Let's suppose you want to solve the inequality

*x*

^{2}-4

*x*+3<0.

**Step 1**. Solve the equation

*f*(

*x*)=

*x*

^{2}-4

*x*+3=0.

In this case, we can "factor by guessing":

*x*

^{2}-4

*x*+3=(

*x*-1)(

*x*-3),

so the roots of the equation

*f*(

*x*)=0 are

*x*=1 and

*x*=3. Draw a picture of the

*x*-axis and mark these points.

**Step 2.**Our solutions partition the

*x*-axis into three intervals. Pick a point (your choice!) in each interval. Let me take

*x*=0,

*x*=2 and

*x*=4. Compute

*f*(

*x*) for these points:

These three points are representative for what happens in the intervals they are contained in:

Since

*f*(0)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . Similarly, since

*f*(2)<0,

*f*(

*x*) will be negative for

**all**

*x*in the interval (1,3). Since

*f*(4)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . You can indicate this on the

*x*-axis by inserting plus or minus signs on the

*x*-axis. I use color coding instead: blue for positive, red for negative:

**Step 3.**We want to solve the inequality

*x*

^{2}-4

*x*+3<0,

so we are looking for all

*x*such that

*f*(

*x*)<0. Consequently, the interval (1,3) contains all solutions to the inequality.

**Why does this work?**Let's look at the graph of

*f*(

*x*):

Note the pivotal role played by the "yellow dots", the

*x*-intercepts of

*f*(

*x*).

*f*(

*x*) can only change its sign by passing through an

*x*-intercept,

*i.e.,*a solution of

*f*(

*x*)=0 will always separate parts of the graph of

*f*(

*x*) above the

*x*-axis from parts below the

*x*-axis. Thus it suffices to pick a representative in each of the three intervals separated by "yellow dots", to test whether

*f*(

*x*) is positive or negative in the interval .

This nice property of polynomials is called the

**Intermediate Value Property**of polynomials; your teacher might also refer to this property as

**continuity**.

Here is another example: Find the solutions of the inequality

For our method to work it is essential that the right side of the inequality equals zero! So let's change our inequality to

**Step 1**. Solve the equation

*f*(

*x*)=

*x*

^{2}-3

*x*+2=0.

Again, we can "factor by guessing":

*x*

^{2}-3

*x*+2=(

*x*-1)(

*x*-2),

so the roots of the equation

*f*(

*x*)=0 are

*x*=1 and

*x*=2. Draw a picture of the

*x*-axis and mark these points.

**Step 2.**Our solutions partition the

*x*-axis into three intervals. Pick a point (your choice!) in each interval. Let me take

*x*=0,

*x*=1.5 and

*x*=3. Compute

*f*(

*x*) for these points:

These three points are representative for what happens in the intervals they are contained in:

Since

*f*(0)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . Similarly, since

*f*(1.5)<0,

*f*(

*x*) will be negative for

**all**

*x*in the interval (1,2). Since

*f*(3)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . You can indicate this on the

*x*-axis by inserting plus or minus signs on the

*x*-axis. I use color coding instead: blue for positive, red for negative:

**Step 3.**We want to solve the inequality

so we are looking for all

*x*such that . Consequently, the set contains all solutions to the inequality. (Since our inequality only stipulates that ,

*x*=1 and

*x*=2 are solutions, so we include them. "" and "" are only symbols; they will never be included as solutions.)

Our next example: Solve

*x*

^{3}>2

*x*. Do

**not**divide by

*x*on both sides! If you do so, you will never be able to arrive at the correct answer. Repeat the pattern instead; make one side of the inequality equal zero:

*x*

^{3}-2

*x*>0.

**Step 1**. Solve the equation

*f*(

*x*)=

*x*

^{3}-2

*x*=0.

We can factor rather easily:

so the roots of the equation

*f*(

*x*)=0 are ,

*x*=0 and . Draw a picture of the

*x*-axis and mark these points.

**Step 2.**Our solutions partition the

*x*-axis into four intervals. Pick a point (your choice!) in each interval. Let me take

*x*=-2,

*x*=-1,

*x*=1 and

*x*=2. Compute

*f*(

*x*) for these points:

These four points are representative for what happens in the intervals they are contained in:

Since

*f*(-2)<0,

*f*(

*x*) will be negative for

**all**

*x*in the interval . Similarly, since

*f*(-1)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . Since

*f*(1)<0,

*f*(

*x*) will be negative for

**all**

*x*in the interval. Since

*f*(2)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . You can indicate this on the

*x*-axis by inserting plus or minus signs on the

*x*-axis. I use color coding instead: blue for positive, red for negative:

**Step 3.**We want to solve the inequality

*x*

^{3}-2

*x*> 0,

so we are looking for all

*x*such that

*f*(

*x*)> 0. Consequently, the set contains all solutions to the inequality. (Since our inequality stipulates that

*f*(

*x*)>0, do not qualify as solutions, so we exclude them. "" and "" are only symbols; they will never be included as solutions.)

Here is my last example: Solve

**Step 1**. Solve the equation

*f*(

*x*)=

*x*

^{3}+3

*x*

^{2}+

*x*+3.

We can factor either by finding a rational zero, or by clever grouping:

*x*

^{3}+3

*x*

^{2}+

*x*+3=(

*x*

^{3}+3

*x*

^{2})+(

*x*+3)=

*x*

^{2}(

*x*+3)+1 (

*x*+3)=(

*x*

^{2}+1)(

*x*+3),

so there is only one real root of the equation

*f*(

*x*)=0, namely

*x*=-3. Draw a picture of the

*x*-axis and mark this point.

The polynomial

*x*

^{2}+1 is irreducible, it does not have real roots. Its complex roots are irrelevant for our purposes.

**Step 2.**Our solution partitions the

*x*-axis into two intervals. Pick a point (your choice!) in each interval. Let me take

*x*=-4 and

*x*=0. Compute

*f*(

*x*) for these points:

These two points are representative for what happens in the intervals they are contained in:

Since

*f*(-4)<0,

*f*(

*x*) will be negative for

**all**

*x*in the interval . Similarly, since

*f*(0)>0,

*f*(

*x*) will be positive for

**all**

*x*in the interval . You can indicate this on the

*x*-axis by inserting plus or minus signs on the

*x*-axis. I use color coding instead: blue for positive, red for negative:

**Step 3.**We want to solve the inequality

so we are looking for all

*x*such that . Consequently, the set is the set of solutions to the inequality. (Since our inequality stipulates that ,

*x*=-3 is also a solution, so we include it. "" and "" are only symbols; they will never be included as solutions.)