Let's try a rational roots example: so factor
this polynomial as a product of irreducible polynomials.
Actually there's a type of here I'm going to fix this typo and then come back.
Okay turns out there was no typo, it's just my imagination,
but let's factor this polynomial as a product of irreducible polynomials
over the real numbers, R.
So take a minute, try to see if you can factor this.
Pause the video and then come back and hopefully
you have a little bit of an idea of what's going on here.
The first thing I'd like to note is that we do have this
denominator here
and it's a little bit complicated to deal with.
So what I'd like to do is I’d actually like to get rid of the denominator. Well how do I do that?
I can common factor out...
I can common factor out the denominator. So I'm going to common factor out
1 over 15.
So that'll leave me with a new polynomial with all integer coefficients.
And why do I do this? Well once I do that, now I have an integral polynomial
that I can now use the Rational Roots Theorem to try to factor.
So let's do that, we factor out the 1 over 15 to get this polynomial.
By the Rational Roots Theorem, we know that the numerator must divide 2,
and the denominator must divide 15.
So the factors of the denominator are
plus minus 1, 3, 5, and 15,
and the factors of 2 are plus minus 1 or 2.
That leaves us with these 16 possible roots.
That's a lot to try, but the way I would
go about this is I would start by attempting the first…
the integer ones. So try plus or minus 1 and try plus or minus 2.
Sum of the coefficients isn’t 0 if I plug in minus 1, I see that I get
negative 15 minus 32, that's already pretty negative and this is not going to cancel that
So I know plus or minus 1 doesn't work.
Let's try plus or minus 2. If try 2 here, I'm going to get…
this one's going to be my biggest term, it's negative 128.
That's 6 plus 2 that's 8, so I need this to be 120 if it's going to work.
2 cubed is 8, 8 times 15 is 120,
so 120 plus 8 is 128, these two cancel, I found a root, x equals 2.
So hence, by the Factor Theorem, x minus 2 is a factor.
Now you have two options at this point:
you could keep trying rational roots,
or you can do the long division and I'm going to recommend here doing the long division.
Why do I recommend doing that?
The reason why is that if I do
the long division here, I know that I'm going to be left with a quadratic,
and I can factor quadratics. I have the quadratic formula so in the worst case,
I can always factor a quadratic polynomial.
So I have no problems actually doing the long division here
because I know if I do that, I'm going to get to something that I can solve quickly.
So do the long division, just roll up your sleeves and do it.
x minus 2 into 15x cubed minus 32x squared plus 3x plus 2.
So x goes into 15x cubed, 15x squared times.
Multiply 15x squared by x minus 2, I get this.
Take the difference, I get minus 2x squared.
x goes into minus 2x squared minus 2x times, so on and so forth…
and keep going.
That gives me 15x squared minus 2x minus 1. I know my remainder here is 0,
I didn't actually write the last line because I don't really need to,
right, I know that 1 times this is going to give me exactly this,
and it's supposed to because I know it's a factor.
So there we go, there's my quotient.
So if I factor this, I get 1 over 15 times
x minus 2, times 15x squared minus 2x minus 1.
Now this is a quadratic polynomial, worst case I plug it into the quadratic formula.
I'm going to try to…
the first thing I'm going to try to do though is I'm always going to try to factor,
right? If I can actually factor this, then I'm done.
1 doesn't have too many roots, it's got to be
plus 1 and minus 1 in some order.
15, it could be 1 or 15, but that's not going to give me minus 2.
5 and 3 in some combinations should give me minus 2, so if I put
5x and 3x, I should be able to multiply them together to get 15,
and I get the minus 2 in the middle because it's 5x times minus 1 and 3x times 1.
Beautiful, okay.
So I factored this without too much stress at the end, and that's it.
Once I have this factorization I'm done. So notice that, a couple of things,
you could have found the rational roots 1/3,
or you could have found the rational roots negative 1/5 above,
and if you've done that you would have…
you could have factored everything quickly if you found all three using the Rational Root Theorem,
or if you could have found - if you found one of these other ones, then you could have
used long division and then you would have to factor the remaining quadratic,
which is perfectly fine, either way is fine.
This just gives you an example of how to do this.
A conjugate roots example, so how about something a little bit different?
So factor the following polynomial
over the complex numbers
into a product of irreducible polynomials
given that 2 minus 3i is a root.
So here the problem’s a little bit different. Instead of telling you
nothing, we actually tell you a root
and say, “Hey, can you find the other ones?”
Pause the video, give it a shot. Try to see if you can
at least come up with a couple of roots
and see if you can simplify this.
Okay, hopefully you've given this a shot.
One thing to note that when we have a real polynomial,
so this polynomial is real, and we have a complex root, strictly complex root, 2 minus 3i,
we know that
this is one root and another root is given by its conjugate.
That's the Conjugate Roots Theorem, okay
So by the Factor Theorem, we know that z minus those two roots must be factors,
we multiply them together and we get a quadratic factor of this quartic.
So if I do the long division and divide this quartic by the quadratic,
then I'm left with another quadratic, and I know I can factor those.
Notice that this is over C, so the Fundamental Theorem of Algebra,
or CPN, whichever theorem you want to use
tells us that this will factor as a product of four linear things
Four linear terms…
products…elements… polynomials, there we go.
Okay, here we go.
So we do that, we find our quadratic factor,
z minus 2 minus 3i and z minus 2 plus 3i,
that gives us z squared minus 4z plus 13.
We do the long division, I didn't do it
in this slide. You should do it, make sure that you know how to do this.
We already saw one example, there’s two on the website.
I don't feel like I need to do many more.
That gives us the following though: z squared minus 4z plus 13,
times z squared minus z minus 1.
This last polynomial doesn't factor, but we can use the quadratic formula,
right? We have a quadratic,
we know that we can find the roots by using the quadratic formula.
Plug it in, simplify, we get 1 plus or minus root 5 over 2,
and if we simplify this what do we get?
Okay, so sorry, we have [these] roots, this is all there is to it.
So now we have two more roots, we have the two roots from before, we know all the factors,
right, just takes z minus root 1, z minus root 2,
z minus root 3, and z minus the 4th root.
And that's it.
That gives us the factorization of this polynomial.
So there's an example with the conjugate roots.
Again so if you're given a root you have a little bit of information,
use that to your advantage. Try to find another root using, let's say, the Conjugate Roots Theorem
or Rational Roots Theorem, something like this.
Worst case, you divide out and hope for the best,
but hopefully you don't divide out by z minus...
hopefully you don't just divide out by this term because it's going to be a little bit ugly
and you're going to have a lot of complex numbers floating around.
Because you divided out by the product,
we actually ended up with a very, very nice term at the end.
So something to think about.