Date:
Topic: Solving Inequalities in One Variable
(2.9)
Definition of a Polynomial Inequality
A polynomial inequality is any inequality that
can be put in one of the forms
f(x) < 0, f(x) > 0, f(x) ≤ 0, or f(x) ≥ 0
where f is a polynomial function.
Determine the x values that cause the polynomial
function f(x) = (x + 3)(x2 + 1)(x - 4)2 to be zero,
positive, and negative.
real zeros? -3 and 4
Evaluate f at one number
within each interval
create a sign chart:
(-)(+)(-)2
-5 -4 -3
negative
(+)(+)(-)2
-2
-1
0
1
2
positive
(+)(+)(+)2
x
3
4
5
6
positive
positive? 3, 4  4,  
negative? , 3
You can also view this on a graphing calculator.
Determine the x values that cause the polynomial
function f(x) = (x + 3)(x2 + 1)(x - 4)2 to be zero,
positive, and negative.
(-)(+)(-)2
-5 -4 -3
negative
(+)(+)(-)2
-2
-1
0
1
2
positive
(+)(+)(+)2
x
3
4
5
6
positive
Solve the function for the four inequalities:
(x + 3)(x2 + 1)(x - 4)2 > 0
(x + 3)(x2 + 1)(x - 4)2 > 0
(x + 3)(x2 + 1)(x - 4)2 < 0
(x + 3)(x2 + 1)(x - 4)2 < 0
3, 4  U 4,  
3, 
, 3
, 3U 4
Procedure for Solving Polynomial Inequalities
• Express the inequality in the standard form
f(x) < 0 or f(x) > 0
• Find the zeros of f, the boundary points.
• Locate these boundary points or zeros on a
number line (or sign chart), thereby dividing
the number line into intervals.
• Evaluate f at one representative number
within each interval.
• Write the solution set; selecting the interval(s)
that satisfy the given inequality.
Example
Solve the inequality and support it graphically:
2x2 – 3x > 2.
Step 1
Write the inequality in standard form. We can write by
subtracting 2 from both sides to get zero on the right.
2x2 – 3x – 2 > 2 – 2
2x2 – 3x – 2 > 0
Step 2
Solve the related quadratic equation to find the zeros
(boundary points).
2x2 – 3x – 2 = 0
(2x + 1)(x – 2) = 0
2x + 1 = 0 or x – 2 = 0
x = -1/2 or x = 2
This is the related quadratic equation.
Factor.
Set each factor equal to 0.
Solve for x.
The boundary points or zeros are –1/2 and 2.
Example cont.
Solve the inequality and support it graphically:
2x2 – 3x > 2.
Locate the boundary points or zeros on a number
line (or sign chart). The number line with the boundary points is
shown as follows:
Step 3
-1/2
2
x
-5 -4 -3
-2
-1
0
1
2
3
4
5
6
The boundary points or zeros divide the number line into three
test intervals. Including the boundary points (because of the given
greater than or equal to sign), the intervals are
(-∞, -1/2], [-1/2, 2], [2, ∞)
Solve the inequality and support it graphically:
2x2 – 3x > 2.
Step 4
Take one representative number within each test interval and
substitute that number into the inequality.
-1/2 (+)(-) 2
(-)(-)
(+)(+)
x
-5 -4 -3
-2
(-∞, -1/2],
-1
0
1
2
3
4
[-1/2, 2],
5
6
[2, ∞)
(2(1)  1)(1  2)  0
(2(0)  1)(0  2)  0
(2(3)  1)(3  2)  0
3 0
2  0
False or Negative
70
True or Positive
[-1/2, 2] does not belong
to the solution set.
[2, ∞) belongs to
the solution set.
True or Positive
(-∞, -1/2] belongs to
the solution set.
Example cont.
Solve the inequality and support it graphically:
2x2 – 3x > 2.
Step 5
Write the solution set; selecting the interval(s) that satisfy the
given inequality:
(-∞, -1/2] or [2, ∞)
Support it graphically.
Text Example
Solve and graph the solution set:
x 1
2
x3
Step 1
Express the inequality so that one side is zero and the other side
is a single quotient. We subtract 2 from both sides to obtain zero on the right.
x 1
2
x3
x 1
20
x3
x  1 2 x  3

0
x  3 x  3
x  1 2 x  3
0
x3
x  1 2x  6
0
x3
x  5
0
x3
This is the given inequality.
Subtract 2 from both sides, obtaining 0
on the right.
The least common denominator is x=3.
Express 2 in terms of this denominator.
Subtract rational expressions.
Apply the distributive property.
Simplify.
Text Example cont.
Solve and graph the solution set:
x 1
2
x3
Step 2
Find boundary points by setting the numerator and the
denominator equal to zero.
x  5
0
x 3
-x - 5
= 0
x = -5
x+3 = 0
x = -3
Set the numerator and denominator equal
to 0. These are the values that make the
previous quotient zero or undefined.
Solve for x.
The boundary points are -5 and -3. Because equality is included in the given lessthan-or-equal-to symbol, we include the value of x that causes the quotient to be
zero. Thus, -5 is included in the solution set. By contrast, we do not include -3 in the
solution set because -3 makes the denominator zero or undefined.
Step 3
Locate boundary points on a number line (or sign chart).
x
-9 -8 -7 -6 -5 -4 -3 -2 -1
0 1
The boundary points divide the number line into three test intervals, namely
(-∞, -5], [-5, -3), (-3, ∞)
Text Example cont.
Solve and graph the solution set:
x
-9 -8 -7 -6 -5 -4 -3 -2 -1
x 1
2
x3
0 1
Step 4
Take one representative number within each test interval and
substitute that number into the inequality.
(-∞, -5],
[-5, -3),
(-3, ∞)
(4)  5
(0)  5
(6)  5
0
0
0
4  3
03
6  3
1
5
false or
true
or
0
1 0
 0 False or
negative
3
3
positive
negative
(-∞, -5] does not belong
to the solution set.
Step 5
[-5,-3) belongs to the
solution set.
(-3, ∞) does not belong
to the solution set.
The solution set are the intervals that produced a true
statement. Our analysis shows that the solution set is [-5,
-3).
Complete Student Checkpoint
Solve and support graphically.
x 3  3x 2  x  3
x  3  x  3
,3  4  34  14  1 0
True

x 3  3x 2  x  3  0


15  0

 3,1  2  3 2  1 2  1  0
x2 x  3  1 x  3  0
False
3 0
 1,1  0  3 0  1 0  1  0
x  3 x2  1  0
3  0
True
x  3 x 1 x 1  0
1,  2  3 2  1 2  1  0
x  3 or x  1 or x  1
15  0
False
,3 or  3,1 or  1,1 or 1,

  
  
   


  
 
  

,3
(-)(-)(-)
U  1,1
(+)(-)(-)(+)(+)(-) (+)(+)(+)
-5 -4 -3 -2 -1
0
1
negative positive negative
2
3
4
positive
x
5
Complete Student Checkpoint
Solve and support graphically:
2x
1
x 1
1 1
(2)  1
 ,1 
0
2  1
True
3 0
(0)  1
1,1 
0
0 1
False
1 0

2x
1 0
x 1
2x 1 x  1

0
x 1
x 1
x 1
0
x 1
x  1 or x  1



 
 ,1 U 1, 

 ,1 or 1,1 or 1,
(-)/(-)
-5
-4
(2)  1
1, 
0
2 1
True
1
0
3

(-)/(+)
-3 -2 -1
0
1
positive
negative
(+)/(+)
2
3
4
positive
x
5
Solving Inequalities
in One Variable