divided by the 2nd. So now it looks like what we've been doing. We're going to take that 16Y cubed and put it over two y ^2 + -9 Y squared over two y ^2 + -8 Y over 2 I squared. When we simplify it, two goes into sixteen 8 * y ^3 / y ^2 is Y2 goes into -9 not at all and Y squareds cancels. So we get -9 halves. Two goes into -8 negative 4 * y / y ^2 leaves us AY in the denominator. Now if it's not a monomial, we have to do what's called polynomial long division. We're going to put y ^2 -, 10 Y -25 within the division box and y -, 5 on the outside. Polynomial long division is going to be very similar to numerical division. So once we get it set up, we're going to think about Y times, what gives me y ^2, and it's going to be y * y gives me y ^2. I would recommend highly keeping all of your YS on top of each other, all of your constants on top of each other, all of your Y squareds on top of each other. Just like with numerical division, we keep all of our 10s on top of our 10s, all of our ones on top of our ones. So we're going to have this y * y -, 5. I'm going to write it off to the left to make it easier to see that I now need to distribute. So I get y ^2 -, 5 Y on regular division. What we do next is we subtract, and we know that subtracting something could be really thought as adding its opposite. So if we went through and we changed all the signs, then we could just add them y ^2 and negative y ^2. Cancel -10 Y and +5 Y is going to give us a -5 Y. We're going to bring down the next term. So now we want to think about the Y times. What gives us -5 Y? And the answer is -5 So I'm going to have a -5 * y -, 5 when I distribute -5 Y plus 25. But now we need to subtract so we can think of that as changing all of our signs. The five YS will cancel and we get -50 We always write our remainder in this case -50 over what we were dividing by. So our final answer is y - 5 + -50 / y - 50. And in this case, instead of a plus a negative, you could just write -50 / y - 5. When we look at the next one, we have no a ^2. I would highly recommend writing in 0A squareds as a placeholder so that then when we have a squared terms or if we do, we have a place to put them. So our first step is going to be a times what gives me a cubed and the answer is going to be a ^2. So I'm going to take that a ^2 and I'm going to multiply it by A + 4. Doing the distributive property, we get a ^3 plus 4A squared. Our next step, change the signs and add. So we're going to get -4 A squared and we're going to bring down the -10 A. So now a here times what gives us -4 A squared and our answer is going to be -4 A. So I'm keeping all my AS lined up. So -4 A times a + 4, that's going to give us negative four a ^2 -, 16 A. We're going to change those signs so that we can add. We're going to get 6A and we're going to bring down the 24 A times what gives us 6 A +6. So 6 * a + 4 is going to give us 6A plus 24. If we change our signs, we're going to get a remainder of 0 in this one. So we know that a + 4 divides evenly into a ^3 -, 10. A+ 24 or a + 4 is a factor of it. If we look at this next one, we have a coefficient in front of the Y this time. So we're going to set it up the same way. Ten y ^3 plus six y ^2 -, 9 Y plus 10 / 5 Y -2. So now it's 5Y times what gives US10Y cubed, and the answer is going to be two y ^2. So two y ^2 * 5 Y -2 ten y ^3 -, 4 Y squared. If we change our signs and add, we're going to get 10 Y squared and we're going to bring down the next term. So now 5 Y times what gives us this ten y ^2, 5 Y times 2Y. So we're going to take 2Y and multiply it by 5 Y -2 and we get ten y ^2 -, 4 Y. If we change our signs, we're going to get -5 Y and we're going to bring down the 10 five Y times. What gives us -5 Y -1? So we get -5 Y +2. If I change my signs, we get a remainder of eight. We're going to have plus 8 / 5 Y -2. So that's our final answer. If you look, you can actually see our final answers over here. Also two y ^2 + 2 Y -1 and then the remainder. It works the same way if we have something besides a linear term. So this one, we're going to have 3X to the fourth plus X ^3 -, 8 X squared -3 X -3 X squared. We didn't have any plain XS, so I'm going to actually put in a 0X there, and then I'm going to do this one the same way we've done the others, X ^2 times. What gives me 3X to the 4th? The answer is going to be 3 X squared, so three X ^2 * X ^2 + 0 X -3. When I distribute, we get 3X to the fourth zero X ^3 -, 9 X squared. If we go through and we change all the signs and atom, we're going to get X ^3 + X ^2, and we're going to bring down the next term X ^2 times what gives us X ^3? That's a single X. So we're going to have X times that X ^2 + 0 X -3 X cubed plus zero X ^2 -, 3 X. If we change our signs so we can add, we're going to get X ^2 and we're going to get 0X's because -3 X and positive 3X. And then we're going to bring down the next term X ^2 times. What gives me X ^2 a positive 1. So we have 1 * X ^2 + 0 X minus three X ^2 + 0 X -3. If we change our signs so we can add, we see that we get a remainder of 0. So that tells me that X ^2 -, 3 divides evenly in to the original polynomial, hence it's a factor. And our final answer is three X ^2 + X + 1, which we can also see over here. Our last example is going to be if we have an F of X equation and AG of X equation and they want us to find F of X / g of X. The first thing we need to do is we need to make sure that we have our polynomial in descending order. So 2X to the 5th -3 X to the fourth minus two X ^3 + 8 X squared -5. Then we're going to divide by X ^2 + 0 X -1 X ^2 * 2 X cubed will give me two X to the 5th. So two X ^3 * X ^2 + 0 X -1 two X to the 5th, 0X to the 4th -2 X cubed. We're going to go through and we're going to change all the signs so we can add. And we end up with -3 X to the 4th plus 0X cubed, bringing down the next term plus eight X ^2 -3 X to the 4th. We have to have an X ^2 * -3 X squared. So -3 X squared times the X ^2 + 0 X -1 gives us -3 X to the fourth plus zero X ^3 +3 X squared. We're going to then change our signs so we can add and we're going to end up with five X squared. We're going to bring down the next term. Oh, I forgot to leave a 0X. I'm going to put a plus zero X -, 5 and I'm actually going to bring down both those terms because I had two terms that canceled so X ^2 times what gives me 5 X squared. And the answer is a +5 S 5 * X ^2 + 0 X -1 five X ^2 + 0 X -5. If we change our signs and add, we're going to get a remainder of 0 SO2X cubed minus three X ^2 + 5. Now this one actually asked for a little more originally, which was what can we never ever have in the denominator? Well, if we think about this as a division problem, our G of X is really in the denominator. So the G of X portion or the X ^2 - 1 can never ever, ever equal 0. So we know X ^2 - 1 can't equal 0. That's difference of squares. So X - 1 X plus one can't equal 0. So X isn't going to be one or X isn't going to be -1 for the domain restrictions. Thank you and have a wonderful day.