In this article, we discuss inequality of the form $p(x)\ge 0$ or $p(x)>0$ , where $p(x)$ is a polynomial with real coefficients such that $p(x)$ can be expressed as product of linear factors: $$p(x)=(x-a_1)(x-a_2)\cdots (x-a_n)$$ where $a_i$ are real numbers (in the language of field theory, this means that $p(x)$ splits in the field of real numbers).

When we plot the polynomial $y=p(x)$ , whenever it crosses the $x$ -axis, the crossing point is a root of $p(x)$ . On either side of the crossing point, the values of $p(x)$ may be negative or positive. The way to solve the inequality $p(x)>0$ or $p(x)\ge 0$ is to look at how $p(x)$ crosses the $x$ -axis, and to realize that when $x$ is very large, $p(x)$ will be positive. This idea can be illustrated in the following figure:

With this idea in mind, the steps are devised when solving inequalities of this type:

1. arrange $a_i$ so that they are in the ascending order: $a_1\ge a_2\ge \cdots \ge a_n$
2. plot $a_i$ on the number line (the real axis), so each $a_i$ is now a point on the line
3. label above the interval $i_1:=(a_1,\infty)$ to the right of $a_1$ positive
4. go to point $a_2$
5. if $a_2\ne a_1$ , label above the interval $i_2:=(a_2,a_1)$ adjacent to $i_1$ negative
6. if $a_2=a_1$ , label above $a_2$ negative
7. go to $a_3$ , and iterate the labeling processes 4-6
8. stop when $(-\infty,a_n)$ is labeled.
9. if we are trying to solve $p(x)\ge 0$ , then all intervals that are labeled positive, including the end points, are solutions to the inequality
10. if we are solving $p(x)>0$ , then all intervals labeled positive, excluding the end points, are solutions to the inequality.

Remark. After all the labeling is done, there should a total of $n+1$ labels, one over each interval, including the null intervals (the points).

To see how this works, let us look at some actual examples.

• Solve $(x-2)x(x+3)\ge 0$ .
  1. Plot $2$ , $0$ , $-3$ on the number line.
  2. The intervals separated by these points are $(2,\infty),(0,2),(-3,0),(-\infty,-3)$ .
  3. Since no two points are the same, the intervals that are labeled positive are $(2,\infty)$ and $(-3,0)$ .
  4. The solutions to the inequality are $[2,\infty) \cup [-3,0]$ , the square brackets signify that the end points are included in the solutions.
  
  
  
• Solve $(x-8)^7>0$ .
  1. Plot $8$ on the number line.
  2. The intervals separated by these points are $(8,\infty),(-\infty,8)$ , since $8$ is repeated $7$ times.
  3. Start with labeling $(8,\infty)$ positive (the first label)
  4. The point $8$ is then labeled alternately $-(2), +(3), -(4), +(5), -(6), +(7)$ .
  5. The last label goes to $(-\infty,8)$ , which is negative.
  6. Therefore, the solution set is $(8,\infty)$ (excluding $8$ ).
  
  
  
• Solve $(x-1)(x+1)^4\ge 0$ .
  1. Plot $1,-1$ on the number line.
  2. The intervals separated by these points are $(1,\infty),(-1,1)$ , and $(-\infty,-1)$ , since $-1$ is repeated $4$ times.
  3. Start with labeling $(1,\infty)$ positive (the first label), followed by $(-1,1)$ as negative.
  4. The point $-1$ is then labeled alternately $+(3), -(4), +(5)$ .
  5. The last label goes to $(-\infty,-1)$ , which is negative.
  6. Therefore, the solution set is $[1,\infty) \cup \lbrace -1\rbrace$ .
  
  
  

Remarks.

1. In the last two examples, we observe that a simplification can be made when solving the inequality: whenever we have repeating roots ($a_i=a_{i+1}$ ). Depending on the parity of the number of repeating roots, we have two situations:
   • If the number $n_i$ of repeating root, say $a_i$ , is odd, then solving inequality involving $(x-a_i)^{n_i}$ is the same as solving the same inequality with $(x-a_i)^{n_i}$ replaced by $(x-a_i)$ . In other words, their solution sets are the same. For example,
     solving $(x-8)^7>0$ is the same as solving $(x-8)>0$ .
   • If the number $n_i$ of repeating root $a_i$ is even, then we look at whether the inequality is strict or not.
     • If the inequality is strict, then we can completely eliminate $(x-a_i)^{n_i}$ from the inequality without altering the solution set. For example,
       solving $(x-1)(x+1)^4>0$ is the same as solving $(x-1)>0$ .
     • Otherwise, we need to remember the roots themselves as solutions. Therefore, the solution set for $(x-1)(x+1)^4\ge 0$ is the same as the solution set of $(x-1)\ge 0$ together with the root $-4$ .
2. Using the rules above, we may also solve inequalities $p(x)<0$ or $p(x)\le 0$ . The solution set for $p(x)<0$ is the complement of the solution set for $p(x)\ge 0$ , and the solution set for $p(x)\le 0$ is the complement of the solution set for $p(x)>0$ .