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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