The Factor Theorem

A theorem means a proposition that is proved by a logical chain of reasoning. We’re going to prove a theorem here, the factor theorem, that is extremely useful for finding the roots of polynomials. We’re also going to see how the structure of certain proofs works. So here is the statement of the theorem. You’re given a polynomial, P, and you’re given a real number, a. And the theorem says that a is a root of P– that is, P of a is 0– if and only if the polynomial P of x is divisible by the polynomial x minus a. A restatement of the theorem would be the following.

In order for a to be a root of P, it is both necessary and sufficient that P be divisible by x minus a. Now, the proof will be in two parts. Because there are two implications to prove. We must prove the left to right and the right to left, as we sometimes say. The first implication to prove is that if a is a root of P, then necessarily, P is divisible by x minus a. That’s called the necessity part of the proof. The second implication, the converse, is in the reverse order. We want to prove that in order for a to be a root of P, it suffices that P be divisible by x minus a.

That’s called the sufficiency part of the proof. To do the proof, let’s note this initial step. When we apply the Euclidean division to divide P by x minus a, we can write, naturally, P in the form Q times x minus a plus the remainder. Now, the degree of R, the remainder, is strictly less than the degree of the polynomial x minus a, which has degree 1. Therefore, it follows that the degree of R is 0. That is, R is a constant. So we have this conclusion, that P can be written as Q times x minus a plus a constant. I’ve called that conclusion star, just to have it on hand during the proof. And now, the end of the proof.

First implication was to show that if a is a root, then P is divisible by x minus a. Well, you look at star. And you see that if a is a root, then putting x equal to a in star immediately tells you that R is 0. When R is 0, that means P is divisible by x minus a, by definition. Second implication. If P is divisible by x minus a, then a must be a root. Proof? If P is divisible, then in star, R is 0. Well, then, putting x equal a clearly gives you P of a equals 0, which shows that a is a root of P. We’ve finished a proof both ways.

That allows us to position our favourite symbol, which as you know, stands for QED. How can a root be a multiple root of a polynomial? Multiplicity is the word that applies here. Let’s see what is meant by that term. Suppose that a is a root of a polynomial, P. As you know, by the factor theorem, that means that P can be written in the form Q times x minus a. That is, we can pull the factor of x minus a out of P. Now, it might be that the polynomial Q that we’ve produced this way also admits the value a as a root.

In that case, we could write Q as M times x minus a for some other polynomial, M. And going back to P, that would mean that P is equal to M times two powers of x minus a. And we could go on this way, possibly, until at some point, we’d have to stop. There would be a highest integer, K, having the property that P of x produces K factors of x minus a, and the other part, N, no longer has a root at a. We can’t go on. In that case, we say that the root a is of multiplicity K. Here’s an example to make this clear. Here’s a polynomial.

I ask you to check that it’s a polynomial of degree 9. This polynomial has been factored as far as you can do so. Because all the terms you see in it are linear or powers of linear. And at the end, the quadratic x squared plus 1 is clearly irreducible. So you can’t go on any further. What are the roots of this polynomial? Well, they jump to the eye. They’re 1, 3, and minus 8, evidently. What are the multiplicities of those roots? 1 is a root of multiplicity 1. 3 of multiplicity 2. And the last root, minus 8, is of multiplicity 4.

Now, we’re going to develop a strategy exactly with that in mind, of producing this kind of factorization of a polynomial to make the roots obvious. That will be in the next segment.