The word “root” has several meanings. Even in mathematics, we’ve used it, for example, in the term square root. Now, here we’re going to use it in a different sense. It’s going to be a number or a point where a function equals 0. We’re going to look particularly at the roots of polynomials. But it’s true for any function that the root of a function f means a value of x for which f of x is 0. We also call such an x a 0 of the function. To put it another way, such an x is a solution of the equation f of x equals 0.
It is a fundamental fact about polynomials that a polynomial of degree n has, at most, n roots. For example, the quadratic polynomial– that means a polynomial of order 2– x squared plus k has either 2, 1, or 0 roots, depending on the value of the parameter k. That’s easy to see. If k is 0, then the function x squared has exactly one point where it’s equal 0– namely, x equals 0, one root. If k is positive– for example, 1– then there are no points for which x squared plus 1 is 0, and therefore, the function has zero roots. And if k is negative, we see that the graph of the function intercepts the horizontal axis in two points.
Each of those is a root, so the function has two roots.
We’re now going to prove a very interesting fact– that if you have a polynomial of odd degree, it must have at least one root. You see that’s false for a polynomial of even degree, as we just saw by example. Now we’re going to give an informal proof of this, and I’m going to focus on the case of degree 3. So I take a polynomial p of degree 3, like this one, and I’m going to prove that it has at least one root. Here’s how the argument goes. I think it will help us understand the nature of functions a little better.
The first observation is that when x is a very large positive number, the term x to the power 3 in the polynomial overwhelms the others. It dominates, and it will give a positive value to p of x. What we’re doing here is we’re talking about the end behaviour of the function as x gets very large, goes to plus infinity. Concretely, what that means is that in the Cartesian plane, on the graph of the function p, you will find a point that is in the first quadrant. If you go sufficiently far to the right, p is positive. You’re in the first quadrant. That’s the reasoning that gave us this conclusion. And now a similar sort of reasoning.
When x is very large in absolute value but negative, the term x cubed will be negative and will dominate the others. And the same sort of reasoning shows that the graph of p will contain a point that’s in the third quadrant down and to the left. It is a fact that the graph of p is a curve that has no breaks in it. It is sometimes said you can trace the graph of the function without lifting your pencil from the paper. That means that when you join the two points that are on the graph to see what it looks like in between, there will necessarily be a point at which you cross the horizontal axis.
Therefore, the fact that the graph crosses the horizontal axis means that there’s a point where the function p is 0– that is, the polynomial p has a root. Now, why wouldn’t this whole argument– which is now complete, by the way– why wouldn’t this argument work for a polynomial of even degree? Reason– in the case of an even degree, we would only be able to assert that there are points on the graph in the first and the second quadrants. And that’s not enough to say that when you join them, there will be some point where the horizontal axis is intercepted. Here are some examples of polynomials and their roots.
The polynomial that you see, x minus 1, x squared plus 1, is of degree 3. Because if you removed a parentheses and worked it out, you’d get an x to the power of 3. Isn’t it clear that it has exactly one root– namely, when x is 1, because then the polynomial is 0? The term x squared plus 1 can never vanish, so it has one root. Similarly, here’s a polynomial which is given to you in what is called factored form as the product of x minus 1, x minus 2, and x minus 7. It’s also of degree 3. And clearly, it has the roots that jump out at you– namely, x equals 1, x equals 2, x equals 7.
In fact, this is a good way to find the roots of a polynomial, is to have it in factored form this way. To be honest, though, polynomials are not given to you, usually, pre-factored, and we will have to discuss how to factor them in order to identify the roots. Here’s an example– the polynomial x cubed plus 6x squared plus 9x. Let’s find its roots by factoring. We first observe that there’s a common factor x in each of the three terms of our polynomial. So I factor it out, and I’m left, in the parentheses, with x squared plus 6x plus 9.
Now, because I’m very sensitive to notable identities, I realise that that term in the parentheses is nothing more than x plus 3 squared. So I have factored our original polynomial in a product of two terms– x and x plus 3 squared. How can this be 0? Answer– it can be 0 if x is 0 or if x is minus 3, and only then. So those are the two roots. Another example– here’s a polynomial of degree 4. Can you find its roots? One way to do it would be to change the order of the terms a little bit. Because if we do that, we can now proceed to factor two out of the last two terms.
We can factor an x out of the first two terms, and now we have revealed a common factor in these two additive terms, x squared minus 1. We factor it out, and that leaves us with x plus 2. And now the roots are obvious. Because how can this product equal 0? Well, either x plus 2 equals 0, which gives minus 2 is a root, or x squared minus 1 equals 0– pardon me, x cubed minus 1 equals 0, which means that x must be 1. We have therefore found all the roots of the polynomial. Now, you could argue that this argument, this clever factoring argument, depended on being pretty ingenious. And that’s true.
We sometimes have to be a little bit ingenious. We will develop our ingenuity somewhat by experience, as you will see. In fact, in the near future, we’ll even allow some guessing to take place– of course, some educated guessing.