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Section 1.6
Polynomial and Rational
Inequalities
Polynomial Inequalities
We said that we can find the solutions
(a.k.a. zeros) of a polynomial by setting
the polynomial equal to zero and solving.
 We are going to use this skill to solve
inequalities such as:

x  x  12  0
2
Solving Quadratic Inequalities
x 2  x  12  0
x  4x  3  0
x  4x  3  0
x4
x  3
Factor
Identify the zeros
(critical points)
There are now 3 intervals: (-∞,-3), (-3,4), and (4,∞).
We will test these three intervals to see which parts of this function
are less than (negative) or greater than (positive) zero.
Testing Intervals


To test, pick a number from each interval and
evaluate
Instead of evaluating, we can also just check the
signs of each factor in our factored form of the
polynomial.
x  4x  3  0
Solution: (-∞,-3) U (4,∞)
Recap of Steps
Factor and solve the quadratic to find the
critical points
 Test each interval
 Determine if (+) or (-) values are desired

Solve the Inequality
3m 2  5m  2
3m 2  5m  2  0
3m 1m  2  0
Solution:
1
m  and  2
3
x2 – 2x ≥ 1
x 2  2x  1  0
x
2
 2  41 1
21
2
2 8
x
2
22 2
x
2
x  1 2
x  2.4and  0.4
Solution:
 ,1 
 
2  1  2 , 
x2 + 2x ≤ -3
x2  2x  3  0
 2  22  413
x
21
 2 8
x
2
x  1  i 2
Test any number to find out if all
numbers are true or false.
No Real Solutions
Solve the rational inequality
2m  5
0
6m
Restrictions?
2m  5  0
6m  0
5
m
2
m6
Solving Rational Inequalities
x 1
0
2
64  x
x 1
0
8  x 8  x 
x  1
x 8
x  8
Solution: (-∞,-8) U (-1,8)
Restrictions?
x  8
v2 1
0
v 1
v  1v  1  0
v 1
v  1 0
v 1
Restrictions?
v  1
x2  8
0
2
x 4
x2  8  0
x  8
2
x  i 8
x2  4  0
x 2  4
x  4i
Test any number… it’s either all positive, or all negative.
Solve
1
1

0
t 2 t 2
t 2
t2

0
t  2t  2 t  2t  2
t 2t 2
0
t  2t  2
2t
0
t  2t  2
Restrictions?
t  2,2
t  0,2,2
Solve
x 2  3x
6
3
x 2  3x
6  0
3
x 2  3x 18
 0
3
3
x 2  3x  18
0
3
x  6x  3  0
3
x6
x  3
No Restrictions
Solve
 t 2  2t
t
4t
 t 2  2t 4t  t 2

0
4t
4t
 t 2  2t
t  0
4t
 t 2  2t  4t  t 2
0
4t
 t 2  2t t 4  t 

0
4t
4t
 6t
0
4t
t 0
t 4
Restrictions?
t4