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So as an example, prove that a real 
polynomial of odd degree has a real root.

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So again, pause the video. 
It's good to actually try this.

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The first step in actually 
attacking this problem is

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really getting a handle on what

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kind of proof technique to use, and I find it easiest 
to actually do this using a proof by contradiction.

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00:00:19,900 --> 00:00:23,900
So if we actually assume that a real polynomial 
of odd degree does not have a [real] root,

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then we actually know from the previous 
theorem what all the factors must be.

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The previous theorem tells us 
that all the factors must be quadratic.

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If all the factors are quadratic, then

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the total degree must be even. It's got to 
be 2 times the number of quadratic factors,

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and that's the contradiction since 
we have a polynomial of odd degree.

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So let's just actually write this up. So 
assume towards a contradiction that

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p of x is a real polynomial of odd 
degree without a [real] root.

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By the Factor Theorem, it can't 
have any real linear factors,

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so something to think about here, 
let me pause at the sentence.

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The Factor Theorem what does 
it really tell you? It tells you…

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00:00:55,100 --> 00:00:57,000
it's sort of like a change in dictionary.

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[Real] roots of a polynomial are the same -

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or correspond to linear factors 
of a polynomial, and vice versa.

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So you have this duality and this change in 
dictionary is kind of what's happening here.

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This is a very common 
theme throughout algebra

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that you often look at 
something in a different light.

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By changing the name, it seems 
like you're not doing anything,

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00:01:16,600 --> 00:01:19,600
but by looking at it in a 
different viewpoint, you actually

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can get more information.

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So we can't have a real linear factor, 
so by Real Factors of Real Polynomials,

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we see that all the factors 
must be quadratic.

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So what is the degree of 
this? Well this is 2 times k,

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and this is something that's 
supposed to have odd degree.

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2k is always even, that's a contradiction.

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So hence the polynomial 
must have a real root,

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and that's it.

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00:01:41,300 --> 00:01:43,800
So again, if these types of 
questions interest you,

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00:01:43,800 --> 00:01:47,600
one of my favorite courses in undergrad 
was a course in Galois Theory that I took.

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So if you like dealing with polynomials and talking 
about roots in different fields and things like that,

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I recommend highly a course in Galois Theory.

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Okay.

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That being said, let's do 
one more final example.

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00:02:00,700 --> 00:02:02,733
So something that I don't do 
in class a lot of is actually

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00:02:02,733 --> 00:02:04,500
trying to find square roots 
of complex numbers,

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00:02:04,500 --> 00:02:07,500
which might come in handy 
if you're trying to solve

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00:02:07,500 --> 00:02:11,500
for factors of a quadratic 
polynomial that are not nice.

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00:02:11,500 --> 00:02:12,900


50
00:02:12,900 --> 00:02:14,500
So as an example,

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00:02:14,500 --> 00:02:16,300


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00:02:16,300 --> 00:02:20,300
I'm going to - let's talk about this one: 
find the square root of 5 plus 2i.

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00:02:20,300 --> 00:02:23,500
Now I've used this symbol here 
which is actually a little bit…

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55
00:02:25,100 --> 00:02:26,333
a little bit bad.

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00:02:26,333 --> 00:02:29,500
I mean we kind of understand what this 
means, this really means find the values z

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such that z squared is equal to 5 plus 2i.

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59
00:02:32,833 --> 00:02:35,233
So, again, pause the video 
and actually try to solve that.

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61
00:02:36,600 --> 00:02:40,000
Hopefully you've actually 
tried, given this a shot,

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00:02:40,700 --> 00:02:43,000
and it's actually not bad when you 
reword it the way I've worded it, right?

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00:02:43,000 --> 00:02:47,400
Find number z squared - find number z 
so that z squared is equal to 5 plus 2i.

65
00:02:47,400 --> 00:02:49,000


66
00:02:49,000 --> 00:02:50,500
How do we…

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00:02:50,500 --> 00:02:53,700
so how do we do that? Well we 
can convert 5 plus 2i to polar form,

68
00:02:53,700 --> 00:02:58,000
and then actually take…
and then use the Complex

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00:02:58,000 --> 00:02:59,833
nth Roots Theorem.

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00:02:59,833 --> 00:03:01,833


71
00:03:01,833 --> 00:03:06,300
So as I said, we're seeking values such 
that z squared is equal to 5 plus 2i.

72
00:03:06,300 --> 00:03:09,000
There's a couple ways to do this, you 
can actually write z as x plus y i,

73
00:03:09,000 --> 00:03:11,500
plug it through, and try to solve it,

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00:03:11,500 --> 00:03:15,133
or you can actually write 
this in polar form and then

75
00:03:15,133 --> 00:03:18,000
try to…and then just use CNRT.

76
00:03:18,000 --> 00:03:18,900


77
00:03:18,900 --> 00:03:20,700
And that's what we're going to use here.

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00:03:20,700 --> 00:03:23,333
So I'm going to write 5 plus 2i in polar form.

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00:03:23,333 --> 00:03:27,500
So the modulus of 5 plus 2i is 5 
squared plus 2 squared, that's 29,

80
00:03:27,500 --> 00:03:30,700
and take the square root of 
that, so square root of 29.

81
00:03:30,700 --> 00:03:33,600
And what's my angle here? My angle 
is not nice, it's going to be y over x.

82
00:03:33,600 --> 00:03:36,633
This lives in the first quadrant, 
so arctan of 2 over 5 is

83
00:03:36,633 --> 00:03:38,633
the correct answer here.

84
00:03:38,633 --> 00:03:41,200
It's not arctan 2 over 5 plus pi.

85
00:03:41,200 --> 00:03:43,733
If you don't understand that you should 
go back a couple videos and look at

86
00:03:43,733 --> 00:03:47,200
when we talked about converting from 
polar to standard form and vice versa.

87
00:03:47,200 --> 00:03:48,800


88
00:03:48,800 --> 00:03:51,700
So thus, once we know this 
is the answer, so this is like

89
00:03:51,700 --> 00:03:56,633
29 to the 1/2, so I take another 1/2, and 
that will give me the 4th root of 29,

90
00:03:56,633 --> 00:03:59,833
times e to the i theta over 2, that's 
going to be a solution to this, right?

91
00:03:59,833 --> 00:04:03,700
If I square this, I get exactly this 
number, which is what I wanted.

92
00:04:03,700 --> 00:04:06,100
And the other solution is given by CNRT.

93
00:04:06,100 --> 00:04:10,733
The other solution is you just 
add even multiples of 2 pi over

94
00:04:10,733 --> 00:04:13,700
n, where n is the degree 
that we're looking for.

95
00:04:13,700 --> 00:04:16,500
Here it's 2 so 2 pi over 2 is just pi.

96
00:04:16,500 --> 00:04:18,800
So we add an i pi here.

97
00:04:18,800 --> 00:04:22,533
So we're adding pi to theta over 2, which is 
the same as adding i pi to the whole thing

98
00:04:22,533 --> 00:04:25,300
Either way you can use brackets 
or do it like this, it’s fine.

99
00:04:25,300 --> 00:04:27,000


100
00:04:27,000 --> 00:04:30,366
Either way, we see our two answers 
are plus minus the 4th root of 29,

101
00:04:30,366 --> 00:04:32,833
e to the i theta over 2.

102
00:04:32,833 --> 00:04:35,033
You could try to convert back 
to standard form. I don't

103
00:04:35,033 --> 00:04:37,833
think the answer is going to be 
nice, maybe it is I'm not sure.

104
00:04:37,833 --> 00:04:39,700
I think it'll be fairly ugly.

105
00:04:39,700 --> 00:04:40,600


106
00:04:40,600 --> 00:04:43,800
Since we didn't tell you 
which form to put it in,

107
00:04:43,800 --> 00:04:45,800
leave it in this form.

108
00:04:45,800 --> 00:04:49,033
Oftentimes, you'll know the angle 
theta and it'll be one of those

109
00:04:49,033 --> 00:04:52,733
common angles like pi over 3 or pi 
over 6 and it's actually not too bad.

110
00:04:52,733 --> 00:04:55,733
This one I just picked a number out of my hat

111
00:04:55,733 --> 00:04:58,000
and went with it. But hopefully

112
00:04:58,000 --> 00:05:00,633
this gives you a little bit of an

113
00:05:00,633 --> 00:05:04,000
idea. Hopefully it helps you to 
solve these types of problems.

114
00:05:04,000 --> 00:05:05,733
So again, where's this most useful? 
It's most useful when you're

115
00:05:05,733 --> 00:05:08,233
trying to use the quadratic 
formula on a complex polynomial.

116
00:05:08,233 --> 00:05:09,133


117
00:05:09,133 --> 00:05:10,333


118
00:05:10,333 --> 00:05:12,000
And I think that's basically it. So,

119
00:05:12,000 --> 00:05:14,333
you know, if you've watched the previous 11 videos

120
00:05:14,333 --> 00:05:16,933
and you somehow made it to video 12,

121
00:05:16,933 --> 00:05:21,000
thank you very much for listening. I 
think it's been - hopefully this has been

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00:05:21,000 --> 00:05:22,433
a good video series.

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00:05:22,433 --> 00:05:26,000
If you have any questions or comments, 
as always feel free to email me

124
00:05:26,000 --> 00:05:28,000
cbruni@uwaterloo.ca,

125
00:05:28,000 --> 00:05:28,833


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00:05:28,833 --> 00:05:30,600
and that's it.

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Thank you very much. Good luck.