OK, to do this long division, any place that I don't have a term, I'm going to put in a placeholder with A0X. So there's my X ^2. I didn't have any XS and my constant down here I didn't have any XS or constant. So I'm going to put a 0X and a zero. So my first thought process should be two X squared times. What gives me 8X to the 4th and my answer is going to be four X ^2. So if I have four X squared and I multiply it by that two X ^2 + 0 X plus one. If I write it over here, I can see that all I'm really doing is the distributive property 8X to the 4th plus zero X ^3 + 4 X squared. We're going to then subtract or I prefer and students have less errors made if they just change all the signs and add them. So at this point, we're going to get fourteen X ^3 -, 2 X squared plus 0X. I'm going to look at the furthest to the left again. So this two X squared now times what gives me 14X cubed and our answer is going to be 7X. So 7X, 7X times two X ^2 + 0 X plus one fourteen X ^3 plus zero X ^2 + 7 X. So that when I now subtract or change the sign so that I can add change all the signs so that I could add them. I'm going to get the 14X cubes to cancel negative two X ^2 - 7 X and I'm going to bring down the zero. So once again, farthest left. So this two X squared times, what gives me this -2 X squared? And the answer is going to be -1 So I'm going to have that -1 times the two X ^2 + 0 X plus one. And I'm going to get negative two X ^2 + 0 X -1. I'm going to go through and I'm going to change all my signs so that I can then add. So I'm going to get -7 X plus one. So my remainder always goes over what I was dividing by. So seven X + 1 divided by two X ^2 + 1.