# Polynomial Inequalities

These are my notes and worked examples on polynomial inequalities.

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We solve inequalities that involve polynomials of degree 3 and higher, as well
as some that involve rational expressions. To solve such inequalities, we use
the information obtained previously about the graph of polynomial and rational
functions.

Suppose that the polynomial or rational inequality is of the form:
$$f(x)<0,f(x)>0,f(x)\leq0,f(x)\geq0$$
Locate the zeros of f if f is a polynomial function, and locate the zeros of the
numerator and the denominator if f is a rational function. If we use these zeros
to divide the real number line into intervals, we know that on each interval the
graph of f is either above the x-axis or below the x-axis. In other words, we
have found the solution of the inequality.

Steps for Solving Polynomial Inequalities
1. Write the inequality so that a polynomial or rational expression is on the
left side and zero is on the right side. For rational expressions, be sure that
the left side is written as a single quotient and find its domain.

2. Determine the real numbers at which the expression on the left side equals
zero and, if the expression is rational, the real numbers at which the
expression on the left side is undefined.

3. Use the numbers found in step 2 to separate the real number line into
intervals.

4. Select a number in each interval and evaluate the function at the number.
A. If the value of f is positive, then f(x)>0 for all numbers in the interval
B. If the value of f is negative then f(x)<0 for all numbers in the interval
If the inequality is not strict, include the solutions of f(x)=0 in the solution
set.

Example 1
Solving a Polynomial Inequality
$$x^4 \leq 4x^2$$

Solution:
Rearrange the inequality so that 0 is on the right side.
$$x^4 \leq 4x^2$$
$$x^4-4x^2 \leq 0$$
This inequality is equivalent to the one that we wish to solve.
Find the real zeros of $$f(x)=x^4-4x^2$$ by solving its equation.
$$x^4-4x^2=0$$
$$x^2(x^2-4)=0$$
$$x^2(x+2)(x-2)=0$$
x=0 or x=-2 or x=2
We use these zeros to separate the real number line into four intervals.
select a number in each interval and evaluate $$f(x)=x^4-4x^2$$ to determine if
f(x) is positive or negative.
We know that f(x)<0 for x in the intervals (-2,0) and (2,0), for all x such that
-2<x<0 or 0<x<2. however, because the original inequality is not strict, numbers
that satisfy the equation are also solution of the inequality. So, we include 0,
-2, and 2. The solution set of the given inequality is $${x|x-2 \leq x \leq 2}$$.

Example 2
Solving a polynomial inequality
$$x^4 > x$$

Solution:
Rearrange the inequality so that 0 is on the right side.
$$x^4-x > 0$$
This inequality is equivalent to the one that we wish to solve.
Find the real zeros of $$f(x)=x^4-x$$ by solving $$x^4-x=0$$
$$x^4-x=0 \longrightarrow x(x^3-1)=0 \longrightarrow x(x-1)(x^2+x+1)=0$$
x=0 or x=1 The squared term is a quadratic and of course does not have any real
solutions.
We use the zeros 0 and 1 to separate the real number line into three intervals.
We choose a number in each interval and evaluate $$f(x)=x^4-x$$ to determine if
the solution is positive or negative.
We know that f(x)>0 for all x in negative infinity to 0 and from 1 to infinity.
Because the original inequality is strict, the solution set of the given
inequality is $${x|x,0,x>1}$$.

Example 3
Solving a rational inequality
Solve $$\frac{(x+3)(2-x)}{(x-1)^2}>0$$ and graph the solution set.

Solution:
The domain of the variable x is $${x|x\not=1}$$. The inequality is already in a
form with 0 on the right side.
Let $$f(x)=\frac{(x+3)(2-x)}{(x-1)^2}$$. The real zeros of the numerator of f
are -3 and 2. The real zero of the denominator is 1.
We use the zeros -3, 1 and 2 to separate the real number line into four
intervals.
Select a number in each interval and evaluate the number in
$$f(x)=\frac{(x+3)(2-x)}{(x-1)^2}$$ to determine in f(x) is positive or
negative.
We know that f(x)>0 for all x in (-3,1) to (1,2). Because the original
inequality is strict, the solution set of the given inequality is
$${x|-3<x<2,x\not=1}$$. Notice the hole at x=1 to indicate that 1 is to be
excluded.

Example 4
Solving a rational inequality
$$\frac{4x+5}{x+2} \geq 3$$

Solution:
The domain of the variable x is $${x|x\not=-2}$$. Rearrange the inequality so
that 0 is on the right side. Then express the left side as a single quotient.
$$\frac{4x+5}{x+2}-3 \geq 0$$
Multiply 3 by $$\frac{x+2}{x+2}$$
$$\frac{4x+5}{x+2}-3(\frac{x+2}{x+2} \geq 0$$
Write as a single quotient
$$\frac{4x+5-3x+6}{x+2} \geq 0$$
Combine like terms
$$\frac{x-1}{x+2} \geq 0$$
The domain of the variable is $${x|x\not=-2}$$.
Let $$f(x)=\frac{x-1}{x+2}$$. The zero of the numerator is 1. The zero of the
denominator is -2.
We use the zeros -2, and 1 to separate the real number line into three
intervals.
Select a number in each interval and evaluate $$f(x)=\frac{4x+5}{x+2}-3$$ to
determine if f(x) is positive or negative.
We know that f(x)>0 for all x from negative $$(-\infty,-2)$$ and $$1,\infty)$$.
Because the original inequality is not strict, numbers that satisfy the equation
$$f(x)=\frac{x-1}{x+2}=0$$ are also solutions of the inequality. Since
$$\frac{x-1}{x+2}=0$$ only if x=1, we conclude that the solution set is
$${x|x<-2,x\geq1}$$.