00:01
This problem gives us a function, and it wants us to do a few things, given a few factors, and first is to verify the given factors of f of x by completing the synthetic division that we see, and then using our results to find the remaining factors of f of x, and writing the complete factorization, and then the zeros of the function.
00:16
So to finish the synthetic division to prove these are both factors, we will pick up with following their steps, and then fill in the blanks as we go.
00:23
They drop down the 2 to get 2, and then 2 times negative 3 gives us negative 6, 1 plus negative 6 gives us negative 5.
00:30
Negative 5 times negative 3 gives us a positive 15.
00:33
Negative 13 plus 15 gives us 2.
00:36
2 times negative 3 gives us negative 6.
00:38
6 plus negative 6 gives us 0.
00:40
So since the remainder is 0, that proves that x plus 3 was a factor.
00:45
For x minus 2, same thing.
00:47
We bring down the 2 first.
00:48
2 times 2 gives us 4.
00:50
1 plus 4 gives us 5.
00:51
5 times the 2 gives us 10.
00:53
Negative 13 plus 10 gives us negative 3.
00:56
Negative 3 times 2 gives us negative 6.
00:58
And once again, 6 plus negative 6 gives us 0.
01:01
So the remainder 0, that means this is a factor.
01:04
What we can do now to be able to factor is to, we could look at the results of factoring our two factors out...