Hello everyone. In this video, we continue our
voyage through the Fall 2000 exam,
and now we've reached question 7. So
part 1 is state the Rational Roots Theorem.
First part’s actually not too, too bad.
Now of course, on our exam, we won't have this issue
because we'll be given all the theorems,
but let's state it anyways. So let f of x be a polynomial with integer coefficients.
If r equals a over b is a rational root of f of x, then
the numerator must divide a naught
and the denominator must divide the leading coefficient a n.
So clearly this is going to be a question about the Rational Roots Theorem based on the first part.
Part 2: let f of x be the polynomial
4x to the power of 4 plus 8x cubed plus 9x squared plus 5x plus 1.
Determine the rational roots of f of x.
Okay.
So we'd like to determine the rational roots of f of x. So
by the previous theorem we know that the only possibilities are
when a divisor of 1 is in the numerator, and a divisor of 4 is in the denominator.
This gives us 6 possibilities: plus minus 1, plus minus a half, and plus minus a quarter.
Now in order to save time, and as you can see this question is very lengthy,
I'm just going to plug them in and go with it.
So f at 1 gives me 27, f at minus 1 gives me 1,
f at a half gives me 7, f at negative a half gives me 0,
f at a quarter is 189 over 64,
and f at negative 1 quarter is 13 over 64, okay?
Only one of these is 0,
that's the most important part, f at negative 1 half,
so negative 1 half is the only rational root of f of x.
The final question asks us to factor this over four different rings
or four different polynomial rings. So over
Q of x, R of x, C of x, and Z 5 of x, okay?
So clearly we want to use the fact that we have a rational root, namely negative 1 half.
So by the Factor Theorem, we know that if negative 1 half is a rational root,
2x plus 1 must be a factor.
So if we know a factor, we can just divide out
by using long division, and get our other factor.
So let's do our long division, here I've done it in
some paint program.
So 2x plus 1 goes into 4x to the power of 4 plus 8x cubed plus 9x squared plus 5x plus 1.
How do we do this? Well compare leading coefficients, so 2x into
4x to the 4, that's 2x cubed.
So we do the division, and what do we get?
We're going to get - so we multiply 2x plus 1 by 2x cubed,
that gives us 4x to the power of 4 plus 2x cubed.
Subtract we get 6x cubed.
Drop down the other term, 9x squared.
2x goes into 6x cubed 3x squared times.
Write it down.
Okay then simplify, so 9 minus 3 gives us the 6x squared, drop down the 5.
2x goes into 6x squared 3x times. 6x squared plus 3x,
multiply that through, we're going to get 2x plus 1, that's good, we want a remainder 0.
Now when we do this though we should keep note of something, right?
So we do this and we have
that f of x - I'm keeping this hidden for a reason - f of x is equal to 2x plus 1
times 2x cubed plus 3x squared plus 3x plus 1, that's the Division Algorithm.
Now I've always said that the goal of these types of factoring questions
is to get down to a quadratic. When you get down to a quadratic, you know what to do.
We unfortunately still have a cubic,
so if we have a cubic and we still need to factor it,
well we really don't have too many tricks, right? They didn't
give us a factor in this question,
so we're left with only one option
and our one option to attempt to factor this,
and pray to God this is going to work, is that this thing has another factor that's rational,
but we already saw that there's only one possibility for a rational factor,
and that's negative a half.
So negative a half might be a repeated root, that's our only hope at this point.
Let's give it a shot and plug negative a half into this cubic.
This is my comment here: we close our eyes and hope that this is a root.
So this I actually decided to plug in, so plug in negative a half,
simplify, we actually get 0.
2x plus 1 is a repeated factor of f of x.
So let's do another long division.
So if we do our long division, 2x into 2x cubed is x squared,
simplify we get 2x squared plus 3x. Again x
times 2x gives me 2x squared. Do that, simplify, again we get 0.
So now our factorization is
f of x equals 2x plus 1 squared times x squared plus x plus 1.
It should be noted if I expand this out, this is
going to also work over Z mod 5,
which is another field that have to work over so
keep in mind this factorization is good over all four of those rings, okay?
So the roots of this last equation are given by the quadratic formula, right,
I mean, so now we have a quadratic, we know what to do. Plug in the quadratic formula,
and when we plug in we get minus 1 plus or minus root 3i over 2.
These factors are imaginary,
so we know we have the complete factorization over Q
and the complete factorization over R, it's given by this,
and to get the factorization over C, well these are our two roots so just
split them up and write them as factors, okay?
Fantastic.
So now over Z 5, we have to do one little check, right?
So now we have a quadratic polynomial and we want to know if it's irreducible over Z 5.
Well from the last homework question, the last homework of this course, question 7,
tells us that if this thing factors, it must have a root, right?
If it's going to factor as something, it's got to factor as a linear term times a linear term
to get to a quadratic term. So if there's a linear factor, then it must have a root
in Z 5. So just check the five numbers and see if it has a root.
So we plug in 0, 1, 2, 3, 4, we clearly don't get 0 in either case,
and we're done. So thus this has no roots, and hence the factorization
over Z 5 is the same over Q and as over R.
Pretty crazy question, there's a ton of work that went into this question.
If you notice, I think this took three pages, but that's okay.
As long - I mean, you're making progress towards the goal.
It's a tougher question, it’s a longer question, it's a lot of parts, but that's okay.
Again, keep at it, don't give up,
and you have to notice a couple of things to hit the Graceland, you know, to get there, but
it can be done, okay, that's kind of the point that I'm making.
Again, trickier question, a little harder than the previous ones, but
it can still be done.
It helps to read questions beforehand it's sort of a good tip in any exam.
Reading questions beforehand definitely helps and I hope that this video helps
solve this problem, give you a little bit of insight as what to do when you're trying to factor things.
Good luck.