2.8: Polynomial Inequalities
F(x)
2(x – 2)²(x + 3)³
And polynomials… and graphing… and designing boxes….
Objectives
 Be able to find the positives, negatives, and zeros of a polynomial
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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).
What is an inequality?
It’s not social justice.
Symbols:
≤ Less Than Or Equal To
≥
<
> Greater Than
Greater Than Or Equal To
Less Than
Like Pacman!
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
- - -4
-
++
+
0
- -
5
++
+
…Which can help figure out which part of
the number line is positive or negative.
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)
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!
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
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.
The first zero
occurs at 4
4 2 -7 -10 24
8 4 -24
__________________________________________
2 1 -6 0
Continued
Continue to factor after you’ve completed the synthetic
division
F(x)= (x-4)(2x² + x – 6)
1st zero
= (x-4)(2x-3)(x-2)
Zeros are: 4, 3/2, and -2.
After Using
Rational Zeros…
• You may now plug in
Synthetic Division Example by http://taylorlstormberg.wordpress.com
the possible rational
using synthetic
division.
• After finding a “true
zero”, find out where
the polynomial is
positive or negative, as
previously done
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.algebratutoring.com
Examples: Solving Graphically
 x³ - 6x² ≤ x-8x
Solution:
(-∞, 0.32] U [1.46, 4.21]
Rewrite!
x³ - 6x² + 8x – 2 ≤ 0
F(x) = x³ - 6x² + 8x – 2
 Zeros approx at:
0.32, 1.46, and 4.21
GRAPH!
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
or above the x-axis
on
However...
there are some “empty” inequalities that have no solutions
Impossible Inequalities
(x² + 7)(2x² + 1) < 0
(x² + 7)(2x² + 1) ≤ 0
 Square roots cannot have
 Again, when solving for
negative numbers
 When solving for zero,
x² would equal -7, which
would be impossible,
unless imaginary
numbers are present.
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.
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
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
-4
- -
++
0
undefined
+
1
3
++
+
Undefined zeros can be found by finding the zero of the denominator of the
function
Examples: Sign Charts for a
Rational Function
 R(x) = (2x +1) / ( (x+3)(x-1))
Solution
Find the real zeros. ( in numerator only)
(-3, -1/2) U (1,∞)
X = -1/2
and
If x is equal to -3 or 1, r(x) is undefined.
R(x) is positive if -3 ˂ x ˂ -1/2 or x ˃ 1. (-∞, -3) U (-1/2, 1)
R(x) is negative if x ˂ -3 or -1/2 ˂ 1 ˂ 1
(-)
------(+) (-)
(-)
-----(-) (-)
Negative
-3
(+)
------(+) (-)
Positive
Negative
-1/2
(+)
------(+) (+)
Positive
1
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.
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)
Applications – Box Problem
1)
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?
2)
Volume V of a box is V(x) = x(y-2x)(y-2x)
3)
Y represents the side-lengths of cardboard, so Y would go into the volume equation.
4)
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.
5)
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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]
Applications – Can Problem
1.
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.
2.
Surface Area = 2πr² + 4000/r
3.
The inequality for this equation is 2πr² + 4000/r < 1000
4.
To solve this equation plug the equation from step two into the graphing calculator.
Then put 1000 in the second “y=“ area below that.
5.
Find out where the lines cross via graphing. The surface area should be 4.62 < r <
9.65 to be less than 1000 cm³.
6.
The volume equation is V = πr²h and V = 2000.
7.
Using this equation, we can find out that H = 2000/πr²
8.
To find values of h we build a double inequality for 2000/(πr²)
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
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Mrs. Barber’s Math practice tests
 gallippi-advfunctions12.blogspot.com
 http://www.regentsprep.org/Regents/math/algtrig/ATE11/RationalInequalitiesLes.htm
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