00:01
In this problem, we consider the polynomial f of x equals x cubed minus 2x minus 2.
00:06
Our first job is to show that the polynomial is irreducible over the rationals.
00:11
And then the second part is to let theta be a root of the polynomial to simplify those expressions in q adjoin theta.
00:20
So for the first task in part a, we note that the polynomial is cubic.
00:26
Cubic.
00:27
So if it has any reducible factors, if it can be factored over q, then one of them has to be linear.
00:36
And so we are going to use the rational roots theorem and say that the only possible rational roots or zeros would be plus or minus 1 or plus or minus 2, because again the constant term is negative 2 and the leading coefficient is 1.
01:02
And so so one of those has to be a root to produce a linear factor.
01:09
It turns out that none of these work because f does not equal 0, f does not equal 0, and neither f nor f will be 0.
01:31
Just a quick calculation there shows that f cannot have a linear factor over the rationals, and therefore since it's cubic it cannot be reduced.
01:46
So just based on that calculation in the rational roots or rational zeros theorem we conclude that the polynomial is irreducible over the rationals.
01:57
It is irreducible.
01:59
I can spell that.
02:00
There we go.
02:06
All right so any zero or root of f will be irrational.
02:11
So in part b we let theta be such a root.
02:23
And since f of theta is 0, that tells us that theta cubed minus 2 theta minus 2 is 0, and therefore, any time we see theta cubed in our calculations, we can replace that by 2 theta plus 2.
02:41
That becomes one of the key elements in the simplifications we're going to do.
02:47
Well, using that, the product we want to simplify is pretty straightforward.
02:51
Forward, we're going to multiply out 1 plus theta times 1 plus theta plus theta squared.
02:58
When we hit that theta cubed, we will replace it with 2 theta plus 2.
03:05
So using the usual distributive property, we get 1 plus 2 theta plus 2 theta squared plus theta cubed.
03:15
Now we're going going to replace that by 2 theta plus 2, and then we will simplify, and we see that we have 3 plus 4 theta plus 2 theta squared.
03:42
So there's the first simplification we're asked to do.
03:48
Now the second calculation is a little bit more involved because we have a quotient.
03:52
We are trying to simplify of 1 plus theta over 1 plus theta plus theta squared.
04:00
So what we want to do first is the 1 over 1 plus theta plus theta squared.
04:06
That is, we want to find the inverse of that element in q theta.
04:13
Well, that takes some calculating.
04:17
We're going to use the euclidean algorithm for polynomials.
04:22
We're going to take the original polynomial, f of theta.
04:27
Theta, we're going to divide it by our, we'll just turn it around, theta squared plus theta plus 1, and get a quotient and a remainder.
04:39
So using long division, you can show that the quotient is theta minus 1, and the remainder is minus 2 theta minus 1.
04:53
All right, so the the euclidean algorithm then says we should take, we're keeping track of this, our dividend, if you will, the divisor and the remainder.
05:06
So the old divisor becomes the new dividend and the old remainder becomes the new divisor.
05:12
That is, we're going to take theta squared plus theta plus one and now divide by negative two theta minus one.
05:23
So now we get a quotient of negative one -half theta minus one -fourth plus three -fourths.
05:35
We got a constant remainder this time.
05:40
All right, so how does that help us with this constant remainder? so let's turn this around and let's express this remainder as a linear combination of the element whose inverse we're trying to find and this element.
05:54
Of course, this element, the theta cubed minus 2 theta minus 2 will collapse to 0 in q theta.
06:01
So we're going to work backwards here.
06:03
We're going to take that last statement we have and write 3 fourths and the form theta squared plus theta plus 1 minus, whoops, get that plus sign there, minus the rest.
06:28
So i'm just solving for 3 fourths in that previous equation.
06:33
And now what we want to do again is try to keep track of where we were with our remainders...