2.8: Polynomial Inequalities

1. 2.8: Polynomial Inequalities F(x) 2(x – 2)²(x + 3)³ And polynomials… and graphing… and designing boxes….

4. Objectives • Be able to find the positives, negatives, and zeros of a polynomial inequality via line method. • Be able to graph polynomial inequalities. • Be able to solve inequalities using the analytical method (synthetic division, rational zeros theorem…) • Know what an “impossible inequality” looks like and understand why it’s impossible. • Be able to solve rational inequalities. • Be able to combine functions with rational inequalities. • Apply this information to real-life problems (box method, cylinder method, etc).

5. What is an inequality?

6. It’s not social justice.

7. Symbols: ≤ Less Than Or Equal To ≥ Greater Than Or Equal To < Less Than > Greater Than Like Pacman!

8. Polynomial inequalities! To solve f(x) > 0, find the values of x that make f(x) positive To solve f(x) < 0, find the values of x that make f(x) negative To solve f(x) = 0, find the values of x that make f(x) zero …Which can help figure out which part of the number line is positive or negative. + + + • - + + + • - • 4 0 5

9. Examples: Finding Zeros and Multiplicity F(x)= (x+2)(x+1)(x-5) Zeros at: -2, -1, and 5 (each multiplicity 1) F(x)= (x+3)²(x+6)(x-9)³ Zeros at: -3 (multiplicity 2), -6(multiplicity 1), 9 (multiplicity 3). F(x)= (4x+8)(x-2)(x+7)² Zeros at: -2, 2( multiplicity 1), -7(multiplicity 2)

10. However… some polynomial inequalities do not appear in a factored form and some don’t have zero on one of the sides. But they can be solved in various ways!

11. Solving Them Analytically To do this, use the rational zeros theorem, finding the possible factors of the constant coefficient over the leading coefficient. How to find possible factors of a polynomial, statisticslectures.com

12. Examples: Solving a Polynomial Analytically F(x)= 2x³-7x²-10x+24 In this example, enter the equation on your calculator, and find the first zero. The next step is synthetic division. • 2 -7 -10 24 8 4 -24 __________________________________________ 2 1 -6 0 • The first zero occurs at 4

13. Continued Continue to factor after you’ve completed the synthetic division F(x)= (x-4)(2x² + x – 6) = (x-4)(2x-3)(x-2) Zeros are: 4, 3/2, and -2. 1st zero

14. After Using Rational Zeros… You may now plug in the possible rational using synthetic division. After finding a “true zero”, find out where the polynomial is positive or negative, as previously done Synthetic Division Example by http://taylorlstormberg.wordpress.com

15. Graphing! Explained Yet another way to solve the polynomial inequality is graphically. To do this, make sure that if there are two polynomials on either side of the inequality get moved to one side. Then we make the inequality equal to zero. After plugging the equation into a graphing calculator, find the zeros of the polynomial. Graphing polynomials from www.algebra-tutoring.com

16. Examples: Solving Graphically • x³ - 6x² ≤ x-8x Solution: Rewrite! (-∞, 0.32] U [1.46, 4.21] x³ - 6x² + 8x – 2 ≤ 0 F(x) = x³ - 6x² + 8x – 2 • Zeros approx at: 0.32, 1.46, and 4.21 GRAPH!

17. In Addition... Depending on the inequality, the solutions of the equation may vary. • If it’s x… < 0, then the solutions are below the x-axis • If it’s x… > 0, then the solutions are above the x-axis • If it’s x… ≤ 0, then the solutions are on or below the x-axis • If it’s x… ≥ 0,then the solutions are on or above the x-axis

18. However... there are some “empty” inequalities that have no solutions

19. Impossible Inequalities (x² + 7)(2x² + 1) < 0 (x² + 7)(2x² + 1) ≤ 0 Square roots cannot have negative numbers When solving for zero, x² would equal -7, which would be impossible, unless imaginary numbers are present. Again, when solving for zero, x² would be equal to either -7 or -1/2. Square roots of negative numbers would still be impossible unless imaginary numbers are present.

20. Examples: Unusual Answers (2x² + 7)(- x² + 1) ˃ 0 solution is (-∞, ∞ ) also, (2x² + 7)(- x² + 1) ≥ 0 solutions is (-∞, ∞ ) BUT, (2x² + 7)(- x² + 1) ˂ 0 (2x² + 7)(- x² + 1) ≤ 0 Both solutions are empty

21. Rational Inequalities Along with polynomial inequalities, there are rational inequalities as well! Not only can the zeros to rational inequalities be zero, positive, or negative, but zeros can also be undefined. Example Equation: y = (x+4)(x+3)/(x+-1) • - • - + + + + + + 0 0 undefined • 4 1 3 Undefined zeros can be found by finding the zero of the denominator of the function

22. Examples: Sign Charts for a Rational Function • R(x) = (2x +1) / ( (x+3)(x-1)) Find the real zeros. ( in numerator only) X = -1/2 If x is equal to -3 or 1, r(x) is undefined. R(x) is positive if -3 ˂ x ˂ -1/2 or x ˃ 1. R(x) is negative if x ˂ -3 or -1/2 ˂ 1 ˂ 1 Solution (-3, -1/2) U (1,∞) and (-∞, -3) U (-1/2, 1) (-) ------- (+) (-) (-) ------ (-) (-) (+) ------- (+) (-) (+) ------- (+) (+) Negative Positive Negative Positive 1 -3 -1/2

23. Combining Functions With Rational Inequalities You can also solve a rational inequality by combining functions. To do this, combine the two functions on the left-hand side by using the least common denominator. (multiply the two denominators together). Then you add the fractions, use distributive property, simplify, and divide.

24. Examples: Combining Fractions • Solve 5 / (x+3) + 3/ (x-1) ˂ 0 Combine the two fractions: 5(x-1)/(x+3)(x-1) + 3(x+3)/ (x+3)(x-1) 5(x-1) + 3(x+3) / (x+3)(x-1) 5x-5 + 3x +9 / (x+3)(x-1) 8x +4 / (x+3)(x-1) (divide both sides by 4) 2x +1 / (x+3)(x-1)

25. Applications – Box Problem • A packaging company wishes to design boxes with a volume of at least 600 in³. Squares are to be cut from the corners of a 20-in by 25-in piece of cardboard, with the flaps folded up to make an open box. What size squares should be cut? • Volume V of a box is V(x) = x(y-2x)(y-2x) • Y represents the side-lengths of cardboard, so Y would go into the volume equation. • X represents the side length of the removed squares and the height of the box. • Volume is now V(x) = x(25-2x)(20-2x) • To obtain a volume of at least 600 in³, we make the equation into an inequality. • 600 ≤ x(25-2x)(20-2x) 6) To solve the application graphically, plug the equation from step four into the calculator and plug 600 into the second y= spot below the first equation. 7) Make the window 0 ≤ x ≤ 10 (because the width of the cardboard is 20) and 0≤y≤1000 8) Find the intercepts between the two graphs. The solution is [1.66, 6.16]

26. Applications – Can Problem • A cannery wants to package cans in a 2-liter(2000cm³) cylindrical can. Find the radius and height of the cans if the cans have a surface area that is less than 1000 cm². • Surface Area = 2πr² + 4000/r • The inequality for this equation is 2πr² + 4000/r < 1000 • To solve this equation plug the equation from step two into the graphing calculator. Then put 1000 in the second “y=“ area below that. • Find out where the lines cross via graphing. The surface area should be 4.62 < r < 9.65 to be less than 1000 cm³. • The volume equation is V = πr²h and V = 2000. • Using this equation, we can find out that H = 2000/πr² • To find values of h we build a double inequality for 2000/(πr²)

27. References • Precalculus 8th edition • http://taylorlstormberg.wordpress.com • statisticslectures.com • Scarseth’s In-Class Notes • www.algebra-tutoring.com • Youtube.com • Quickmeme.com • Deviantart.com • Hark.com • Sosmath.com • http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html • https://www.desmos.com/calculator • Mrs. Barber’s Math practice tests • gallippi-advfunctions12.blogspot.com • http://www.regentsprep.org/Regents/math/algtrig/ATE11/RationalInequalitiesLes.htm