We’re going to develop a general strategy for finding the roots of a polynomial. It will be based on the factor theorem. And it will involve finding the root factorization of the polynomial. Now in order to reveal, fully, the structure of a polynomial, we would really like to factor it into the following form– a product of linear factors of the form x minus r and then a final term Q of x where Q is irreducible, has no roots. A general strategy for doing this is the following. First, we find a root r of the polynomial P by some means. We then use polynomial division and the factor theorem to divide P by x minus r.
And we know that it will be divisible, so we’ll have P equal x minus r times another polynomial P1. P1 will be of lower degree than p. We now repeat the process. We look at P1. We find a root of P1. We take out a linear factor. And we go on until we have the root factorization. So you see that we’re using so-called order reduction. At each step, the new polynomial will have degree less than the previous one. Let’s illustrate this with the following example. We want to find the root factorization of a certain polynomial of degree 4. First step, we observe that P(1) equals 0, as you can easily check. It follows that 1 is a root of P.
Therefore, by the factor theorem, P will be divisible by x minus 1. Now we can perform this long division. I will spare you the gruesome details. But here’s the answer and a certain polynomial of degree 3 is the quotient. We now proceed to study that polynomial Q of degree 3. We observe that Q vanishes at minus 3. That is Q of minus 3 is 0. That is minus 3 is a root. That means that x plus 3 will divide Q by the factor theorem. When we do that long division, then a certain quadratic x squared plus 2 appears. That quadratic is clearly irreducible. It can never be 0. It has no roots. So the procedure has ended.
And now we get the root factorization of P. Now when you look at this example and the way we’ve solved it, surely that comes to mind to ask but how did we find the root 1 for the polynomial P? How did we find the root minus 3 for the polynomial Q? To be perfectly honest, we guessed. However, we guessed in an educated way. I will explain what I mean. Consider the following example. You have a polynomial of degree 3 and you would like to guess its roots.
Well you decide that you are feeling lucky and you try x equals 1 minus 1, 2 minus 2, 3 minus 3– that is you plug them into P and see if you get 0, actually none of these give you 0. None of these values are roots. Well, you didn’t guess in an educated way because it turns out we could have known a priori that the only possible rational roots of this polynomial are plus and minus 1 and plus and minus 1/2. Why is that? It’s because of something called the rational root theorem which I now state. Here it is. Suppose we have a polynomial with integer coefficients whose constant term is non-zero. It might have rational roots.
Any rational root can be expressed in the form p over q, where p and q are integers and the fraction is irreducible. In that case, the theorem says, the p must divide the constant term and the q must divide the leading coefficient. If you apply that statement to the example above, you will see that the q must divide 2 and the p must divide 1. That leads to 4 possibilities in all– plus and minus 1, plus and minus 1/2. When you try these four possibilities, it turns out that exactly one of them is a root, namely x equals a 1/2. Now note an important fact about this theorem.
This theorem does not say there will be a rational root of the polynomial. It says if there is a root, then the p over q will be such that and so on. So it’s not a guarantee that there will be a rational root. It’s simply a good way to produce educated guesses. Here’s an example. We want to find– we have found, actually already, that x equals 1/2 is a root of this polynomial, how do we go on to go towards the root factorization? Well we divide by x minus 1/2. After the long division, we get a certain quadratic. Can we go on? Can we factor that quadratic? No, it’s irreducible. Why is it irreducible? Because its discriminant. It’s strictly negative.
Therefore we have achieved the root factorization of p and p has exactly one root. Here is another example of a slightly different type. As usual, we want to begin by finding the first root that will give us our entry point to the procedure. What should be our guesses for that first root? By the rational root theorem, we know that any rational root p over q will be such that q has to be plus or minus 1 because the leading coefficient of our polynomial is 1. And the p has to be a divisor of 8. That produces, as you can see, 8 possibilities in all– plus and minus 1, 2, 4, 8. We try them all. Eight times we calculate P.
And we discover that exactly one of these numbers is a root, namely 4. Given that 4 is a root, we divide P by x minus 4. We know it will divide evenly. And a certain quadratic polynomial then results. Is that quadratic polynomial susceptible to being factored further? Well, its discriminant is strictly positive. So, yes, it has two roots. And therefore, that quadratic polynomial will be the product of two linear terms, namely the two that you see. The root factorization of P is now complete. P has been factored down as far as it can be. A cubic polynomial can have no more than three roots.