00:01
Given the function of f and two factors, we first want to verify that those really are factors.
00:07
We can do this with synthetic division.
00:09
We can synthetically divide the negative 2, which would be the k value from the first factor, from the coefficients of 2, 1, negative 5, and 2.
00:18
And we should expect a 0 for a remainder.
00:21
Drop down the 2, multiply, add, multiply, add, multiply, add, multiply add.
00:28
Get a zero for remainder.
00:30
Do this again with one for the k value.
00:34
Same coefficients.
00:36
Drop down the two, multiply add, multiply add, multiply add.
00:41
We get it again, both are factors.
00:46
Now, in part b, we're asked to find the remaining factor.
00:54
And so there really will only be two pieces of degree three.
01:00
We're already given two.
01:01
But what we can do is take one of the remainders and divide the second one out of what's left over...