1 00:00:00,000 --> 00:00:04,000 So the theorem here, Rational Roots 
Theorem, if we have a polynomial of degree n 2 00:00:04,000 --> 00:00:07,233 that's integral, so with integer coefficients, 3 00:00:07,233 --> 00:00:11,766 and we have a rational root, r is 
equal to s over t in lowest terms, 4 00:00:11,766 --> 00:00:16,133 then the numerator of that rational 
root must divide the constant term, 5 00:00:16,133 --> 00:00:18,533 that's what this is saying, 
s divides the constant term, 6 00:00:18,533 --> 00:00:23,433 and the denominator of that rational 
root must divide the leading coefficient. 7 00:00:23,433 --> 00:00:26,366 I won't do an example of this 
theorem here, that will be in part 2 8 00:00:26,366 --> 00:00:28,300 so if you want to check that out, 9 00:00:28,300 --> 00:00:31,200 here I'm just going to do the proof. 
Again, pause the video here. 10 00:00:31,200 --> 00:00:34,233 This is a proof that you should 
be able to do, it's not too bad. 11 00:00:34,233 --> 00:00:39,033 The only trick would be - the only trick while 
doing this proof is to actually get - once you 12 00:00:39,033 --> 00:00:41,866 do step 1, actually get an integer 13 00:00:41,866 --> 00:00:43,533 object. 14 00:00:43,533 --> 00:00:45,600 15 00:00:45,600 --> 00:00:47,233 So hopefully you’ve paused 
the video, come back, 16 00:00:47,233 --> 00:00:49,866 when I said step 1, what 
I really meant was well 17 00:00:49,866 --> 00:00:52,266 when you plug in the rational root, clearly 
that's something that you should do. 18 00:00:52,266 --> 00:00:54,366 You have a root you know 
if you plug it in you get 0, 19 00:00:54,366 --> 00:00:57,066 that sounds like a reasonable 
first step in this problem, 20 00:00:57,066 --> 00:00:59,033 21 00:00:59,033 --> 00:01:00,933 but then once you do that, you get like 22 00:01:00,933 --> 00:01:03,466 a n times s over t to the power of n 23 00:01:03,466 --> 00:01:05,733 and it's like well okay I have 
these rational numbers, the sum 24 00:01:05,733 --> 00:01:07,833 of these rational numbers 
is an integer, it's 0. 25 00:01:07,833 --> 00:01:09,333 How do I deal with that? 26 00:01:09,333 --> 00:01:12,000 And the correct way is to actually 
get rid of the fraction part. 27 00:01:12,000 --> 00:01:15,533 How do we do that? Well we multiply 
by the denominator, which is t to the n. 28 00:01:15,533 --> 00:01:16,400 29 00:01:16,400 --> 00:01:18,566 So if you look at all these terms, so 
I've only written a couple of terms 30 00:01:18,566 --> 00:01:21,133 out here but we also have like 31 00:01:21,133 --> 00:01:26,333 a n minus 1, times s to the n 
minus 1 over t to the n minus 1, 32 00:01:26,333 --> 00:01:29,100 so if I multiply that by t to the n, 
I'm just going to get a t term, 33 00:01:29,100 --> 00:01:33,400 and I'm left with the numerator which 
stays untouched, s to the n minus 1. 34 00:01:33,400 --> 00:01:35,533 And we keep going all the way down. 35 00:01:35,533 --> 00:01:37,066 36 00:01:37,066 --> 00:01:40,000 So what do I do at this point? Well 
now I know I look at this and I have 37 00:01:40,000 --> 00:01:43,700 s divides a, and I want to 
show that s divides a naught. 38 00:01:43,700 --> 00:01:46,366 All of these terms depend on s. 39 00:01:46,366 --> 00:01:46,866 40 00:01:46,866 --> 00:01:49,966 So…by the way somebody in class 
said, “Well you should use DIC,” 41 00:01:49,966 --> 00:01:52,766 and I said, “Well you can, but 
DIC is really only for two terms, 42 00:01:52,766 --> 00:01:56,100 so you have to use some sort of 
like DIC plus induction type thing.” 43 00:01:56,100 --> 00:01:59,200 I don't really want to do that 
so I'm just going to write it out 44 00:01:59,200 --> 00:02:00,533 as a factor like this. 45 00:02:00,533 --> 00:02:03,133 By the way, I'm using a “dot dot dot” 
argument here, which is kind of 46 00:02:03,133 --> 00:02:06,466 contradicting what I said earlier. 
If you really want to be like 47 00:02:06,466 --> 00:02:09,966 super, super formal with this, you 
can write this as a summation. 48 00:02:09,966 --> 00:02:14,100 The reason why I don't is because I think that this 
is a little bit easier to understand and cope with, 49 00:02:14,100 --> 00:02:17,200 but if you want, write it as a 
summation, the proof will pan through. 50 00:02:17,200 --> 00:02:17,900 51 00:02:17,900 --> 00:02:20,000 So I'm going to rewrite 
this by factoring out the s 52 00:02:20,000 --> 00:02:22,200 and bringing all of the first… 53 00:02:22,200 --> 00:02:23,633 54 00:02:23,633 --> 00:02:26,433 how many terms is this? n minus 
1 terms to the other side. 55 00:02:26,433 --> 00:02:28,866 So I'm going to leave a naught 
times 2 to the n on one side, 56 00:02:28,866 --> 00:02:31,766 you know flip the sides because 
it's a little easier to see. 57 00:02:31,766 --> 00:02:36,266 Negative s will factor out of all these 
terms, and I'm left with this integer. 58 00:02:36,266 --> 00:02:37,900 Okay? 59 00:02:37,900 --> 00:02:40,633 Therefore s divides a naught t to the n. 60 00:02:40,633 --> 00:02:41,800 61 00:02:41,800 --> 00:02:45,233 We're almost done, we want to show s 
divides a naught, we got this t to the n here 62 00:02:45,233 --> 00:02:47,266 that's bothering us, 63 00:02:47,266 --> 00:02:50,133 and the trick now to solving 
this is to recall oh well 64 00:02:50,133 --> 00:02:52,900 s over t was a root in lowest terms, 65 00:02:52,900 --> 00:02:56,300 and because it’s a root in lowest 
terms, the gcd of s and t must be 1. 66 00:02:56,300 --> 00:03:00,800 So therefore the gcd of s and t to 
the n must also be 1 by, let's say, 67 00:03:00,800 --> 00:03:02,766 GCDPF that's perfectly reasonable, right? 68 00:03:02,766 --> 00:03:07,366 If s and t don't share a prime factor, then surely 
s and t to the n don’t share a prime factor, 69 00:03:07,366 --> 00:03:09,166 so they must be co-prime. 70 00:03:09,166 --> 00:03:12,200 And if s and t to the n are co-prime, then 
we can use Coprimeness and Divisibility 71 00:03:12,200 --> 00:03:14,200 and say hey well s 
doesn't divide t to the n, 72 00:03:14,200 --> 00:03:16,766 but s divides this product, 
so therefore s must divide 73 00:03:16,766 --> 00:03:17,266 74 00:03:17,266 --> 00:03:19,300 a naught. 75 00:03:19,300 --> 00:03:21,533 And I only proved the 
first part. Why? Well 76 00:03:21,533 --> 00:03:23,833 the proof is similar 
for t divides a to the n. 77 00:03:23,833 --> 00:03:26,933 Take the last… 78 00:03:26,933 --> 00:03:28,900 n minus 1 terms, factor out a t, 79 00:03:28,900 --> 00:03:31,266 bring it to the other side, 
it's the same argument. 80 00:03:31,266 --> 00:03:34,600 So here “similarly” actually is 
valid as a proof technique. 81 00:03:34,600 --> 00:03:37,633 If you need to see it, write it out. If you want 
some practice, you should write out that proof. 82 00:03:37,633 --> 00:03:40,000 That's a very easy proof to 
write out the other direction, 83 00:03:40,000 --> 00:03:41,800 but I won't do it here. 84 00:03:41,800 --> 00:03:42,733 85 00:03:42,733 --> 00:03:45,600 So the Rational Roots Theorem 
gives us one way to help to factor, 86 00:03:45,600 --> 00:03:47,933 and the last theorem that 
we're going to talk about, 87 00:03:47,933 --> 00:03:50,866 I believe unless I'm mistaken, is 
the Conjugate Roots Theorem, 88 00:03:50,866 --> 00:03:54,133 and that will also help us 
try to factor polynomials. 89 00:03:54,133 --> 00:03:57,066 It's a really, really cute theorem, it's 
actually not hard to prove either 90 00:03:57,066 --> 00:03:59,766 so I do encourage you to 
pause the video and try it. 91 00:03:59,766 --> 00:04:02,933 If c is a complex root 
of a real polynomial, 92 00:04:02,933 --> 00:04:04,633 93 00:04:04,633 --> 00:04:08,000 then c bar is a root of p of x. 94 00:04:08,000 --> 00:04:11,533 So if c is - so if we take a 
polynomial with real coefficients 95 00:04:11,533 --> 00:04:15,066 and we try to factor it over C 
and we know that c is a root, 96 00:04:15,066 --> 00:04:18,100 then we also know that 
c bar is a root of p of x 97 00:04:18,100 --> 00:04:20,433 It's kind of neat, and the 
proof really isn't that hard. 98 00:04:20,433 --> 00:04:23,300 You really should pause the video, 
make sure that you can do this proof. 99 00:04:23,300 --> 00:04:24,933 100 00:04:24,933 --> 00:04:28,766 It turns out all you have to do to 
prove this is actually write p of x 101 00:04:28,766 --> 00:04:31,400 as a polynomial a to the n x to the n 102 00:04:31,400 --> 00:04:34,166 plus dot dot dot plus a1 x plus a0, 103 00:04:34,166 --> 00:04:37,433 plug in c, and then take 
the conjugates of both sides. 104 00:04:37,433 --> 00:04:39,700 105 00:04:39,700 --> 00:04:41,700 So here we go, that's exactly 
what I'm doing here. So 106 00:04:41,700 --> 00:04:44,366 write p of x as a polynomial 
with p at c equals 0, 107 00:04:44,366 --> 00:04:48,000 then let's look at p at c bar. Well if 
I take p at c bar, I'm going to just 108 00:04:48,000 --> 00:04:52,000 plug in c bar into x 
for every possible x. 109 00:04:52,000 --> 00:04:52,633 110 00:04:52,633 --> 00:04:54,933 Since my coefficients are real 111 00:04:54,933 --> 00:04:58,300 and by Properties of Conjugates, I can… 112 00:04:58,300 --> 00:04:59,000 113 00:04:59,000 --> 00:05:01,600 so if I take the conjugate of c to the 
power of n, that's the same as taking 114 00:05:01,600 --> 00:05:04,400 c the n and the conjugate of 
that, those are the same. 115 00:05:04,400 --> 00:05:05,266 116 00:05:05,266 --> 00:05:07,200 Since a n is real, 117 00:05:07,200 --> 00:05:10,833 it's the same as a n bar. So 
I'm writing this as all of this. 118 00:05:10,833 --> 00:05:13,900 Now I'm adding a bunch of terms 
that I've taken the conjugate of, 119 00:05:13,900 --> 00:05:18,366 so it's the same as adding all the 
terms then taking the conjugate. 120 00:05:18,366 --> 00:05:21,100 So the sum of the conjugates 
is the conjugate of the sum, 121 00:05:21,100 --> 00:05:23,733 that's by PCJ again, 122 00:05:23,733 --> 00:05:28,833 and this just becomes…this is just 
the polynomial now when I plug in c. 123 00:05:28,833 --> 00:05:29,700 124 00:05:29,700 --> 00:05:32,300 So it’s p of c and I'm just 
taking the conjugate of it, 125 00:05:32,300 --> 00:05:35,266 but p of c is 0, c was a root. 126 00:05:35,266 --> 00:05:38,533 So the conjugate of 0 is 
still 0, we're done. So 127 00:05:38,533 --> 00:05:42,733 c bar is also a root of a real 
polynomial when c is a root. 128 00:05:42,733 --> 00:05:45,566 This is not true if your 
polynomial is not real, right? 129 00:05:45,566 --> 00:05:48,233 If your polynomial has complex 
coefficients, this proof breaks down. 130 00:05:48,233 --> 00:05:52,366 I can't just take the bar of each 
coefficient, that doesn't work. 131 00:05:52,366 --> 00:05:54,433 132 00:05:54,433 --> 00:05:56,833 But when it is real, you can do that. 133 00:05:56,833 --> 00:06:00,866 So something…it's pretty cool. I know 
I think it's a really neat theorem, 134 00:06:00,866 --> 00:06:02,700 and you can use this - so for example 135 00:06:02,700 --> 00:06:06,500 if you can find a complex 
root, a strictly complex root, 136 00:06:06,500 --> 00:06:09,566 then you actually know 2 roots, 137 00:06:09,566 --> 00:06:13,266 and when you know 2 roots, then you can 
multiply, you know, their factors together and 138 00:06:13,266 --> 00:06:17,400 divide out and get some new 
polynomial that’s degree 2 less. 139 00:06:17,400 --> 00:06:18,000 140 00:06:18,000 --> 00:06:20,400 By doing that, you reduce 
your computation and 141 00:06:20,400 --> 00:06:24,466 you reduce your search space and so on and 
so forth. So these give you ways to factor 142 00:06:24,466 --> 00:06:25,933 these polynomials. 143 00:06:25,933 --> 00:06:27,300 144 00:06:27,300 --> 00:06:29,300 I think that's all I want to 
talk about in this video. 145 00:06:29,300 --> 00:06:31,766 I'm just trying to scroll and I don't think I have anything 146 00:06:31,766 --> 00:06:33,100 else to talk about. 147 00:06:33,100 --> 00:06:36,300 So again this was the theoretical video. 148 00:06:36,300 --> 00:06:40,266 If you want to see applications of 
this, check out part 2 of Week 11. 149 00:06:40,266 --> 00:06:42,733 Hopefully this gave you an 
idea of what we did this week. 150 00:06:42,733 --> 00:06:44,866 Hopefully this helps you to try to factor 151 00:06:44,866 --> 00:06:46,866 polynomials, gives you some ideas, 152 00:06:46,866 --> 00:06:51,100 and hopefully it helps clear up some of the proofs 
that you might have been struggling with this week. 153 00:06:51,100 --> 00:06:54,333 Okay that's all I have to say, thank 
you very much and all the best.