1 00:00:00,566 --> 00:00:04,433 Hello everyone. Welcome to Week 11 
part 1 of Carmen's Core Concepts, 2 00:00:04,433 --> 00:00:06,033 my name is Carmen Bruni, 3 00:00:06,033 --> 00:00:08,766 and in these videos we talk about the 4 00:00:08,766 --> 00:00:11,566 current week in Math 135. 5 00:00:11,566 --> 00:00:12,366 6 00:00:12,366 --> 00:00:14,533 This time I've decided to 
break it up into two parts, 7 00:00:14,533 --> 00:00:17,700 part 1 and part 2, and I've decided to make part 1 8 00:00:17,700 --> 00:00:19,966 a more theoretical part where 
we talk more about the 9 00:00:19,966 --> 00:00:23,133 proofs related to the theorems 
that we did this week, 10 00:00:23,133 --> 00:00:25,500 and in part 2 we talk about 11 00:00:25,500 --> 00:00:27,200 12 00:00:27,200 --> 00:00:29,633 the applications of the theorems. 13 00:00:29,633 --> 00:00:32,000 So in theory, the two parts are independent, 14 00:00:32,000 --> 00:00:35,566 however you should probably watch 
part 1 before you watch part 2 15 00:00:35,566 --> 00:00:39,666 unless you feel very comfortable with the theory, 
then you can go ahead straight to part 2, 16 00:00:39,666 --> 00:00:40,666 17 00:00:40,666 --> 00:00:42,400 either way is fine. 18 00:00:42,400 --> 00:00:45,033 With that being said, let's begin. 19 00:00:45,033 --> 00:00:47,466 So polynomial statements. 20 00:00:47,466 --> 00:00:51,000 So let f and g be non-zero polynomials over a field F 21 00:00:51,000 --> 00:00:55,600 such that they are not additive inverses, and we'll 
see why we need that condition in a second, 22 00:00:55,600 --> 00:00:58,900 then we get the following 
for free. So we get 23 00:00:58,900 --> 00:01:03,400 the degree of f plus g is less than or equal to 
the max of [the] degree of f and the degree of g. 24 00:01:03,400 --> 00:01:07,733 The degree of f times g is the 
sum of the degrees of f and g, 25 00:01:07,733 --> 00:01:10,033 and if f divides g and g divides f, 26 00:01:10,033 --> 00:01:13,400 then f is equal to c g for some 27 00:01:13,400 --> 00:01:15,333 element of our field, F. 28 00:01:15,333 --> 00:01:18,100 I want to talk about each of 
these a little bit individually. 29 00:01:18,100 --> 00:01:20,300 Why do I need f and g to 
not be additive inverses? 30 00:01:20,300 --> 00:01:23,700 Well the first one, then we would 
get the degree of the 0 polynomial 31 00:01:23,700 --> 00:01:26,633 In this course, we say that the [degree 
of the] 0 polynomial is undefined. 32 00:01:26,633 --> 00:01:29,100 Some textbooks, and it's not 
unreasonable to do so, but 33 00:01:29,100 --> 00:01:31,466 it does require a little bit of understanding, 34 00:01:31,466 --> 00:01:34,333 define the degree of the 0 
polynomial to be negative infinity, 35 00:01:34,333 --> 00:01:37,200 and it’s so that statements like 
this actually hold for free, right? 36 00:01:37,200 --> 00:01:40,766 Negative infinity should, morally 
speaking, be less than any integers, so 37 00:01:40,766 --> 00:01:42,000 38 00:01:42,000 --> 00:01:44,033 that's perfectly fine, 39 00:01:44,033 --> 00:01:47,766 but we're not going to do that. We just say 
the degree of the 0 polynomial is undefined. 40 00:01:47,766 --> 00:01:48,566 41 00:01:48,566 --> 00:01:52,000 That being said, it's always good to 
double check some of these theorems. 42 00:01:52,000 --> 00:01:54,766 So why is this a less than 
or equal to statement? Well 43 00:01:54,766 --> 00:01:57,366 the leading terms of f and g 
could cancel each other out, 44 00:01:57,366 --> 00:02:02,200 that's possible, so you actually might 
get the strictly less than case, okay? 45 00:02:02,200 --> 00:02:03,966 Usually though, as long as those two 46 00:02:03,966 --> 00:02:06,700 terms don't cancel each other 
out, this will be an equality, 47 00:02:06,700 --> 00:02:10,700 that's something you can check. These aren’t 
hard to prove, but I'm not going to prove them here. 48 00:02:10,700 --> 00:02:13,566 We proved the last one in class, I 
mentioned the second one in class, 49 00:02:13,566 --> 00:02:16,066 and the first one is 
actually not too hard either. 50 00:02:16,066 --> 00:02:18,300 Just write out f and g 
in terms of polynomials, 51 00:02:18,300 --> 00:02:21,100 add them up, and then argue why it 
must be less than or equal to the - 52 00:02:21,100 --> 00:02:23,800 and then argue what the 
degree of the sum must be. 53 00:02:23,800 --> 00:02:25,300 54 00:02:25,300 --> 00:02:26,533 55 00:02:26,533 --> 00:02:28,766 That's I think - oh and the last one, 
let me talk a little about the last one. 56 00:02:28,766 --> 00:02:30,766 The second one is also fairly 
clear once you do it. You 57 00:02:30,766 --> 00:02:33,833 just multiply the leading two 
terms and that'll be the sum. 58 00:02:33,833 --> 00:02:38,166 You never get 0 when you multiply the 
two terms because you're over a field, 59 00:02:38,166 --> 00:02:40,666 so you can't get cancellation 
so this is actually okay, 60 00:02:40,666 --> 00:02:42,866 but this does require that we have a field. 61 00:02:42,866 --> 00:02:44,133 62 00:02:44,133 --> 00:02:47,333 f divides g and g divides f, the 
way I like to talk about this one - 63 00:02:47,333 --> 00:02:49,566 or what I’d like to talk about this one 64 00:02:49,566 --> 00:02:52,000 is I'd like to relate back to the integer case. 65 00:02:52,000 --> 00:02:55,266 So if I have a divides b and b 
divides a, then what do we know? 66 00:02:55,266 --> 00:02:55,866 67 00:02:55,866 --> 00:02:57,633 As integers… 68 00:02:57,633 --> 00:03:00,900 we think about this then we 
say, “Okay well we know that 69 00:03:00,900 --> 00:03:03,133 a could be equal to b,” that’s a possibility. 70 00:03:03,133 --> 00:03:07,000 and the case that a lot of us forget is that a 
could also be equal to the negation of b. 71 00:03:07,000 --> 00:03:09,866 So realistically, a is equal to plus or minus b, 72 00:03:09,866 --> 00:03:10,800 73 00:03:10,800 --> 00:03:12,633 and it turns out that 
plus or minus 1 happens 74 00:03:12,633 --> 00:03:17,300 to be the only two elements that 
are invertible in the integers, 75 00:03:17,300 --> 00:03:21,066 and that actually matters a little bit. You'll 
see where that comes up in the proof, 76 00:03:21,066 --> 00:03:24,666 but I won't talk about that anymore, 
but this is what I like to - or 77 00:03:24,666 --> 00:03:28,066 what I'm trying to get at here is the analogy 
here, right? So we have these polynomials, 78 00:03:28,066 --> 00:03:31,466 we have this notation for integers, 
how do these two things relate? 79 00:03:31,466 --> 00:03:35,500 Well here we don't get just plus or minus 1, 
we get any invertible element in our field. 80 00:03:35,500 --> 00:03:37,733 81 00:03:37,733 --> 00:03:40,566 Turns out that 0 won't work 
because if f is 0 and g is 0, 82 00:03:40,566 --> 00:03:44,500 we didn't define 0 dividing 0 as a 
polynomial, but that's not a big deal. 83 00:03:44,500 --> 00:03:45,500 84 00:03:45,500 --> 00:03:49,500 Oh we can't have them as 0 because f 
and g are non-zero, either way is fine. 85 00:03:49,500 --> 00:03:51,600 But yeah, so something to 
think about here, right? 86 00:03:51,600 --> 00:03:54,000 When you see a new problem 
in an unfamiliar situation, 87 00:03:54,000 --> 00:03:56,300 reduce back to a situation 
that is familiar to you, 88 00:03:56,300 --> 00:04:00,333 and try to reason out what the correct 
statement and the correct proof should be there. 89 00:04:00,333 --> 00:04:01,166 90 00:04:01,166 --> 00:04:05,200 It’s a good trick in mathematics. If the problem is 
too hard, get rid of it and solve an easier one. 91 00:04:05,200 --> 00:04:08,833 This works well for, you know, 
assignments and when you're 92 00:04:08,833 --> 00:04:11,833 doing work on napkins and 
[stuff] like that, but you know 93 00:04:11,833 --> 00:04:13,900 be careful on a test, right? Don't
just change the problem, 94 00:04:13,900 --> 00:04:16,000 make it easier, and then 
solve the easier problem. 95 00:04:16,000 --> 00:04:17,500 96 00:04:17,500 --> 00:04:20,733 It is a good tactic to get a feeling 
for what the problem's saying. 97 00:04:20,733 --> 00:04:22,766 98 00:04:22,766 --> 00:04:27,466 Alright, Remainder Theorem. So we're going to 
start with our one by one theorem-crunching here, 99 00:04:27,466 --> 00:04:32,766 and then in the next…like I said in the 
next part, we'll talk about the applications. 100 00:04:32,766 --> 00:04:35,100 So what does the Remainder Theorem say? Well 101 00:04:35,100 --> 00:04:36,166 102 00:04:36,166 --> 00:04:40,233 it gives us a way to divide polynomials 
when we have a very specific form; 103 00:04:40,233 --> 00:04:43,566 if we just care about the remainder. 
So if we're dividing a polynomial by 104 00:04:43,566 --> 00:04:46,866 a linear term x minus c, so 
this specific form x minus c, 105 00:04:46,866 --> 00:04:49,566 then the remainder must be f at c. 106 00:04:49,566 --> 00:04:52,766 This will become important for us when 
we talk about the Factor Theorem. 107 00:04:52,766 --> 00:04:57,133 It turns out that we actually divide by these 
types of polynomials more than you might imagine. 108 00:04:57,133 --> 00:04:59,733 109 00:04:59,733 --> 00:05:04,000 That being said, pause the video, see if you 
can give this proof a shot. It's not too bad, 110 00:05:04,000 --> 00:05:07,700 none of these are too, too bad. Some of 
them are a little bit more involved, but 111 00:05:07,700 --> 00:05:10,000 the proof technique 
part of it is not too bad. 112 00:05:10,000 --> 00:05:10,733 113 00:05:10,733 --> 00:05:15,166 So hopefully after you've paused it, you've 
given this a shot, you've tried it out, 114 00:05:15,166 --> 00:05:17,900 how do we go through this? 115 00:05:17,900 --> 00:05:20,133 Well this is the Remainder Theorem, 
so we need to talk about a remainder, 116 00:05:20,133 --> 00:05:23,300 so we should actually just 
divide f of x by this x minus c 117 00:05:23,300 --> 00:05:25,666 using the Division Algorithm for Polynomials. 118 00:05:25,666 --> 00:05:28,900 That gives us some quotient 
and some remainder, 119 00:05:28,900 --> 00:05:31,033 and then we need to argue 
whether [the] remainder is f at c. 120 00:05:31,033 --> 00:05:35,133 Well if we're dividing by a linear polynomial, 
we know the remainder must be constant, 121 00:05:35,133 --> 00:05:37,300 and the question is, 
“Well what constant is it,” 122 00:05:37,300 --> 00:05:39,866 and that's what we can try to figure out. 123 00:05:39,866 --> 00:05:43,366 So using the Division Algorithm gives 
us these polynomials q and x… 124 00:05:43,366 --> 00:05:45,266 or q of x and r of x. 125 00:05:45,266 --> 00:05:48,000 They turn out to be unique 
when we require that r of x is 0 126 00:05:48,000 --> 00:05:51,266 or the degree of r of x is less 
than the degree of x minus c, 127 00:05:51,266 --> 00:05:54,166 and the degree of x minus 
c, we know that, that's 1. 128 00:05:54,166 --> 00:05:59,300 So what does that mean? Well that means 
that the degree of r of x is 0, or r of x is 0, 129 00:05:59,300 --> 00:06:01,466 but the only degree 0 
polynomials are constants, 130 00:06:01,466 --> 00:06:06,133 so therefore r of x is equal to some 
constant k for some k inside the field F, 131 00:06:06,133 --> 00:06:09,700 and here k can be 0 which also 
encompasses the first remainder case. 132 00:06:09,700 --> 00:06:11,066 133 00:06:11,066 --> 00:06:15,100 So how do we determine what k is? Well 
we have this relationship between f, and 134 00:06:15,100 --> 00:06:16,900 q, and r, and x minus c, 135 00:06:16,900 --> 00:06:19,166 so let's actually substitute a value in for x. 136 00:06:19,166 --> 00:06:21,466 And what value are we going to substitute in? 137 00:06:21,466 --> 00:06:24,666 We're going to substitute in 
the value x is equal to c. 138 00:06:24,666 --> 00:06:27,700 That makes sense, c is inside my 
field, this is perfectly reasonable, 139 00:06:27,700 --> 00:06:29,900 and when I do that, the 
middle term goes away 140 00:06:29,900 --> 00:06:32,666 and I'm left with f at c is equal to r at c, 141 00:06:32,666 --> 00:06:36,166 but r at c is just k, right, r at 
c is a constant polynomial. 142 00:06:36,166 --> 00:06:38,366 So k is equal to f at c 
and we know that r of x 143 00:06:38,366 --> 00:06:41,000 must be the constant polynomial 
f at c, and we're done. 144 00:06:41,000 --> 00:06:42,833 145 00:06:42,833 --> 00:06:45,133 That's the Remainder Theorem, and it's 146 00:06:45,133 --> 00:06:49,233 particularly useful for us to talk about this in 
the context of the Factor Theorem, right? 147 00:06:49,233 --> 00:06:53,600 So what - or how does this help us with 
the Factor Theorem? Well it says okay well, 148 00:06:53,600 --> 00:06:57,166 I know what the remainder is and I 
want to know that x minus c is a factor, 149 00:06:57,166 --> 00:06:59,233 well then I want a 0 remainder. 150 00:06:59,233 --> 00:07:01,333 So therefore I would want 
f at c to be 0, and if I 151 00:07:01,333 --> 00:07:03,800 get f at c to be 0, then 
I know that c is a root 152 00:07:03,800 --> 00:07:06,166 and all of this kind of ties 
together to the Factor Theorem. 153 00:07:06,166 --> 00:07:06,966 154 00:07:06,966 --> 00:07:08,966 So suppose that f of x is a 155 00:07:08,966 --> 00:07:11,000 polynomial over the field F 156 00:07:11,000 --> 00:07:13,666 and c is some constant in F, then 
the polynomial x minus c is a 157 00:07:13,666 --> 00:07:16,566 factor of f of x if and 
only if f at c is 0, 158 00:07:16,566 --> 00:07:18,500 that is, c is a root of f of x. 159 00:07:18,500 --> 00:07:20,200 I've kind of already given 
you the proof, but if you 160 00:07:20,200 --> 00:07:23,466 don't see it yet, you should 
pause the video and try it, 161 00:07:23,466 --> 00:07:24,133 162 00:07:24,133 --> 00:07:25,966 and hopefully you've come back after trying it. 163 00:07:25,966 --> 00:07:28,000 Let's see the proof. It's 
actually pretty quick. 164 00:07:28,000 --> 00:07:29,466 165 00:07:29,466 --> 00:07:31,933 So when is x minus c a 
factor of f of x? Well if I 166 00:07:31,933 --> 00:07:33,733 divide by x minus c, 167 00:07:33,733 --> 00:07:36,000 if it’s a factor, then the 
remainder must be 0. 168 00:07:36,000 --> 00:07:37,300 169 00:07:37,300 --> 00:07:39,166 That's just by the Division 
Algorithm for Polynomials, right, 170 00:07:39,166 --> 00:07:42,466 the quotient and the remainder must be unique 
so if I divide by x minus c and it's a factor, 171 00:07:42,466 --> 00:07:44,000 then it must have a 0 remainder. 172 00:07:44,000 --> 00:07:44,866 173 00:07:44,866 --> 00:07:46,433 And when does this have 
a 0 remainder? Well 174 00:07:46,433 --> 00:07:49,800 the Remainder Theorem says that 
r of x is actually equal to f at c, 175 00:07:49,800 --> 00:07:52,666 but r of x is 0, so therefore f at c is 0, 176 00:07:52,666 --> 00:07:56,966 and this goes in both directions, right? 
So the remainder is always f at c 177 00:07:56,966 --> 00:07:57,700 178 00:07:57,700 --> 00:08:00,000 and f at c is 0 if and only if r of x is 0 179 00:08:00,000 --> 00:08:03,466 because they're equal to each other and 
that's actually it, it's actually pretty quick. 180 00:08:03,466 --> 00:08:05,266 181 00:08:05,266 --> 00:08:08,066 I've written it out pretty compactly 
and pretty quickly, but 182 00:08:08,066 --> 00:08:09,966 the idea I think is pretty clear. 183 00:08:09,966 --> 00:08:10,733 184 00:08:10,733 --> 00:08:12,666 And this gives us the Factor Theorem. 185 00:08:12,666 --> 00:08:13,133 186 00:08:13,133 --> 00:08:15,799 This is useful for finding roots 
which we'll see in the next video.