Example 1 Solve
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.
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
Step 3 : Determine
where the polynomial is zero. Notice
that these points won’t make the inequality true (in this case) because 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 and .
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 (
and ) 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 that 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
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
Also, note that we plugged the test points into the
factored from 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,
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
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,
Notice that we do need to exclude the endpoints since we
have a strict inequality (< in this case) in the inequality.