Algebra - Polynomial Inequalities - Pauls Online Math Notes

There is a fairly simple process to solving these. If you can remember it you’ll always be able to solve these kinds of inequalities.

Step 1 : Get a zero on one side of the inequality. It doesn’t matter which side has the zero, however, we’re going to be factoring in the next step so keep that in mind as you do this step. Make sure that you’ve got something that’s going to be easy to factor.

\[{x^2} - 3x - 10 < 0\]

Step 2 : If possible, factor the polynomial. Note that it won’t always be possible to factor this, but that won’t change things. This step is really here to simplify the process more than anything. Almost all of the problems that we’re going to look at will be factorable.

\[\left( {x - 5} \right)\left( {x + 2} \right) < 0\]

Step 3 : Determine where the polynomial is zero. Notice that these points won’t make the inequality true (in this case) because \(0 < 0\) is NOT a true inequality. That isn’t a problem. These points are going to allow us to find the actual solution.

In our case the polynomial will be zero at \(x = - 2\) and \(x = 5\).

Now, before moving on to the next step let’s address why we want these points.

We haven’t discussed graphing polynomials yet, however, the graphs of polynomials are nice smooth functions that have no breaks in them. This means that as we are moving across the number line (in any direction) if the value of the polynomial changes sign (say from positive to negative) then it MUST go through zero!

So, that means that these two numbers (\(x = 5\) and \(x = - 2\)) are the ONLY places where the polynomial can change sign. The number line is then divided into three regions. In each region if the inequality is satisfied by one point from that region then it is satisfied for ALL points in that region. If this wasn’t true (i.e it was positive at one point in the region and negative at another) then it must also be zero somewhere in that region, but that can’t happen as we’ve already determined all the places where the polynomial can be zero! Likewise, if the inequality isn’t satisfied for some point in that region then it isn’t satisfied for ANY point in that region.

This leads us into the next step.

Step 4 : Graph the points where the polynomial is zero (i.e. the points from the previous step) on a number line and pick a test point from each of the regions. Plug each of these test points into the polynomial and determine the sign of the polynomial at that point.

This is the step in the process that has all the work, although it isn’t too bad. Here is the number line for this problem.

Now, let’s talk about this a little. When we pick test points make sure that you pick easy numbers to work with. So, don’t choose large numbers or fractions unless you are forced to by the problem.

Also, note that we plugged the test points into the factored form of the polynomial and all we’re really after here is whether or not the polynomial is positive or negative. Therefore, we didn’t actually bother with values of the polynomial just the sign and we can get that from the product shown. The product of two negatives is a positive, etc.

We are now ready for the final step in the process.

Step 5 : Write down the answer. Recall that we discussed earlier that if any point from a region satisfied the inequality then ALL points in that region satisfied the inequality and likewise if any point from a region did not satisfy the inequality then NONE of the points in that region would satisfy the inequality.

This means that all we need to do is look up at the number line above. If the test point from a region satisfies the inequality then that region is part of the solution. If the test point doesn’t satisfy the inequality then that region isn’t part of the solution.

Now, also notice that any value of \(x\) that will satisfy the original inequality will also satisfy the inequality from Step 2 and likewise, if an \(x\) satisfies the inequality from Step 2 then it will satisfy the original inequality.

So, that means that all we need to do is determine the regions in which the polynomial from Step 2 is negative. For this problem that is only the middle region. The inequality and interval notation for the solution to this inequality are,

\[ - 2 < x < 5\hspace{0.5in}\left( { - 2,5} \right)\]

Notice that we do need to exclude the endpoints since we have a strict inequality (< in this case) in the inequality.