00:01
A couple of questions to get through here.
00:02
First of which is to take x to the 4th plus 5x to the 3rd minus 13x minus 4 and divided by x plus 2 using synthetic division.
00:11
So to divide by x plus 2, i'll throw a negative 2 in the box.
00:17
And all these numbers i'm writing here are representative of the coefficients of each of these terms.
00:25
So as x to the 4th, x to the 3rd, i'm putting in a placeholder for x squared term.
00:31
This is my linear term and this is the constant.
00:37
Drop the 1, multiply, add, multiply, add, multiply, add, multiply, add, and multiply.
00:51
We end up with a remainder of negative 2.
00:59
So that means when we divided x plus 2 out, that left us with the polynomial represented by these four numbers.
01:10
Took out a degree, so that is a 1x to the 3rd, plus 3x squared, minus 6x, minus 1, plus a remainder of negative 2.
01:55
That is our divisor times our quotient plus our remainder.
02:02
In this next one, we are going to find a polynomial p of x, so it has degree 3, zeros at negative 4, positive 4, and 1, and p of 0 equals 8.
02:14
So if we have zeros at negative 4, positive 4, and positive 1, we know that's going to give us degree 3.
02:23
So with zeros there, that means we have factors.
02:28
If x equals negative 4 is a 0, x plus 4 is a factor.
02:32
If x equals positive 4, x minus 4 is a factor.
02:35
If x equals positive 1, x minus 1 is a factor.
02:41
When we multiply these out, we would do so so that we can double check this piece of information.
02:46
If p of 0 equals 8, that means that is our constant.
02:53
So what we're working with here gives us x squared minus 16 times x minus 1, which is x cubed minus x squared minus 16x plus 16.
03:13
So like i said, that constant is supposed to be an 8, not a 16.
03:17
So what we need to consider is that there is a potential for there to be a value other than 1.
03:24
And that value is not going to affect the zeros or, in turn, the factors that we have.
03:29
So what we have to consider is that what would we multiply this entire polynomial by so that the 16 became an 8? what we could use is 1 .5.
03:40
We multiply by 1ā2.
03:42
We still have us our degree 3 polynomial.
03:48
It still has 0s at 4, negative 4, and positive 1.
03:53
And now it has that p of 0 equals positive 8.
03:57
In this last one, we are going to make a list of all possible rational zeros of p of x equals 2x to the third minus 4x squared minus 7x plus 3...