00:01
In this question, fs is equals to 2s minus 1 upon s square minus 3s plus 2.
00:12
Now, by middle term split method, this can be written as 2s minus 1 upon s square minus 2s minus s plus 2.
00:24
So, from here fs will be equals to 2s minus 1 upon s minus 2 and s minus 1.
00:36
So, now 2s minus 1 upon s minus 2 s minus 1 can be written as a upon s minus 2 plus b upon s minus 1.
00:55
So, after simplifying this will be a s minus 1 plus b s minus 2 divided by s minus 2 here it is s minus 1 and here it is 2s minus 1 upon s minus 2 s minus 1.
01:17
So, from here here the here a s minus a plus b s minus 2 b divided by s minus 2 s minus 1 and this is 2s minus 1 divided by s minus 2 s minus 1.
01:39
So, 2s minus 1 upon s minus 2 s minus 1 equals to coefficient of s will be a plus b in minus a plus 2 b and whole upon s minus 2 s minus 1.
02:00
So, on comparing coefficients on comparing on comparing coefficients here it will coefficient of s a plus b equals to 2 and a plus 2 b equals to 1.
02:20
So, after simplification here a plus b equals to 2...