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So let's see an example.

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My example here is: 
factor this polynomial

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over Z mod 7.

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Take a minute, pause the video, 
to try to see if you can factor this. 

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And now hopefully if you're 
back, we can give this a shot.

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So what should we try to do? 
Well let's try to find a factor.

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We don't have too many 
techniques at this point,

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so let's just guess and check.

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Now if I plug in the number - so 
if I'm guessing and checking, 

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some simple numbers to plug in should 
be 0 and plus or minus 1, for sure to start,

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and then get a little bit higher, and 
then hopefully you find one by then.

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Here we notice that 1 minus 2 plus 
3 minus 4 plus 2 that actually is 0, 

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and so 1 is a root.

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So the sum of the coefficients 
is 0 if and only if 1 is a root,

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it's an interesting little fun exercise to play 
with yourself, but it's not too hard to prove.

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Okay so we have a root, so let's

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take that root and let's factor it out. 

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So we know that x minus 1 is a factor, we can 
use long division to get that factor out of there,

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and so we have f of x is equal to 
x minus 1 times this polynomial.

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So we'll do a long division example later, I don't 
want to do one now. There's also lots on the

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Math 135 Resources Page that you can 
check out. I think I do at least two there.

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So here we have that 
f of x is equal to this.

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Now we look at this new polynomial, 
and now it's of smaller degree

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and we repeat the same process. We say, 
“Okay well can I factor this polynomial?”

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And the answer is yes we can. 

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The sum of the coefficients is still 0, 
1 minus 1 plus 2 minus 2, it's still 0. 

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So now what we can do is we 
can either do long division again,

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or we can actually group this.

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So take the first two terms,

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and we have a common factor of 
x squared which you can pull out, 

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and we're left with x minus 1. The second 
two terms, they have a common factor of 2,

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which we can pull out and 
we're left with x minus 1, 

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and so both these terms have x 
minus 1, we can factor them out

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and we would get this polynomial here.

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So we have f of x is equal to x minus 
1 squared, times x squared plus 2.

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And now we have to look at this polynomial 
and say, “Well can it be factored?”

46
00:02:02,933 --> 00:02:05,333
Well if it can be factored, again 
just like what we talked about

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in the previous slides, if you can factor a 
quadratic into irreducible polynomials,

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it must be a factor of two - there 
must be two linear factors

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that multiply together to 
give me x squared plus 2.

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But if I have two linear factors 
and they correspond to roots.

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So if this factors over Z mod 7, 
then there must be roots in Z mod 7.

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So just check all the values 
to see if there are any roots. 

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There's only 7 values to check, and the 
operation is done modulo 7, it's pretty quick.

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So check 0, 1, 2, 3, 4, 5, 6,

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we plug them into x squared plus 2 and 
we get the values 2, 3, 6, 4, 4, 6, 2.

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In particular, none of these are 0.

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Actually this last one should be a 3, I apologize,

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but yes, nonetheless, none of these are 0

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so therefore x squared plus 
2 has no root in Z mod 7,

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and the above form is completely 
factorized. And that's it.

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So it gives us a nice little example, there's 
a couple of things to note here though.

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So notice that we knew that x equals 
1 is a root of this polynomial,

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but we didn't know how many 
times x minus 1 was a factor,

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and the way that we did that was we 
took that root, or we took that factor,

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and we factored it out. We got this 
cubic and then we do this again, okay,

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and that gave us an x minus 1 
squared. So we saw that 1 was a

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root, but we didn't see how 
many times 1 was a root.

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And that leads to the concept of multiplicity.

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So we say that this 
root 1 has multiplicity 2

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because x minus 1 squared 
is a factor of my polynomial,

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and that gives us this definition.

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00:03:27,600 --> 00:03:31,000
So the multiplicity of a 
root c inside a field F

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of a polynomial f of x over the field F

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is the largest natural number k such that

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x minus c to the power 
of k is a factor of f of x.

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Notice that we can take k to 
be a natural number because

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we already know that c is 
a root of my polynomial,

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and so k must be at least 1.

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So for example, the multiplicity of the 
root 1 in the last example was 2.

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Okay, so that's multiplicity of roots.

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We've seen an example of how to factor,

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but in this example we just basically guessed 
and hoped that we could find a root.

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00:04:04,666 --> 00:04:08,733
Now one thing to notice that - I’m 
going backwards in the slides,

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but one thing to note is that 
in this particular example,

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we got lucky that we 
found a root. It's possible

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for a quartic polynomial to factor as the 
product of two quadratic polynomials,

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and in which case how 
would we do that? And

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it turns out this question, we're not going 
to talk about this much in this class

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at least over finite fields, but I mean the 
question gets a little bit complicated, right?

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How do I factor this as the product of 
two irreducible quadratic polynomials.

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00:04:32,000 --> 00:04:33,233
It's going to take a little bit of work.

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00:04:33,233 --> 00:04:37,100
You could try a guess and check method, 
and that's probably enough for us

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00:04:37,100 --> 00:04:40,533
but in general, you can sort of see oh 
this question can be really complicated

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00:04:40,533 --> 00:04:43,000
Life is pretty good when you 
have roots but what if this

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factored as, let's say, x 
squared plus 2 squared,

101
00:04:45,700 --> 00:04:48,233
then you might not know 
what the factors are.

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Anyway something to think about.

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So what techniques do we have for finding 
roots? We've already seen an example

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and now I'm going to talk about 
the techniques in general. Well

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one thing we can do is we can try to use the 
Rational Roots Theorem to guess a rational root,

107
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depending on what field were in, right?

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00:05:02,633 --> 00:05:04,833
So in the previous example, 
we were in Z mod 7.

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Rational Root Theorem doesn't 
make much sense there, but

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we can still try to trial and 
error and try to guess a root.

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That's still does make sense.

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We could use the Conjugate Roots 
Theorem, that we'll see in a little bit.

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We've seen an example already, I 
talked about it I didn't show you, but 

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we did factoring and grouping. You'll see 
maybe a couple examples of this later. 

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We can use long division once we find a root 
to reduce our problem to something smaller,

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and if we get down to 
a quadratic, life is good.

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We know the quadratic formula, and 
hopefully we can use that to find roots.

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So these are the techniques that I would 
have if I'm trying to factor something.

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00:05:35,733 --> 00:05:38,000
So keep these in mind when 
you approach a question

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of factoring polynomials.