Solving Polynomial Inequalities

1. 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!

2. Solving Polynomial Inequalities • Set equal to zero (Move everything to one side.) • Factor • The solution from each factor becomes a CRITICAL NUMBER • Plot critical numbers on a number line • 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

3. Example with Two Factors • Set inequal to zero in general form x2 – x > 6 x2 – x – 6 > 0

4. -2 3 Next, set up intervals on a SIGN GRAPH 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 (Critical #s are where the product would be exactly equal to zero!)

5. 3 -2 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

6. 3 -2 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 Test each region of the graph: Where will both factors be negative? Where will both factors be positive?

7. x= 0 Test: x = -10 x=10 3 -2 Example with Two Factors x2 – x – 6 > 0 (x – 3)(x + 2) > 0 Critical #s: -2, 3 (-)(-) (-)(+) (+)(+) Pick any # in the interval and plug it into x. For example, test x = -10, x=0 and x=10

8. 3 -2 Example with Two Factors x2 – x – 6 > 0 (x – 3)(x + 2) > 0 Critical #s: -2, 3 (-)(-) (-)(+) (+)(+) Solution Set: {x: x < -2 or x > 3}

9. Take Special Note: • Whether endpoints are included or open. >OR< open ≤OR closed circle • The number of regions to test is one more than the number of critical numbers. • With single power factors, intervals will generally alternate. • When you cross a double root, two factors change sign at the same time so the intervals will not alternate there.

10. 3 -2 0 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 (-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+) Signs by region:

11. 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 (-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+) Signs by region: 3 -2 0

12. 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 Solution Set: {x: -2 ≤ x ≤ 0 or x ≥ 3} 3 -2 0

13. 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 (-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+) Signs by region 3 -2 0

14. 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 (-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+) Signs by region 0 3 -2

15. 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 (-)(-)(-)(-)(-) (-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+) Signs by region 4 7 -5 0

16. 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 (-)(-)(-)(-)(-) (-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+) Signs by region -5 0 7 4

17. 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 (-)(-)(-)(-)(-) (-)(+)(+)(-)(-) (+)(+)(+)(-)(-) (+)(+)(+)(+)(-) (+)(+)(+)(+)(+) Solution Set: {x: x < -5 or -5 < x < 0 or 4 < x < 7} -5 0 4 7

18. Watch for Special Cases! • Could get {all reals}! (x)(x)(x2+5) 0 Critical #s: 0 d.r. (-)(-)(+) (+)(+)(+) Solution Set: {reals} 0

19. Watch for Special Cases! • Could get ! (x)(x)(x + 5) (x + 5) < 0 Critical #s: -5 d.r., 0 d.r. (-)(-)(-)(-) (-)(-)(+)(+) (+)(+)(+)(+) Solution Set: or { } -5 0

20. 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. (-)(-)(-)(-) (-)(-)(+)(+) (+)(+)(+)(+) Solution Set: {-5, 0} can’t get a negative product, but the critical #s do produce a zeroproduct 0 -5