Download Lesson 2.1 - Elgin Local Schools

Lesson 3-2
Polynomial Inequalities in
One Variable
http://www.mathsisfun.com/algebra/inequality-quadratic-solving.html
Objective:
Objective:
To solve and graph
polynomial inequalities in
one variable.
Polynomial Inequalities:
Polynomial Inequalities:
P(x) > 0 or P(x) < 0
Polynomial Inequalities:
P(x) > 0 or P(x) < 0
a) Use a sign graph of P(x).
Polynomial Inequalities:
P(x) > 0 or P(x) < 0
b) Analyze a graph of P(x).
Polynomial Inequalities:
P(x) > 0 or P(x) < 0
b) Analyze a graph of P(x).
i. P(x) > 0 when the graph is above the x-axis.
Polynomial Inequalities:
P(x) > 0 or P(x) < 0
b) Analyze a graph of P(x).
i. P(x) > 0 when the graph is above the x-axis.
ii. P(x) < 0 when the graph is below the x-axis.
Example:
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 1: Find the zeros of the polynomial.
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 1: Find the zeros of the polynomial.
P(x) = x3 – 2x2 – 3x
P(x) = x(x2 – 2x – 3)
P(x) = x(x – 3)(x + 1)
Zeros: x = 0, x = 3, x = - 1
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
3
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
Now, these 3 zeros separate this
number line into 4 areas.
3
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
3
All the values less than -1,
all the values between -1 and 0,
all the values between 0 and 3, and
all the values greater than 3.
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
Now, pick a number in an area,
like 1. Substitute 1 in for x.
1(1-3)(1+1)  1(-2)(2)  -4
3
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
Now, pick a number in an area,
like 1. Substitute 1 in for x.
1(1-3)(1+1)  1(-2)(2)  -4
3
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
So, that means when x = 1, P(x),
which is y, is negative.
3
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
3
Now, using our rule for roots we
discovered a chapter ago, the
graph of P(x) will change signs
after it goes through each root,
since they are all single roots.
Example:
Solve x3 – 2x2 – 3x < 0 by using a sign graph.
Step 2: Plot the zeros on a number line.
-1
0
3
Using this sign graph and the fact
that we are dealing with P(x) < 0
(which means negative numbers)
the answer to the problem is:
x< -1 U 0 < x < 3 or (- 8, -1) U (0,3)
Example:
Solve (x2 – 1)(x – 4)2 > 0.
Example:
Solve
( x + 2) ( x - 5)
x-4
2
£0
Example:
Use your grapher to solve 2x + x -8x + 3 > 0.
3
2
Assignment:
Pg. 103
C.E. 1-6 all
W.E. 1-19 odd