00:01
On the top of that, we have a problem.
00:04
We're going to solve this problem using synthetic division.
00:08
And our goal today is going to be to provide it in the following forms, a and b.
00:15
So we can get started immediately.
00:17
The first step in solving a synthetic division problem is to zero our divisor.
00:23
Our divisor is x minus 4, and we need to equal that to 0.
00:27
We can get x by itself by adding 4 to both sides, and we get x is equal to 4.
00:33
After that our next step is to set up and solve the problem this is where all the synthetic division comes in four goes on the outside of this half rectangle and on the inside the coefficients of our dividend go in there but only the coefficients so for example this would be two because in our dividend the first term is two x cubed but we only take the coefficient so two would be there then we have negative five negative 11 and negative 17.
01:12
After the initial setup, we can start solving.
01:15
And the first step is to bring down the leading coefficient.
01:19
We always need to do that.
01:21
Every time we're going to start, that leading coefficient needs to come down.
01:25
After that, we can start to multiply by our divisor.
01:28
So our resultant, which is 2, needs to be multiplied by our divisor.
01:32
So 4 times 2 is 8, and that will be put in the next column, and then we can add.
01:38
Negative 5 plus 8 is 3.
01:42
Now we repeat the process again.
01:44
Our resultant times our divisor, which is 12, and we put that in the next column, and we add.
01:51
Negative 11 plus 12 is 1, and one more time, we multiply our resultant with our divisor.
01:59
1 times 4 is 4.
02:02
And then we can add.
02:03
And then our resultant is going to be negative 13.
02:08
This last number is always our remainder.
02:11
Always.
02:12
So i'm mark that for you guys.
02:20
Now that we have our answer, we need to put back our x's.
02:24
Because notice when we set up the problem, we only took the coefficients.
02:28
Now that we have our answer, we need to put back the x's in there.
02:31
But we need to put them back in increasing order from right to left.
02:35
Here's what i mean.
02:37
So to the left of the remainder is 1.
02:40
And that is always going to be our y intercept.
02:43
It has zero x's.
02:45
The number to the left of that will have 1x.
02:49
So 3x, and then the number to the left of that will have 2x's, so 2x squared.
02:57
Say, for example, we had another number next to the 2, say like 9.
03:02
That number would have 3xs, so it would be 9x cubed.
03:07
Of course, we only have 3 numbers, so i'll get rid of that.
03:15
But you guys get the idea.
03:17
And now this number here, or this, what we came up with here, is now considered our quotient.
03:27
And the quotient is the resultant of dividing 2.
03:31
Numbers.
03:34
So i'll put that there...