00:01
Okay, this question involves the remainder theorem and the factor theorem, which is this, basically.
00:09
Let's just call the function p of x.
00:13
I'll remove that for now.
00:16
I suppose i'm dividing p of x by x minus a.
00:25
Up here i got a quotient, it's called it q of x, and then down here somewhere, a remainder.
00:32
And then i can say a p of x, this item here, p of x, will be this, times this, plus r.
00:47
That's your basic remainder theorem.
00:51
Okay, let's just remove all this a moment.
00:54
Watch it again a bit later.
00:58
There is p of x.
01:00
So what we have then is p of x is always x minus a, my divisor, times.
01:10
The quotient q of x plus r the remainder that learn off by heart so if i'm dividing by x minus 1 which i am here so x minus 1 p of x q of x unknown r i know is 0 which is a factor factor means no remainder so that will apply if i'm dividing by x minus 1.
01:48
Now how would that help? i can see here if i plug in p of 1 in the words make x1 what happens well p of 1 here will be 1 minus 1 times q of 1 which is unknown but i don't care because that item here is 0 so p of 1 1 then must be 0.
02:21
So this, p of 1 is 0.
02:26
Let's plug it in then.
02:28
So here i get 6, i'm green.
02:33
6 times 1 cubed is 6.
02:36
M times 1 squared is m.
02:39
N times 1 is n minus 5.
02:43
That has to be 0.
02:46
Which means i know m plus n has to be minus 1.
02:52
So there is my first equation.
02:56
M plus n is negative 1...