00:01
Given our function f in two factors, x plus, excuse, i wrote that wrong, x plus 1 and x minus 2, for part a, we are asked to, we're asked to verify that those given factors are factors of the function.
00:31
We can do this with synthetic division.
00:33
So k for the first factor is negative 1.
00:37
So i'm going to put negative one outside of the structure, and we've got coefficients of 3x cubed, negative 1x squared, negative 8x, negative 4.
00:51
We should expect a remainder of 0.
00:53
We drop down the first number 3, we multiply to get negative 3.
00:56
Add to get negative 4, multiply to get 4, add to get 0.
01:03
We just confirmed, yes, x plus 1 is a factor.
01:07
And then we continue this process.
01:11
Our other factor, negative x minus 2, would be a k value of 2.
01:17
So i could synthetically divide 2 from what we are quotient from above.
01:23
Co -efficient 3, negative 4, negative 4...