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 x 2 – x > 6 x 2 – x – 6 > 0

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

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,

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

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

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

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” x 3 – x 2 – 6x > 0 x(x – 3)(x + 2) > 0 Critical #s: -2, 0, 3 (-)(-)(-) (-)(-)(+) (+)(-)(+) (+)(+)(+) Signs by region:

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:

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}

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

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

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 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 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} Example with 5 Factors but 4 Critical #s

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

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

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