Excluded Values for Rational
Expressions
Andrew Gloag
Melissa Kramer
Anne Gloag
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Printed: February 18, 2015
AUTHORS
Andrew Gloag
Melissa Kramer
Anne Gloag
www.ck12.org
C HAPTER
Chapter 1. Excluded Values for Rational Expressions
1
Excluded Values for
Rational Expressions
Here you’ll learn how to find the excluded values for a rational expression and how to simplify the expression.
Did you know that Pluto is no longer considered a planet? Even so, we can still calculate the distance between it
and the Sun. All we would have to know is the force of attraction between Pluto and the Sun, the masses of the two
objects, and the value of the gravitation constant. In this Concept, you’ll learn how to simplify rational expressions,
including real-life expressions such as the one we would encounter here.
Guidance
You have gained experience working with rational functions so far. In this Concept, you will continue simplifying
rational expressions by factoring.
To simplify a rational expression means to reduce the fraction into its lowest terms.
To do this, you will need to remember a property about multiplication.
For all real values a and b, and b 6= 0, ab
b = a.
Example A
Simplify
x2 −2x+1
8x−8 .
Solution: Factor both pieces of the rational expression and reduce.
(x − 1)(x − 1)
x2 − 2x + 1
→
8x − 8
8(x − 1)
2
x − 2x + 1 x − 1
=
8x − 8
8
Finding Excluded Values of Rational Expressions
As stated in a previous Concept, excluded values are also called points of discontinuity. These are the values that
make the denominator equal to zero and are not part of the domain.
Example B
Find the excluded values of
2x+1
.
x2 −x−6
Solution: Factor the denominator of the rational expression.
2x + 1
x2 − x − 6
=
2x + 1
(x + 2)(x − 3)
Find the value that makes each factor equal zero.
x = −2, x = 3
1
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These are excluded values of the domain of the rational expression.
Real-Life Rational Expressions
The gravitational force between two objects in given by the formula F =
G(m1 m2 )
.
(d 2 )
The gravitation constant is given
G = 6.67 × 10−11 (N · m2 /kg2 ).
by
The force of attraction between the Earth and the Moon is F = 2.0 × 1020 N (with
masses of m1 = 5.97 × 1024 kg for the Earth and m2 = 7.36 × 1022 kg for the Moon).
Example C
What is the distance between the Earth and the Moon?
Solution:
Let’s start with the Law of Gravitation formula. F = G
Now plug in the known values.
Multiply the masses together.
Cancel the kg2 units.
Multiply the numbers in the numerator.
Multiply both sides by d 2 .
Cancel common factors.
Simplify.
m1 m2
d2
N · m2 (5.97 × 1024 kg)(7.36 × 1022 kg)
.
kg2
d2
N · m2 4.39 × 1047 kg2
.
2.0 × 1020 N = 6.67 × 10−11
kg2
d2
2
N · m2 4.39 × 1047
kg
2.0 × 1020 N = 6.67 × 10−11
·
2
d2
kg
2.93 × 1037
2.0 × 1020 N
N · m2
d2
2.93 × 1037 2
2.0 × 1020 N · d 2 =
· d · N · m2
d2
2.93 × 1037 2
2.0 × 1020 N · d 2 =
· d · N · m2
2
d
2.0 × 1020 N · d 2 = 2.93 × 1037 N · m2
2.0 × 1020 N = 6.67 × 10−11
Simplify.
2.93 × 1037 N · m2
2.0 × 1020 N
2
d = 1.465 × 1017 m2
Take the square root of both sides.
d = 3.84 × 108 m
Divide both sides by 2.0 × 1020 N.
d2 =
Video Review
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79434
2
www.ck12.org
Chapter 1. Excluded Values for Rational Expressions
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79435
Guided Practice
Find the excluded values by simplifying
4x−2
.
2x2 +x−1
Solution:
Both the numerator and denominator can be factored using methods learned in previous Concepts.
4x − 2
2x2 + x − 1
→
2(2x − 1)
(2x − 1)(x + 1)
The expression (2x − 1) appears in both the numerator and denominator and can be canceled. The expression
becomes:
4x − 2
2x2 + x − 1
=
2
x+1
Since x + 1 = 0 ⇒ x = −1, then x = −1 is an excluded value.
Explore More
Sample explanations for some of the practice exercises below are available by viewing the following video. Note
that there is not always a match between the number of the practice exercise in the video and the number of the
practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Ba
sic Algebra: Simplifying Rational Expressions (15:22)
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/48
Reduce each fraction to lowest terms.
1.
2.
3.
4.
5.
6.
4
2x−8
2
x +2x
x
9x+3
12x+4
6x2 +2x
4x
x−2
2
x −4x+4
2
x −9
5x+15
3
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7.
8.
9.
10.
11.
12.
x2 +6x+8
x2 +4x
2x2 +10x
x2 +10x+25
x2 +6x+5
x2 −x−2
x2 −16
x2 +2x−8
3x2 +3x−18
2x2 +5x−3
x3 +x2 −20x
6x2 +6x−120
Find the excluded values for each rational expression.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
2
x
4
x+2
2x−1
(x−1)2
3x+1
x2 −4
x2
x2 +9
2x2 +3x−1
x2 −3x−28
5x3 −4
x2 +3x
9
x3 +11x2 +30x
4x−1
x2 +3x−5
5x+11
3x2 −2x−4
x2 −1
2x3 +x+3
12
x2 +6x+1
23.
24.
25. In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum
of the reciprocals of each resistance: R1c = R11 + R12 . If R1 = 25Ω and the total resistance is Rc = 10Ω, what is
the resistance R2 ?
26. Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between
the two objects is doubled, what is the new force of attraction between the two objects?
27. Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both
objects was doubled, and if the distance between the objects was doubled, then what would be the new force
of attraction between the two objects?
28. A sphere with radius r has a volume of 34 πr3 and a surface area of 4πr2 . Find the ratio of the surface area to
the volume of the sphere.
29. The side of a cube is increased by a factor of two. Find the ratio of the old volume to the new volume.
30. The radius of a sphere is decreased by four units. Find the ratio of the old volume to the new volume.
Mixed Review
31.
32.
33.
34.
35.
36.
37.
4
Name 4p6 + 7p3 − 9.
Simplify (4b2 + b + 7b3 ) + (5b2 − 6b4 + b3 ). Write the answer in standard form.
State the Zero Product Property.
Why can’t the Zero Product Property be used in this situation: (5x + 1)(x − 4) = 2?
Shelly earns $4.85 an hour plus $15 in tips. Graph her possible total earnings for one day of work.
Multiply and simplify: (−4x2 + 8x − 1)(−7x2 + 6x + 8).
A rectangle’s perimeter is 65 yards. The length is 7 more yards than its width.