Often it is easier to use and remember new terms when you have a 'hook' or comparison to a term you know already.
"Polynomial inequality" is a term generally used to refer to inequalities where the x variable has a degree of 3 or greater.
The prefix "poly" means 'multiple' or 'many', and the root word "nomial" means 'name' or 'term'.
Therefore a "polynomial" is literally: "many terms".
The prefix "in" means 'not', and the root word "equal" of course means 'the same'.
Therefore and "inequality" refers to things which are "not same" or "not equal".
Can you use this logic to identify the origin of some of the other terms used in this lesson?
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 KhanAcademy: Rational Inequalities
Guidance
Polynomial Inequalities
Solving polynomial inequalities is very similar to solving quadratic inequalities. The basic steps are the same:
 Set up the inequality in the form \begin{align*}p(x)>0\end{align*} (or \begin{align*}p(x)<0, p(x)\le0,p(x)\ge0\end{align*})
 Find the solutions to the equation \begin{align*}p(x)=0\end{align*}.
 Divide the number line into intervals based on the solutions to \begin{align*}p(x)=0\end{align*}.
 Use test points to find solution sets to the equation.
Rational Inequalities
There is one step added to the process of solving rational inequalities because a rational function can also change signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function \begin{align*}r(x)=\frac{x}{x^{2}9}\end{align*} below.
If we want to solve the inequality \begin{align*}\frac{x}{x^{2}9}>0\end{align*}, then we need to use the following critical points: \begin{align*}x=0, x=3,\end{align*} and \begin{align*}x=3\end{align*}. \begin{align*}x=0\end{align*} is the solution of setting the numerator equal to 0, and this gives us the only root of the function. \begin{align*}x=\pm3\end{align*} are the vertical asymptotes, the \begin{align*}x\end{align*}coordinates that make the function undefined because putting in 3 or 3 for \begin{align*}x\end{align*} will cause a division by zero.
Testing the intervals between each critical point to see if the values in that interval satisfy the function gives us:
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}(\infty,3)\end{align*}  4    no 
(3, 0)  2  +  yes 
(0, 3)  2    no 
\begin{align*}(3,+\infty)\end{align*}  4  +  yes 
Thus, the solutions to \begin{align*}\frac{x}{x^{2}9}>0\end{align*} are \begin{align*}x\in(3,0)\cup(3,+\infty)\end{align*}.
Example A
Solve \begin{align*}x^{3}3x^{2}+2x\ge0\end{align*}
Solution
The polynomial is already in the correct form \begin{align*}p(x)\ge0\end{align*} so we solve the equation
\begin{align*}x^{3}3x^{2}+2x & = 0\\ x(x^23x+2) & = 0\\ x(x2)(x1) & = 0\end{align*}
The zeros are at \begin{align*}x=0, x=1,\end{align*} and \begin{align*}x=2\end{align*}.
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}(\infty, 0)\end{align*}  5    no 
(0, 1)  \begin{align*}\frac{1}{2}\end{align*}  +  yes 
(1, 2)  \begin{align*}\frac{3}{2}\end{align*}    no 
\begin{align*}(2,+\infty)\end{align*}  3  +  yes 
Notice that this inequality is greater than or equal to zero, so we include the zeros in the solution set. Therefore the solutions are \begin{align*}x\in[0,1]\cup[2,+\infty]\end{align*}.
Example B
Solve \begin{align*}6x^{4}+5x^{2}<25\end{align*}.
Solution
First we will change the inequality to \begin{align*}6x^{4}+5x^{2}25<0\end{align*}. Now, solve the equation \begin{align*}6x^{4}+5x^{2}25=0\end{align*}.
\begin{align*}6x^{4}+5x^{2}25 & = 0\\ (3x^25)(2x^2+5) & = 0\end{align*}
The first term yields the solutions \begin{align*}x=\pm\sqrt{\frac{5}{3}}\end{align*} and there are no real solutions for the second term.
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}\left ( \infty,\sqrt{\frac{5}{3}} \right )\end{align*}  3  +  no 
\begin{align*}\left ( \sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}} \right )\end{align*}  0    yes 
\begin{align*}\left ( \sqrt{\frac{5}{3}},+\infty \right )\end{align*}  3  +  no 
Finally, the solution set is \begin{align*}x\in\left ( \sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}} \right )\end{align*}.
Example C
Find the solution set of the inequality
\begin{align*}\frac{4x12}{3x2}<0\end{align*}
Solution
From the numerator we solve \begin{align*}4x12=0\end{align*} or \begin{align*}x=3\end{align*}. In the denominator, solve \begin{align*}3x2=0\end{align*} and we find the critical point \begin{align*}x=\frac{2}{3}\end{align*}.
Making the table
Interval  Test Point  Positive/Negative?  Part of Solution set? 

\begin{align*}\left ( \infty,\frac{2}{3} \right )\end{align*}  0  +  no 
\begin{align*}\left ( \frac{2}{3},3 \right )\end{align*}  1    yes 
\begin{align*}(3,+\infty)\end{align*}  5  +  no 
Therefore, the solution set includes the numbers in the interval \begin{align*}\left ( \frac{2}{3},3 \right )\end{align*}. Or in setbuilder notation, the solution is \begin{align*}\left \{ x\frac{2}{3}<x<3 \right \}\end{align*}.
How many of the terms we have used recently were you able to track down the origins of? A few are listed below, did you find others? Binomial: "twonamed" or "twoterms"  from "bi", meaning 'two' and "nomial", meaning 'name' or 'term'. Quadratic: "related to a square"  from "quadratus", meaning 'square'. Rational: "related to a ratio"  from "ratio", meaning 'reason' (as in "to reason" or "calculate") and "al", meaning "related to". 

Vocabulary
A quadratic inequality is a square function that is specified to be greater or lesser than a given value.
Polynomial Inequality: A term generally used to describe an inequality with an x term coefficient of 3 or greater.
Rational Inequality: A function which may be expressed as a ratio of two polynomials, specified to be greater or less than a given value.
Guided Practice
Questions
1) Solve the inequality \begin{align*}\frac{3x+2}{x^{2}}>3\end{align*}
2) (using technology) The McNeil Surf Company makes wetsuits. For a given number of wetsuits \begin{align*}x\end{align*}, McNeil's profit, in dollars, is given by the function \begin{align*}P(x)=0.01x^{2}+25x3000\end{align*}.
 a) If the manager of McNeil wants the profit to stay above $9,000, what is the minimum and maximum number of wetsuits they can manufacture to maintain that level of profit?
 b) What is the maximum profit McNeil can make?
 c) Can you explain why this shape might make sense for a profit function?
For the rational functions below, Q's 3  6, determine limitations on the domain and the asymptotes, and then sketch the graph.
3) \begin{align*}f(x) \geq \frac{2x+5}{x1}\end{align*}
4) \begin{align*}f(x) \leq \frac{x+2}{x^{2}+1}\end{align*}
Solutions
1) First, we use algebra to rewrite the inequality to get zero on one side
 \begin{align*}\frac{3x+2}{x^{2}} & > 3\\ \frac{3x+2}{x^{2}}  3 & > 0\\ \frac{3x^2+3x+2}{x^{2}} & > 0\\ \frac{3x^23x2}{x^{2}} & < 0\end{align*}
 Notice in the last step we multiplied both sides by 1, so we changed the direction of the inequality. The numerator cannot be factored, so we use the quadratic formula to solve \begin{align*}3x^{2}3x2=0\end{align*}.
 \begin{align*}x & = \frac{ 3 \pm \sqrt{94(3)(2)}}{6}\\ x & = \frac{3 \pm \sqrt{33}}{6}\end{align*}
 So the two zeros of the rational function are \begin{align*}x=\frac{3+\sqrt{33}}{6}\approx 1.457\end{align*} and \begin{align*}x=\frac{3\sqrt{33}}{6}\approx0.457\end{align*}. The final critical point is \begin{align*}x=0\end{align*} because of the \begin{align*}x^{2}\end{align*} term in the denominator of the inequality. Using these values we construct the table:

Interval Test Point Positive/Negative? Part of Solution set? \begin{align*}\left ( \infty,\frac{3\sqrt{33}}{6} \right )\end{align*} 1 + no \begin{align*}\left ( \frac{3\sqrt{33}}{6},0 \right )\end{align*} \begin{align*}\frac{1}{4}\end{align*}  yes \begin{align*}\left ( 0,\frac{3+\sqrt{33}}{6} \right )\end{align*} 1  yes \begin{align*}\left ( \frac{3+\sqrt{33}}{6},+\infty \right )\end{align*} 2 + no  The final solution set is \begin{align*}x\in\left ( \frac{3\sqrt{33}}{6},0 \right )\cup\left ( 0,\frac{3+\sqrt{33}}{6} \right )\end{align*}
 Notice in this example that the intervals of the solution are on both sides of \begin{align*}x=0\end{align*}. You may be tempted to include \begin{align*}x=0\end{align*} in the solution set, but that will not work. The reason is because at \begin{align*}x=0\end{align*} the rational function is undefined, so x = 0 cannot satisfy the inequality.
2) a) Set up the inequality
 \begin{align*}0.01x^{2}+25x3000 & > 9000\\ 0.01x^2 + 25x 12000 & > 0\end{align*}
 With a calculator you can graph the function \begin{align*}Y_{1}=0.01x^{2}+25x12000\end{align*}.
 On a TI83: use the window \begin{align*}[1000,4000] \times [5000,15000]\end{align*}.
 The settings are:
 \begin{align*}Xmin = 1000, Xmax = 4000, Xscl = 500\end{align*}
 \begin{align*}Ymin=5000, Ymax=5000, Yscl=1000 \ xres=1\end{align*}
 On a software grapher, the image looks like this with the window \begin{align*}x : 0 \to 2400\end{align*} and \begin{align*}y : 0 \to 3500\end{align*}
 Using the CALC menu (2ND TRACE), and selecting the option ZEROS, we can see that the zeros of \begin{align*}Y_{1}=0.01x^{2}+25x12000\end{align*} are at \begin{align*}x=647.920\end{align*} and \begin{align*}x=1852.080\end{align*}.
 By inspecting the graph, we can see that the solution set to the inequality \begin{align*}0.01x^{2}+25x12000>0\end{align*} is \begin{align*}x\in(648,1852)\end{align*}.
 Visually that looks like:
 b) Keeping the same graph open, use CALC MAXIMUM to solve for the maximum profit. The maximum is at (1250, 3625), indicating that the maximum profit is $3625 above the minimum we set: of $9000.
 So the actual maximum profit is $12,625 when 1250 wetsuits are produced.
 c) One possible reason the profit function might take this shape is labor costs. If McNeil wants to make a very large number of wetsuits in a short period of time, then that may require paying overtime for workers, and this could reduce the profit margin.
3) To identify the graph of the inequality \begin{align*}f(x) \geq \frac{2x+5}{x1}\end{align*}, first treat it as if it were the equality \begin{align*}f(x) \geq \frac{2x+5}{x1}\end{align*}
 For \begin{align*}f(x) \geq \frac{2x+5}{x1} :\end{align*}
 To find the critical points, identify the value(s) which make the denominator = 0: \begin{align*}x\ne 1\end{align*}
 That gives us a vertical asymptote of \begin{align*}x=1\end{align*}
 The horizontal asymptote becomes apparent as x becomes truly huge and the "+5" and "1" no longer matter. At that point, we have \begin{align*}f(x) = \frac{2x}{x} \to f(x) = 2\end{align*} So the horizontal asymptote is \begin{align*}y=2\end{align*}
 Now that you know the shape of the graph, simply shade the area above the lines, since the original function was f(x) is greaterthan function, and leave the lines solid since it was a greaterthan or equal to.
 The final graph should look like:
4) To graph \begin{align*}f(x) \leq \frac{x+2}{x^{2}+1}\end{align*} first treat it as if it were \begin{align*}f(x) = \frac{x+2}{x^{2}+1}\end{align*}
 To identify domain limitations, find value(s) which make the denominator = 0: In this case, where \begin{align*}x^2\end{align*}, the only variable in the denominator, is added to 0, any value for x will be positive. So the domain is all real numbers.
 With no limitations on the domain, there are no vertical asymptotes.
 The horizontal asymptote: \begin{align*}y=0\end{align*} becomes apparent as x becomes huge and the "+2" and "+1" no longer have an effect, giving: \begin{align*}f(x) = \frac{x}{x^2} \to f(x) = \frac{1}{x} \to f(x) = 0\end{align*} So the horizontal asymptote is 0
 Now that you know the shape of the graph, simply shade the area below the lines, since the original function was f(x) is lessthan function, and leave the lines solid since it was or equal to.
 The final graph should look like:
Practice
Find the solution set of the following inequalities without using a calculator. Display the solution set on the number line.
 \begin{align*}x^{2}+2x3\le0\end{align*}
 \begin{align*}3x^{2}7x+2>0\end{align*}
 \begin{align*}6x^{2}13x+5\ge0\end{align*}
 \begin{align*}\frac{5x1}{x2}>0\end{align*}
 \begin{align*}\frac{1x}{x}<1\end{align*}
 Solve: \begin{align*}4x^3  4x^2  3x > 0\end{align*}
 Solve: \begin{align*}\frac{x^4}{4}  x^2 <0\end{align*}
 Solve: \begin{align*}4x^3  8x^2  x + 2 \geq 0 \end{align*}
 \begin{align*}\frac{n^3  2n^2  n + 2}{n^3 + 3n^2 + 4n + 12} < 0 \end{align*}
 \begin{align*}\frac{n^3 + 3n^2  4n  12}{n^3  5n^2 + 4n  20} \leq 0 \end{align*}
 \begin{align*}\frac{2n^3 + 5n^2  18n  45}{3n^3  n^2 +27n  9} \geq 0 \end{align*}
 \begin{align*}\frac{12n^3 + 16n^2  3n  4}{8n^3 +12n^2 + 10n +15} >0 \end{align*}
Use a calculator to solve the following inequalities. Round your answer three places after the decimal.
 \begin{align*}9.8t^{2}+357.6t \geq 0\end{align*}
 \begin{align*}x^{3}5x+7 \leq 4x^{2}+18\end{align*}
 \begin{align*}\frac{x^{2}2x}{x5} > x^{2}25\end{align*}
 Solve and graph: \begin{align*}f(x) > \frac{9x^{2}4}{3x+2}\end{align*}
 The total resistance of two electronics components wired in parallel is given by \begin{align*}\frac{R_{1}R_{2}}{R_{1}+R_{2}}\end{align*} where \begin{align*}R_{1}\end{align*} and \begin{align*}R_{2}\end{align*} are the individual resistances (in Ohms) of the two components. a) If the resistance of \begin{align*}R_{1}\end{align*} is 20 Ohms, what is the maximum resistance of \begin{align*}R_{2}\end{align*} if the total resistance must be less than 15 Ohms? b) What is the maximum theoretical resistance of this circuit? How do you know?
 A rectangular lot of land has a length that is 7 meters more than twice its width. If the area of the lot is greater than 60 square meters, what are the possible values of the widths of the lot?