Polynomial and Rational Inequalities

Solving polynomial and rational inequalities is a bit harder than solving similar equations. Even in a simple problem where you can factor like

x²-5x+6 > 0

it is not clear what to do after you factor, because the usual principle of zero factors doesn't hold for inequalities. So once you get

(x-2)(x-3) > 0

it is not clear what to do. With the equation

(x-2)(x-3)=0

you can use the principle of zero factors because for a product of two numbers to be zero one or the other has to be zero, so you can set each factor to zero. There is a way to do something similar with the inequality, but it is a bit more complicated.

You see it is not true that if a product of two numbers is greater than or equal to zero that one or the other of them must be greater than or equal to zero, but there is something else that is true, and to think of it you have to think of greater than or equal to zero as positive and think about the way positive and negative numbers multiply. When is the product of two numbers positive? Well, there are two ways for this to happen. They must be either both positive or both negative. You could then work this problem out by using that and take both cases.

Case 1: x-2 > 0, x-3 > 0

This is the case when both factors are positive. It will give us a positive product, because the product of two positive numbers is positive. By adding 2 to both sides for the first inequality and 3 to both sides for the second inequality, this gives us

x > 2 and x > 3.

This is a funny kind of condition, though, because one of the conditions isn't necessary, because it is weaker than the other one. Certainly if x is greater than or equal to 3, then it must be greater than or equal to 2, since 3 is greater than 2, so we can get rid of the first condition and just use the second one, and this reduces to

x > 3.

Case 2: x-2 < 0, x-3 < 0

This is the case when both factors are negative. It will give us a positive product, because the product of two negative numbers is positive. By adding 2 to both sides for the first inequality and 3 to both sides for the second inequality, this gives us

x < 2 and x < 3.

Again one of these isn't necessary, because it is weaker than the other. This time, since 2 is less than 3, certainly if x is less than or equal to 2, then it is less than or equal to 3, so only the first condition is necessary, so this reduces to

x < 2.

Putting the two cases together we get that there are two possible kinds of numbers that will satisfy the inequality, numbers that are less than or equal to 2 and numbers that are greater than or equal to 3. We can write this in interval notation by using a union.

The square brackets indicate that the 2 and the 3 are included. This is true because for x=2 and 3, the left side of the inequality is zero and zero is certainly greater than or equal to zero. The plus and minus infinity have round brackets, because infinity is not a number, so it is not included. Here is a graph on the number line of the solutions to the inequality. Notice how the interval notation is sort of miniature picture of the graph. The intervals represent pieces cut out of the number line.

You could also use this method to solve other kinds of inequalities as well. If it was a less than inequality, you would use the fact that in order to multiply two numbers to get a negative number, one of them has to be positive, and the other has to be negative, and again you could divide into two cases, one where the first factor was positive and the second factor was negative, and the other where the first factor was negative and the second factor was positive. If the inequality was strict, that is < or >, then the only difference would be that you wouldn't include the endpoints, so you would use round brackets in interval notation.

But there is a better way to do this whole thing that is quicker and allows you to deal with more complicated inequalities much easier. Suppose you had a more complicated inequality like

(x-1)(x-2)(x-3)(x-4)(x-5) < 0

To break it down into cases like I did with the equation above would get very time consuming, because there are too many possibilities. In this case, because the product is negative, there must be an odd number of minus signs. But it is really not that complicated, because a lot of them are impossible, like for example, it is not possible for x-1 to be negative and x-5 to be positive, because x-1 will always be bigger than x-5. But here is something interesting that will happen with the signs of the expression on the left side of the inequality as you travel up and down the real line. At any place where one of the factors is zero, there will be a sign change, because there is a sign change of one and only one factor. For example for x=5.1 all of the factors are positive, but for x=4.9, the first four factors are positive, but x-5 is negative. This means that if we can find out the sign of the product for just one number, then we can figure out the sign everywhere else and solve our inequality, because after that it will just flip flop. The easiest way to keep track of this is to plot the places where the product is zero on the number line, and write in the signs. Here are the places where the product is zero plotted in order on the number line.

Think of these numbers as cutting the real line into intervals, with alternating ones making up the solution set of the inequality. The only problem is which ones do we take. One way we could determine this is to simply pick one number on one of the intervals and test out the sign of that interval with it. So, for example, what happens when x=2.5? Then the product is (2.5-1)(2.5-2)(2.5-3)(2.5-4)(2.5-5)=(1.5)(.5)(-.5)(-1.5)(-2.5), which is negative, since there are 3 negatives. Then we could put a minus sign in the interval between 2 and 3 and then alternate to fill in the rest. But we could also do it even easier by just reasoning about it, and the easiest interval for that is the right most one, the one that has the big numbers. For any number bigger than 5 we know that all of the factors must be positive, because x-5 is, and it is the smallest. So we know that the right most interval must be plus and can alternate from there. We can even be a little bit lazier about this and realize that as long as all of the x coefficients are positive, this will always happen, and if we had a problem where one of the x coefficients was negative, we could always multiply by -1 on both sides to get it to be positive, remembering to reverse the direction of the inequality of course. Anyway, however we decide it, we get for this problem the following map of plus and minus signs for the product. The meaning of this picture is that the left side of the inequality is positive, that is greater than zero, wherever the pluses are and negative, that is less than zero, where the minuses are. So since we are trying to find the numbers where it is less than zero, we get for the solution of the inequality. An important thing to remember throughout all of this is that positive numbers are greater than zero and negative numbers are less than zero.

This method can also be applied to rational inequalities as well using all the places where either the numerator or the denominator is zero for the sign changing places, because for quotients as well as products the total number of minus signs determines the sign of the answer. The only difference is that you never include any places where the denominator is zero in your answer even if the inequality is greater than or equal or less than or equal, because a zero in the denominator doesn't make the quotient zero, it makes it undefined.

Example:

x+2 x-3

> 0.

Here the zeros of the numerator and the denominator are -2 and 3. Here they are on the number line.

If we alternate the pluses and minus we get this. This time since the inequality says greater than or equal to zero we use the intervals with the pluses, but unlike with products we don't include the 3, because it make the denominator 0, but we do include the -2, because it is perfectly all right for numerators to be zero and when the numerator is zero the inequality will be satisfied, because the quotient will be zero and zero is greater than or equal to zero. So the final answer is If this all sounds too simple, though, you are right, because it isn't. There is an exception to this alternating signs game, but it isn't really that bad to deal with. The problem occurs when a factor is raised to an even power, like for example

(x-1)²(x-3) > 0.

The problem is that here there won't be a change of signs at 1, because (x-1)²  is always greater than or equal to zero. This is easy to deal with, though, you just throw (x-1)² out of the game, and reduce the inequality to

x-3 > 0.

Except not quite, because you still have to take into account the fact that the expression is going to be equal to zero at x=1 and 0 > 0. So you throw it out and just use that consideration to tweak your answer a bit at the end. So here at first we get that the answer should be

but 1 needs to be included as a solution as well, so we union it as a singleton and get You also need to sometimes watch out with strict inequalities. In that case the problem is that there is a number that you don't want to include because it makes the product 0, because 0 is neither strictly less than 0 nor strictly equal to 0.

More Examples

For each of these the instruction is to solve for x and write the solution set in interval notation.

Example 1:

(x-2)(x+5)>0

Solution:

Lining up the places where the left side is equal to 0 in numerical order on the number line and alternating the pluses and minus we get as a plot of the signs of the left side of the inequality. Since the inequality says greater than zero, we want the pluses, so in interval notation that is

Example 2:

x²+x < 12

Solution:

For this one first we need to get a zero on one side of the inequality, so we subtract 12 from both sides to get

x²+x-12 < 0.

Then factoring this we get

(x+4)(x-3) < 0.

Plotting the zeros on the number line and alternating pluses and minus we get

Since then inequality is less than or equal to zero, we want the minuses and we also want the endpoints, because they are where the left side is zero. So in interval notation the solution set is

[-4,3].

Example 3:

(x-1)³(x+2)⁴(x-3) > 0

Solution:

Here we can partly ignore the second factor, because it has an even exponent. So I am just plotting the 1 and the 3 on the number line. Then alternating pluses and minus we get This would give for the answer except that this would include -2, which we don't want, because the left side is zero there, so not greater than zero, so we need to remove just the single number -2 from the solution. The way to write that in interval notation is to break up the first interval into two pieces using round brackets, so that the -2 will not be included. Doing that the solution can be written like this.

Example 4:

(2x-3)(4x+3) > 0

Solution:

The only thing that is a little difficult about this one is that it involves fractions. To find the zeros we have to set 2x-3 and 4x+3 equal to zero.

2x-3=0
2x=3
x=3/2

4x+3=0
4x=-3
x=-3/4

Now line these two up on the number line and alternate pluses and minuses.

Then again since the numbers that are greater than zero are the positive ones, the solution set is the union of the two plus intervals

Example 5:

(-x+3)(2x-5) < 0

Solution:

To make it easier to tell where the pluses and minuses are, multiply both sides by -1 so that the coefficients on x will all be positive. But when we do that we have to make sure to use the rule that when you multiply both sides of an inequality by a negative number the direction of the inequality flips, so then the inequality becomes

(x-3)(2x-5) > 0

Then set each factor to zero to get 3 and 5/2 for the zeros and switching places. Then plotting these on the number line and alternating the pluses and minuses we get

Since greater than or equal to zero is plus, the solution is

Example 6:

(x-1)(-x+2) -x+3

> 0

Solution:

Since there are two negative coefficients here we don't have to multiply both sides to get rid of them. We can just multiply top and bottom by -1 on the left side to get

(x-1)(x-2) x-3

> 0

Now the zeros in the numerator and denominator are 1,2, and 3. Plotting these on the number line and alternating pluses and minuses we get

and greater than or equal to zero is plus we get the plus intervals. Since the inequality is greater than or equal, we should include the endpoints, but this time with a quotient we don't want to include the 3, because it makes the denominator zero, and that would be undefined, not zero, so the solution set is

Example 7:

(x+1)2 x+2

< 0

For this one since (x+1)² cannot be negative we can temporarily ignore it. -2 is the only place where the sign will change. Plotting it on the number line we get

Since less than zero is minus we get for the solution set, but we have to make sure that -1 is not in it, because at -1 the quotient will be zero, which is not less than zero, but this time we don't have to worry, because -1 is not in our set, because it is greater than -2, so we leave it as is.

Example 8:

x+2 x+3

< 5

Solution:

For this one we have to remember that this whole method depends on having 0 on the right side of the inequality, because we need to reason about signs of numbers which have to do with whether numbers are greater or less than 0, not 5. So we have to do something to get 0 on the right side. To do that we have to subtract 5 from both sides and then turn the left side into a single fraction by finding a common denominator.
 

x+2 x+3

- 5 < 0

x+2 x+3

-

5(x+3) x+3

< 0

x+2-5(x+3) x+3

< 0

x+2-5x-15 x+3

< 0

-4x-13 x+3

< 0

Then we need to multiply both sides by -1 in order to have all positive x coefficients and reverse the sense of the inequality, so this becomes

4x+13 x+3

> 0.

Now the zeros in the numerator and denominator are -13/4 and -3. -13/4 is less than -3 since -12/4=-3, so plotting them on the number line and alternating pluses and minuses it looks like this.

We take the outer intervals because greater than or equal to zero is plus, but we must make sure to not include the -3, because it makes the denominator zero, so our solution set is

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