00:01
This problem wants us to use synthetic division to divide for our first question and also use synthetic division in the remainder theorem to evaluate our second polynomial.
00:09
So the process for doing synthetic division and evaluating is going to look very similar for both.
00:14
There's just one small difference when you start and also what you're looking for at the very end as far as what your answer would be.
00:21
But if we are just dividing using synthetic division, when we look at what we are dividing with, in this case x minus 2, we use the opposite of that number.
00:30
That we see.
00:30
So we're going to evaluate, or not evaluate, we're going to divide with a positive 2 instead of a negative 2.
00:35
And then we list all of our coefficients in order.
00:39
And if we skip a term, then we use a placeholder of 0, but it looks like we have a coefficient for everything here.
00:45
We have 1x to the 4th, negative 1x cubed, 1x squared, negative 1 for x, and then 2 for our constant.
00:53
So to divide, we'll start to process by bringing down our first number, in this case 1.
00:59
And then we multiply by our outside value.
01:01
2 times 1 gives us 2, and then we write the result of that multiplication underneath the next number, and then we add multiply, add multiply, until we get to the end.
01:10
So adding negative 1 plus 2 gives us 1, 1 times 2 gives us 2, 1 plus 2 gives us 3, 3 times 2 gives us 6, negative 1 plus 6 gives us 5, 5 times 2 gives us 10, and then 2 plus 10 gives us 12.
01:25
To write our results, we start with one less exponent for our first coefficient, because synthetic always works with a first power to divide.
01:35
So dividing x the first and x fourth will always result next to the third, or just basically one less, like we said, for the exponent...