00:01
So in part a of this problem, we're first being asked to divide p of x by d of x.
00:05
Now, they're told we can use the long or synthetic division.
00:08
However, because our d of x is a quadratic, we're not going to be able to use synthetic division in this case.
00:13
So we're going to actually have to do this by long division.
00:15
So p of x is going to go inside our symbol.
00:19
So have x to the 5th plus x to the 4, minus 7x to the 3rd.
00:23
Because i'm missing that x squared term, i'll put a place so they're in its place.
00:27
So 0x squared plus 6x plus 6.
00:30
And then we'll put our d of x function on the outside.
00:33
So that's x squared plus x minus 1.
00:36
So step one is we're going to divide our first term, x to the fifth, by the first term in our divisor x squared.
00:42
Well, x to the fifth divided by x squared is x to the third.
00:46
And i always put it above its late term.
00:48
Next, we're going to multiply x to the third by our g of x.
00:51
Well, x to the third times x squared is x to the fifth.
00:54
X to the third times x is positive x to the 4.
00:57
And x to the third times negative 1 is negative 1.
01:00
X to the third.
01:02
Then, just like with regular division, we need to subtract.
01:05
Well, x to the fifth minus x to the fifth is zero.
01:08
That term should cancel.
01:09
Then we have x to the fourth minus x to the four minus x to the four.
01:11
That's also zero.
01:12
So those cancel.
01:14
So then we have negative 7x to the third minus negative one x to the third, which is equal to negative six x to the third.
01:22
And i'm going to bring down all those are all the remaining terms...