1. Polynomial inequalities Objective –To Solve polynomial inequalities.

2. Solving polynomial inequalities Rewrite the polynomial so that all terms are on one side and zero on the other. Factor the polynomial. We are interested in when factors are either pos. or neg., so we must know when the factor equals zero. The values of x for which the factors equal zero are the boundary points, which we place on the number line. The intervals around the boundary points must be tested to find on which interval(s) will the polynomial be positive/negative.

3. Quadratic Inequalities When solving inequalities we are trying to find all possible values of the variable which will make the inequality true. Consider the inequality We are trying to find all the values of x for which the quadratic is greater than zero or positive.

4. Solving a quadratic inequality We can find the values where the quadratic equals zero by solving the equation,

5. Solving a quadratic inequality For the quadratic inequality, we found zeros 3 and –2 by solving the equation. Put these values on a number line and we can see three intervals that we will test in the inequality. We will test one value from each interval. -2 3

6. Solving a quadratic inequality IntervalTest PointEvaluate in the inequalityTrue/False

7. Solving a quadratic inequality You may recall the graph of a quadratic function is a parabola and the values we just found are the zeros or x- intercepts. The graph of is

8. Solving a quadratic inequality Thus the intervals is the solution set for the quadratic inequality,. In summary, one way to solve quadratic inequalities is to find the zeros and test a value from each of the intervals surrounding the zeros to determine which intervals make the inequality true.

9. Solve To solve this inequality we observe that 0 is already on one side so we need to factor it. Use calculator or synthetic division!

10. Solve : (x – 3)(x + 1)(x – 6) < 0 The 3 boundary values are x = 3,-1,6 They create 4 intervals: Pick a number in each interval to test the sign of that interval. If the polynomial is negative there then the interval is in the solution set. -1 3 6

11. Solve : (x – 3)(x + 1)(x – 6) < 0 Pick a number in each interval to test the sign of that interval. If the polynomial is negative there then the interval is in the solution set. 36 (-2,-40) (0,18) (4,-10) (7,32) neg pos neg pos

12. Solve: x 3 +3x 2 ≥ 10x 1.To solve, first we must rewrite the inequality so all terms are on one side and 0 on the other, then factor.

13. Solve: x 3 +3x 2 -10x ≥ 0 1.To solve, first we must rewrite the inequality so all terms are on one side and 0 on the other, then factor. 2.x(x-2)(x+5) ≥ 0 3.Boundary points: 0, 2, -5 -5 0 2

14. Solve: x 3 +3x 2 -10x ≥ 0 1.To solve, first we must rewrite the inequality so all terms are on one side and 0 on the other, then factor. 2.x(x-2)(x+5) ≥ 0 3.Boundary points: 0, 2, -5 4. -5 0 2 5.Solution set: [-5,0] u [ 2, )

15. Solving rational inequalities **VERY similar to solving polynomial inequalities EXCEPT if the denominator equals zero, there is a domain restriction. The function is not defined there. (open circle on number line) Step 1: Rewrite the inequality so all terms are on one side and zero on the other. Step 2: Factor both numerator & denominator to find boundary values for regions to check when function becomes positive or negative. And do as before !

16. Example: Factor numerator and denominator:

17. Solving Rational Inequalities 1. Zeros of the denominator are marked with open circles. -4 4 2. Solutions to the equation are marked as indicated. 0 -3

18. Solving Rational Inequalities 1. Zeros of the denominator are marked with open circles. (-4, -3] 2. Solutions to the equation are marked as indicated. [0, 4) 3. Test any number to determine true or false. Shade where true. Shading alternates (except for repeated roots). 1 or -3.5

19. Solve the following inequalities: 1) 2) 3)

20. Solution: 1).75 1 neg pos neg

21. Solve the following inequalities: 2) -3 0 3 + - + -

22. Solution: 3)