Gelfand: Chunk 16 - Polynomial Division - Remainders, Bezout and Roots

March 7, 2017

There Must be an Easier Way…

All this polynomial division is very exciting and all, but it’s also very long-winded. Surely there’s an easier way? (At least in certain circumstances?)

Reassuringly there is, at least for remainders, and factoring, and this is all up next, courtesy of Bezout and his Polynomial Remainder Theorem.Bezout was a busy guy, and there are other theorems he came up with which you should not confuse with this one. There is even one called “Bezout’s Theorem” which is not this one. Confusing, right?

I’ll let you go to GelfandSection 37. for the simple reason why this works, but in general, if ​$P$ is an arbitrary polynomial (i.e. any polynomial in a single variable, ​$x$ in what follows) which we want to divide by ​$x - a$ then we can use the following equality to give us a jump on things:

This simple formula gives us a bunch of cool stuff and these cool things are the subject of the rest of this post.

Tricks Arising (Pt. 1)

Find the Remainder Quickly and Simply - Bezout’s Remainder Theorem

We can use Bezout’s theorem in a sneaky way. To find the remainder when ​$P$ is divided by ​$x - a$, simply substitute ​$a$ for ​$x$. We don’t need to calculate the quotient to be able to do this either.

E.g. we have a polynomial ​$P$ which is ​$x^4$, and we want to know what the remainder is when we divide it by ​$x - 2$. In this scenario, ​$a = 2$, and we can put this along with ​$x = 2$ into our equality.

This is the aforementoned Remainder Theorem, discovered by Etienne Bezout.Note, this is only a method for finding the remainder. If you need the quotient, you still have to take the long way round as we described in Chunk 14.

Find out if ​$P$ is Divisible by ​$x - a$, aka “Roots” to Factoring Joy

The previous trick was nice, but not too magical. Up next is a consequence of Bezout’s theorem, and it’s going to give us another weapon in our fight against factoring.

Now that’s magical.

First we need to take a step back. Two in fact. Forget about polynomials, and forget about division. We’re going to introduce properly the concept of polynomial roots.Polynomial roots are not roots of numbers such as square roots and cube roots. I’ve written a separate section to disambiguate all this. If you feel the need jump over there now to make sure you’re happy with the difference, and what we are and arent talking about here.

Polynomials have things called roots. If I tell you they are also called a “zero” you might think back to the last time we heard the “zero” word with regards to factoring, in Chunk 11 - Factoring to Zero. In that post we found great solace for our annihilated-terms woes by “looking for the zero result”.

Gelfand is now going to take the finding-the-remainder element of this approach, and bring us up to speed with roots in general (without explicitly calling out the link I’ve just made - I found it by reading around the topic). (Oakley FTW!)

The definition of a root ​$x = a$ of a function ​$f(x)$ from the above linked page is as follows:

I.e. the number represented by ​$a$ is a root of the function represented by ​$f(x)$ because when you make ​$a = x$ the result is ​$0$.

Pause a while for that to sink in. It’s very simple, but powerful. It means if we find an ​$a$ (and there could be more than one, then we have a root of ​$f(x)$.

Now consider the fact that ​$f(x)$ is just another way of representing our polynomial, ​$P$. We have seen that ​$a$ is the root of this polynomial. When this is the case, ​$x - a$ will be a factor (because ​$x - a = 0$, we’re just rearranging our expression using standard laws).

That last bit might have been a bit fast, so lets step up on it slowly, and we’ll do it from the Bezout angle this time.

Remember this from Bezout?

If we state Bezout in a different way we can see why this is a logical consequence.

“To find whether polynomial ​$P$ is divisible by ​$x - a$ (without remainder), test whether it becomes zero after substitution of ​$a$ for ​$x$.”

That means (and it’s not too much of a brain stretch to see it) that when you put a value into your polynomial ​$P$, and the result does become zero, then it is divisible by ​$x - a$, then that means ​$a$ is a root of ​$P$. (We have rooted our polynomial ​$P$.) That also means ​$x - a$ is a factor of ​$P$.

One last pointPerhaps a self-evident one, but I dislike loose ends. How does this relate to division (polynomial or otherwise)? Remember, two or more factors are the result of dividing a polynomial. To get back to the original polynomial, you simply take their productWhich is just a fancy maths word for multiplying them together. Bezout’s formula is simply one which make evident both sides of the situation - a polynomial in entirety, and the same, but broken into pieces by division and about to be multiplied and added up again.

Still Too Magical? Let’s Check With Some Problems

Does that seem a little too magical? Are you feeling that adding strings to factoring bows should be harder-fought? lets tackle some of the followiung problems to check.

Problem 146(a)

Lets start trying values for ​$x$ to see if we get a zero result. ​$0$ first.

Nope. Not the answer we were looking for. How about ​$1$?

Super! That means ​$1$ is a root of ​$x^4 + 5x - 6$, which in turn means that ​$x - 1$ must be a factor of the same polynomial.

To factor from here we need to use the good-ole polynomial division to see what’s left once we factor out ​$x - a$

Awesome. That means our factoring is as follows:

We could go farther, and factor ​$x^3 + x^2 + x - 6$, but you get the gist.

We’ll do one more before moving on, because it introduces another little clue for us to pick up on

Problem 146(b)

First up, let’s try with ​$0$

Thats no good. How about ​$1$?

That’s no good either. And we’re actually moving away from a zero result too. How about ​$-1$?

That’s our root!

We can then take this into our division to get the quotient for our factoring.

Which means our factoring is as follows:

Tricks Arising (Pt. 2)

Nothing Subtracted? Try a Negative

If when you look at your polynomial you see that nothing is subtractedAs in Problem 146(b) above. you know that your root will have to be a negative number (so that your sum will add to zero when you mix in the ​$x$).

How Many Roots?

The gory details of how to get to this fact are in Gelfand, but its a truism that a polynomial of degree ​$x$ will have at most ​$x$ different roots. The “different” is important - remember a polynomial can have two roots which are the same (n.b. ​$(x - 1)(x - 1) = (x - 1)^2$ which translates as two roots which are the same). Mathematicians say this has “two equal roots”.

Further Reading

There is a lot more to know about all this, but we’ve covered what Gelfand needs us to know for now. If you are excited to learn more, I’d recommend the Fundamental Theorem of Algebra on MathsIsFun. It sounds scary, but it is really well written.