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Chapter 4. Rational Equations and Functions
4.4 Rational Expressions
Learning Objectives
• Simplify rational expressions.
• Find excluded values of rational expressions.
• Simplify rational models of real-world situations.
Introduction
A rational expression is reduced to lowest terms by factoring the numerator and denominator completely and divide
out common factors. For example, the expression
x · z1 x
=
y · z1 y
simplifies to simplest form by dividing out the common factor z.
Simplify Rational Expressions.
To simplify rational expressions means that the numerator and denominator of the rational expression have no
common factors. In order to simplify to lowest terms, we factor the numerator and denominator as much as we
can and divide out common factors from the numerator and the denominator of the fraction.
Example 1
Simplify each rational expression.
a)
4x−2
2x2 +x−1
b)
x2 −2x+1
8x−8
−4
c) x2x−5x+6
2
Solution
a) Factor the numerator and denominator completely.
Divide the common term (2x − 1). Answer
2
x+1
b) Factor the numerator and denominator completely.
Divide the common term (x − 1). Answer
(x−1)(x−1)
8(x−1)
x−1
8
c) Factor the numerator and denominator completely.
Divide the common term (x − 2). Answer
2(2x−1)
(2x−1)(x+1)
(x−2)(x+2)
(x−2)(x−3)
x+2
x−3
Common mistakes in simplifying rational expressions:
When simplify rational expressions, you are only allowed to divide out common factors from the denominator but
NOT common terms. For example, in the expression
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4.4. Rational Expressions
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(x + 1).(x − 3)
(x + 2).(x − 3)
we can cross out the (x − 3) factor because
(x−3)
(x−3)
= 1.
We write
(x + 1) · (
x−
3)1 (x + 1)
=
(x + 2) · (
x−
3)1 (x + 2)
However, don’t make the mistake of dividing out common terms in the numerator and denominator. For instance,
in the expression.
x2 + 1
x2 − 5
we cannot cross out the x2 terms.
x2 + 1 x2 + 1
=
x2 − 5 x2 − 5
When we cross out terms that are part of a sum or a difference we are violating the order of operations (PEMDAS).
We must remember that the fraction sign means division. When we perform the operation
(x2 + 1)
(x2 − 5)
we are dividing the numerator by the denominator
(x2 + 1) ÷ (x2 − 5)
The order of operations says that we must perform the operations inside the parenthesis before we can perform the
division.
Try this with numbers:
9 + 1 10
=
= 2.5
9−5
4
CORRECT
But if we divide incorrectly we obtain the following
1
9 + 1
= − = −0.2.
9 − 5
5
INCORRECT
Find Excluded Values of Rational Expressions
Whenever a variable expression is present in the denominator of a fraction, we must be aware of the possibility that
the denominator could be zero. Since division by zero is undefined, certain values of the variable must be excluded.
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Chapter 4. Rational Equations and Functions
These values are the vertical asymptotes (i.e. values that cannot exist for x). For example, in the expression
the value of x = 3 must be excluded.
2
x−3
,
To find the excluded values we simply set the denominator equal to zero and solve the resulting equation.
Example 2
Find the excluded values of the following expressions.
a)
x
x+4
b)
2x+1
x2 −x−6
c)
4
x2 −5x
Solution
a) When we set the denominator equal to zero we obtain. x + 4 = 0 ⇒ x = −4 is the excluded value.
b) When we set the denominator equal to zero we obtain. x2 − x − 6 = 0
Solve by factoring. (x − 3)(x + 2) = 0
⇒ x = 3 and x = −2 are the excluded values.
c) When we set the denominator equal to zero we obtain. x2 − 5x = 0
Solve by factoring. x(x − 5) = 0
⇒ x = 0 and x = 5 are the excluded values.
Removable Zeros
Notice that in the expressions in Example 1, we removed a division by zero when we simplified the problem. For
instance,
4x − 2
2x2 + x − 1
was rewritten as
2(2x − 1)
.
(2x − 1)(x + 1)
This expression experiences division by zero when x =
1
2
and x = −1.
2
However, when we divide out common factors, we simplify the expression to x+1
. The reduced form allows the
1
value x = 2 . We thus removed a division by zero and the reduced expression has only x = −1 as the excluded value.
Technically the original expression and the simplified expression are not the same. When we simplify to simplest
form we should specify the removed excluded value. Thus,
2
1
4x − 2
=
, x =
2x2 + x − 1 x + 1
2
The expression from Example 1, part b simplifies to
x2 − 2x + 1 x − 1
=
, x = 1
8x − 8
8
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The expression from Example 1, part c simplifies to
x2 − 4
x2 − 5x + 6
=
x+2
, x = 2
x−3
Simplify Rational Models of Real-World Situations
Many real world situations involve expressions that contain rational coefficients or expressions where the variable
appears in the denominator.
Example 3
The gravitational force between two objects in given by the formula F = G md1 m2 2 . if the gravitation constant is given
by G ≈ 6.67 × 10−11 (N · m2 /kg2 ). 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).
What is the distance between the Earth and the Moon?
Solution
F =G
Lets start with the Law of Gravitation formula.
Now plug in the known values.
Multiply the masses together.
Divide out the kg2 units.
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
2
kg
N · m 4.39 × 1047
2 · 0 × 1020 N = 6.67 × 10−11
·
2
d2
kg
2.0 × 1020 N = 6.67 × 10−11
2.93 × 1037
N · m2
d2
2.93 × 1037 2
Multiply both sides by d 2 . 2.0 × 1020 N · d 2 =
· d · N · m2
d2
2.93 × 1037 2
Divide out common factors. 2 · 0 × 1020 N · d 2 =
· d · N · m2
2
d
Simplify. 2.0 × 1020 N · d 2 = 2.93 × 1037 N · m2
Multiply the numbers in the numerator.
2.0 × 1020 N =
2.93 × 1037 N · m2
2.0 × 1020 N
d 2 = 1.465 × 1017 m2
Divide both sides by 2.0 × 1020 N.
d2 =
Simplify.
d = 3.84 × 108 m
Take the square root of both sides.
Answer The distance is 3.84 × 108 m.
This is indeed the distance between the Earth and the Moon.
Example 4
The area of a circle is given by A = πr2 and the circumference of a circle is given by C = 2πr. Find the ratio of the
circumference and area of the circle.
Solution
The ratio of the circumference and area of the circle is:
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2πr
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Chapter 4. Rational Equations and Functions
π1r1
We divide out common factors from the numerator and denominator. 2
π1 r2 r
Simplify.
Answer
2
r
Example 5
The height of a cylinder is 2 units more than its radius. Find the ratio of the surface area of the cylinder to its volume.
Solution
Define variables.
Let R = the radius of the base of the cylinder.
Then, R + 2 = the height of the cylinder
To find the surface area of a cylinder, we need to add the areas of the top and bottom circle and the area of the curved
surface.
The volume of a cylinder is the area of the base of the cylinder times its height, so:
The volume of the cylinder is
The ratio of the surface area of the cylinder to its volume is
Distribute to eliminate the parentheses in the numerator.
Combine like terms in the numerator.
Factor common terms in the numerator.
Divide out common terms in the numerator and denominator.
Simplify.
V = πR2 (R + 2)
2πR2 + 2πR(R + 2)
πR2 (R + 2)
2πR2 + 2πR2 + 4πR
πR2 (R + 2)
4πR2 + 4πR
πR2 (R + 2)
4πR(R + 1)
πR2 (R + 2)
4
π1
R1 (R + 1)
π1 R R (R + 2)
2
4(R + 1)
Answer
R(R + 2)
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4.4. Rational Expressions
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Review Questions
Simplify each rational expression to lowest terms.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
4
2x−8
x2 +2x
x
9x+3
12x+4
6x2 +2x
4x
x−2
x2 −4x+4
x2 −9
5x+15
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
2x2 +x+3
12
x2 +6x+1
23.
24.
25. 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?
26. 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?
27. A sphere with radius r has a volume of 43 πr3 and a surface area of 4πr2 . Find the ratio the surface area to the
volume of a sphere.
28. The side of a cube is increased by a factor of two. Find the ratio of the old volume to the new volume.
29. The radius of a sphere is decreased by four units. Find the ratio of the old volume to the new volume.
Review Answers
2
1. x−4
2. x + 2, x = 0
3. 34 , x = − 13
4. 3x+1
2 , x = 0
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5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
Chapter 4. Rational Equations and Functions
1
x−2
x−3
5 , x = −3
x+2
x , x = −4
2x
x+5
x+5
x−2 , x = −1
x−4
x−2 , x = −4
3x−6
2x−1 , x = −3
x
6 , x = −5, x =
4
x=0
x = −2
x=1
x = 2, x = −2
none
x = −4, x = 7
x = 0, x = −3
x = 0, x = −5, x = −6
x ≈ 1.19, x ≈ −4.19
x ≈ 1.54, x ≈ −0.87
none
x ≈ −0.17, x ≈ −5.83
5 Newtons
36 Newtons
3
r
1
8
r3
(r−4)3
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