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Published byMicheal Woolcock Modified over 9 years ago

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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!

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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

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Example with Two Factors Set inequal to zero in general form x 2 – x > 6 x 2 – x – 6 > 0

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Example with Two Factors Factor and solve. Each response is a CRITICAL NUMBER x 2 – x – 6 > 0 (x – 3)(x + 2) > 0 Critical #s: -2, 3 -23 Next, set up intervals on a SIGN GRAPH (Critical #s are where the product would be exactly equal to zero!)

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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. x 2 – x – 6 > 0 (x – 3)(x + 2) > 0 Critical #s: -2, 3 -2 3

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Example with Two Factors In order to get ab > 0, wouldn’t both factors have to be positive or both negative? x 2 – 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? -2 3

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x 2 – 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 Example with Two Factors -2 3 Test: x = -10 x= 0 x=10

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

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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.

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Example with Three Factors You must test a number in each region to determine the sign. Make a “sign graph” x 3 – x 2 – 6x > 0 x(x – 3)(x + 2) > 0 Critical #s: -2, 0, 3 (-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+) Signs by region: -2 3 0

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Example with Three Factors You must test a number in each region to determine the sign. Called a “sign graph” x 3 – x 2 – 6x > 0 x(x – 3)(x + 2) > 0 Critical #s: -2, 0, 3 (-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+) Signs by region: -2 3 0

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Example with Three Factors You must test a number in each region to determine the sign. Called a “sign graph” x 3 – x 2 – 6x > 0 x(x – 3)(x + 2) > 0 Critical #s: -2, 0, 3 Solution Set: {x: -2 ≤ x ≤ 0 or x ≥ 3} -2 3 0

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Example with Four Factors Watch two signs change together when a factor appears twice. x 4 – x 3 – 6x 2 > 0 (x)(x)(x – 3)(x + 2) > 0 Critical #s: -2, 0 d.r., 3 (-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+) Signs by region -2 3 0

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Watch two signs change together when a factor appears twice. x 4 – x 3 – 6x 2 > 0 (x)(x)(x – 3)(x + 2) > 0 Critical #s: -2, 0 d.r., 3 (-)(-)(-)(-) (-)(-)(-)(+) (+)(+)(-)(+) (+)(+)(+)(+) Signs by region Example with Four Factors -2 30

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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 7 0 4 Example with 5 Factors but 4 Critical #s

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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 -570 4 Example with 5 Factors but 4 Critical #s

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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 7 04 Example with 5 Factors but 4 Critical #s

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Could get {all reals}! (x)(x)(x 2 +5) 0 Critical #s: 0 d.r. (-)(-)(+) (+)(+)(+) Solution Set: {reals} Watch for Special Cases! 0

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

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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 zero product Watch for Special Cases! 0 -5

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