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Solving Polynomial Inequalities
Basic Principle:
When will the product “ab” be positive?
Answer:
When a and b are both positive OR
when a and b are both negative!
Basic Principle:
When will the product “abcde” be negative?
Answer: When an odd number of the factors
are negative!
Solving Polynomial Inequalities
1. Set equal to zero (Move everything to one side.)
2. Factor
3. The solution from each factor becomes a
CRITICAL NUMBER
4. Plot critical numbers on a number line
5. Test any one number in each interval
noting each factor as positive or negative.
Find the signs that make the desired
inequality true.
* This is called a SIGN GRAPH
Example with Two Factors
•
Set inequal to zero in general form
x2 – x > 6
x2 – x – 6 > 0
Example with Two Factors
•
•
Factor and solve.
Each response is a CRITICAL NUMBER
x2 – x – 6 > 0
(x – 3)(x + 2) > 0
Critical #s: -2, 3
-2
3
Next, set up intervals on a SIGN GRAPH
Example with Two Factors
•
The critical numbers make this expression
exactly equal to zero so they are not included
as part of the solution to a strict inequality.
x2 – x – 6 > 0
(x – 3)(x + 2) > 0
Critical #s: -2, 3
-2
3
Example with Two Factors
•
In order to get ab > 0, wouldn’t both factors
have to be positive or both negative?
x2 – x – 6 > 0
(x – 3)(x + 2) > 0
Critical #s: -2, 3
-2
3
Test each region of the graph: Where will both factors be
negative? Where will both factors be positive?
Example with Two Factors
x2 – x – 6 > 0
(x – 3)(x + 2) > 0
Critical #s: -2, 3
Test: x = -10
(-)(-)
-2
x= 0
(-)(+)
3
x=10
(+)(+)
Pick any # in the interval and plug it into x.
For example, test x = -10, x=0 and x=10
Example with Two Factors
2
x
–x–6>0
(x – 3)(x + 2) > 0
Critical #s: -2, 3
(-)(-)
-2
(-)(+)
3
(+)(+)
Solution Set: {x: x < -2 or x > 3}
Take Special Note:
1. Whether endpoints are included or open.
> OR <  open ≤ OR  closed circle
2. The number of regions to test is one more
than the number of critical numbers.
3. With single power factors, intervals will
generally alternate.
4. When you cross a double root, two
factors change sign at the same time so
the intervals will not alternate there.
Example with Three Factors
•
You must test a number in each region to
determine the sign. Make a “sign graph”
x3 – x2 – 6x > 0
x(x – 3)(x + 2) > 0
Critical #s: -2, 0, 3
3
-2
0
(-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+)
Signs by region:
Example with Three Factors
•
You must test a number in each region to
determine the sign. Called a “sign graph”
x3 – x2 – 6x > 0
x(x – 3)(x + 2) > 0
Critical #s: -2, 0, 3
3
-2
0
(-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+)
Signs by region:
Example with Three Factors
•
You must test a number in each region to
determine the sign. Called a “sign graph”
x3 – x2 – 6x > 0
x(x – 3)(x + 2) > 0
Critical #s: -2, 0, 3
-2
0
3
Solution Set: {x: -2 ≤ x ≤ 0 or x ≥ 3}
Example with Four Factors
•
Watch two signs change together when a
factor appears twice.
x4 – x3 – 6x2 > 0
(x)(x)(x – 3)(x + 2) > 0
Critical #s: -2, 0 d.r., 3
-2
0
3
(-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+)
Signs by region
Example with Four Factors
•
Watch two signs change together when a
factor appears twice.
x4 – x3 – 6x2 > 0
(x)(x)(x – 3)(x + 2) > 0
Critical #s: -2, 0 d.r., 3
-2
0
3
(-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+)
Signs by region
Example with 5 Factors but 4 Critical #s
• Watch two signs change together
when a factor appears twice.
(x)(x+5)(x+5)(x – 4)(x - 7) < 0
Critical #s: -5 d.r., 0, 4, 7
-5
(-)(-)(-)(-)(-)
0
4
7
(-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+)
Signs by region
Example with 5 Factors but 4 Critical #s
• Watch two signs change together
when a factor appears twice.
(x)(x+5)(x+5)(x – 4)(x - 7) < 0
Critical #s: -5 d.r., 0, 4, 7
-5
(-)(-)(-)(-)(-)
0
4
7
(-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+)
Signs by region
Example with 5 Factors but 4 Critical #s
• Watch two signs change together
when a factor appears twice.
(x)(x+5)(x+5)(x – 4)(x - 7) < 0
Critical #s: -5 d.r., 0, 4, 7
-5
(-)(-)(-)(-)(-)
0
4
7
(-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+)
Solution Set: {x: x < -5 or -5 < x < 0 or 4 < x < 7}
Watch for Special Cases!
• Could get {all reals}!
(x)(x)(x2+5) 0
Critical #s: 0 d.r.
0
(-)(-)(+) (+)(+)(+)
Solution Set: {reals}
Watch for Special Cases!
• Could get  !
(x)(x)(x + 5) (x + 5) < 0
Critical #s: -5 d.r., 0 d.r.
-5
(-)(-)(-)(-)
0
(-)(-)(+)(+)
(+)(+)(+)(+)
Solution Set:  or { }
Watch for Special Cases!
• Now what if we changed < to ≤ ?
(x)(x)(x + 5) (x + 5) ≤ 0
Critical #s: -5 d.r., 0 d.r.
(-)(-)(-)(-)
-5
0
(-)(-)(+)(+)
(+)(+)(+)(+)
Solution Set: {-5, 0}
can’t get a negative product, but the
critical #s do produce a zero product