Polynomial inequalities

• Objective

– To Solve polynomial inequalities.

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.

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.

062 xx

Solving a quadratic inequality

We can find the values where the quadratic equals zero

by solving the equation, 062 xx

023 xx

02or03 xx

2or3 xx

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.

062 xx

x2 −x−6 0

-2 3

Solving a quadratic inequality

Interval Test Point

Evaluate in the inequality True/False

2,

3,2

,3

06639633 2

0 2 − 0 −600−6 −6 0

066416644 2

3x

0x

4x

True

True

False

062 xx

062 xx

062 xx

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

62 xxy

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.

062 xx

−∞,−2( ) or 3,∞( )

Solve

• To solve this inequality we observe that 0 is already on one side so we need to factor it.

x3 −8x2 +9x+18 < 0

• Use calculator or synthetic division!

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.

),6(),6,3(),3,1(),1,(

-1 3 6

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.

(−∞,−1)∪(3,6)

-1 3 6

(-2,-40) (0,18) (4,-10) (7,32)

neg pos neg pos

Solve: x3 +3x2 ≥ 10x

1. To solve, first we must rewrite the inequality so all terms are on one side and 0 on the other, then factor.

Solve: x3 +3x2 -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) ≥ 03. Boundary points: 0, 2, -5

-5 0 2

Solve: x3 +3x2 -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, ) ∞

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 !

Example:

• Factor numerator and denominator:

2

2

30

16

x x

x

Solving Rational Inequalities

2

2

30

16

x x

x

1. Zeros of the denominator are marked with open circles.

2

2

16 0

16

4

x

x

x

-4 4

2. Solutions to the equation are marked as indicated.

2 3 0

3 0

0 3

x x

x x

x x

0-3

Solving Rational Inequalities

2

2

30

16

x x

x

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

2

2

1 3 10

1 164

015

true

or

(−3.5)2 + 3(−3.5)(−3.5)2 −16

−.4666 < 0 true

-3.5

Solve the following inequalities:

1)

2)

3)

043

1

x

x

09

22

xx

2x +6x2 −5x

≤0

Solution:

1)x−13−4x

0

x1, x≠34

.75 1

neg pos neg

(−∞,.75)∪(1,∞)

Solve the following inequalities:

2) 2x

9−x2≥0

x=0,x≠3,−3

-3 0 3

+ - + -

(−∞,−3)∪[0,3)

Solution:

3)2x +6x2 −5x

≤0

x=−3, x≠0,5

(−∞,−3]∪(0,5)