Here’s some good news. There is a whole class of inequalities that we are already experts in. We just have to realize it really. What I mean is polynomial inequalities. Now, if you’re considering an inequality like, say, p of x less than or equals 0, where p is a polynomial, if the polynomial is of degree 1, then you have a linear or affine inequality, easy to solve. And if it’s of degree 2, you have a quadratic inequality. We’ve looked at those in detail. So let’s suppose the polynomial is of degree 3 or more. Now, it could be a different inequality than less than or equal 0. It could be the reverse inequality. It could be the two strict inequalities.
Nothing will change very much relative to the discussion we’re about to undertake. The main fact I wish to present is that if you have complete knowledge of the roots of the polynomial, then you’re very close to being able to write the solution set of your inequality. Let me illustrate that by the following abstract example. We want to find the points where a given polynomial p is strictly positive. Let’s suppose that we know about this polynomial that it has four distinct roots, exactly four roots, and they are the following, the points minus 2, minus 1, 1, and 3. I claim that you can essentially almost right now write down the solution set.
Here’s the thinking, consider the interval between minus 2 and minus 1. There are no roots of the polynomial in that interval, inside the interval because we know that it only has the four roots that we’ve mentioned. That means that in between minus 2 and minus 1 the polynomial must be non-zero. It must be either always above the x-axis or always below. Why? Because the graph of the polynomial is a curve without breaks. It’s something that you can trace, as they say, without lifting your pencil from the paper. So you can’t cross the x-axis. If you can’t cross the x-axis in drawing the curve, that means it must either be always above or always below.
Now, how do you know which one it’s going to be? You can just pick an arbitrary point between minus 2 and minus 1, evaluate the function at that point. And you will see whether it’s positive or negative. Let’s say it’s positive. Therefore, you know that between minus 2 and minus 1 the polynomial is positive. Similarly between minus 1 and plus 1, the polynomial is never 0. So it’s either always positive or always negative, and you can determine which by arbitrarily picking a point in that interval, evaluating the polynomial. And let’s say it’s negative in this example. You continue this. Between 1 and 3, you pick a point, and you find the polynomial happens to be, say, positive.
And then to the right of 3, you pick any point, and it turns out to be positive in this example. And finally to the left of the smallest root, minus 2, you pick a point and evaluate it and it’s negative. Well, now we can write out our solution set. The set of points where p of x is strictly positive will be, you see, the points between minus 2 and minus 1 and then the points between 1 and 3 and also the points to the right of 3. So you get the union of two intervals, open intervals because you’re looking at the strict inequality. Now, here’s the polynomial that was actually behind this example.
It’s a polynomial of degree 5, as you see, with the four roots that we mentioned. You could have solved this inequality by generating the graph of the polynomial and looking at the points where p was positive. But here we’re trying to emphasize logical analysis.
Now, let’s look at this example, which will help us remember how to find roots of polynomials, and the example consists of solving the inequality P (x) greater than 0, where P is a certain cubic polynomial. And a hint is given in the statement of the problem. It is said that the polynomial has an integer root. If it has an integer root, then by the rational root theorem, that integer would have to be a divisor of the constant term 2. So it can only be plus or minus 1 or plus or minus 2, four possibilities. We try all four possibilities; that is, we evaluate P at each of these four numbers to see if we get 0.
And the only one that works turns out to be x equals 2. Given that x equals 2 is the root of the polynomial, we know from the factor theorem that the polynomial is divisible by x minus 2. And when we perform the Euclidean algorithm, we manage to write our cubic polynomial as the product of x minus 2 and then a certain quadratic polynomial. The quadratic polynomial is easily analysed by the quadratic formula, and we find the roots of this quadratic polynomial to be 1 plus or minus root 2. You remember that formula that you will never forget till the day you die. So now we have our three roots, and it is of interest to order them.
It’s easy to see that 1 minus root 2 is strictly less than 2, which, in turn, is strictly less than 1 plus root 2. So these are our three roots in increasing order. Here’s our polynomial written as the product of the linear term and the quadratic term, and we’re now going to construct what is called a table of variations. It’s sometimes also called a table of signs for reasons that you’ll see.
Here are our three roots, and they’re going to determine a subdivision of the reals into four parts, and the value of x will determine whether certain expressions are positive or negative, depending on whether x is to the left of these three roots or in between two of them or to the right. For example, consider the expression x minus 2. We know that x minus 2 will be 0 when x is 2, and it will be negative when x is less than 2, and it will be positive when x is greater than 2. That explains the minus signs and the plus signs and also the 0 that you see in this first real line of our table of variations.
Now, the other expression that we want to study, the other factor of the polynomial P is x squared minus 2x minus 1. We know that this quadratic polynomial is 0 exactly at the two roots. So we put 0 there. It’s strictly negative between the two roots. So we put minus signs. And then to the left or right of the two roots it’s positive, hence the plus signs. And now our polynomial P, which is the product of the preceding two expressions. Using the fact that negative times positive is negative, negative times negative is positive, positive times positive is positive, we then determine the signs in the four regions that we get for P of x. And that’s what we get.
And now once we have our table of variations, we can read off the solutions to the inequality of interest; namely, P of x greater than 0. This will be satisfied when x is between the first two roots because of the plus sign you see there or when x is to the right of the biggest root. So our solution set is the union of two open intervals. The point of this example and what you should see is that at heart it is knowledge of the roots of the polynomial that has led to the answer.