The document provides examples and explanations for rational functions and operations involving rational functions such as addition, subtraction, multiplication, division, and simplifying complex fractions. Some key points include: - Rational functions are defined as the quotient of two polynomials where the denominator is not equal to 0. - To add or subtract rational functions, they are first brought to a common denominator by factoring the denominators to find the least common multiple (LCM). - To multiply rational functions, the numerators are multiplied and the denominators are multiplied. Common factors in the numerator and denominator can then be cancelled. - To divide rational functions, the division is changed to multiplication by the reciprocal of the denominator rational function.

1. 𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 7
Rational Functions
"There are 10 types of people in the
world: those who know binary, and
those who do not.“ – Anonymous -

2. Lehman College, Department of Mathematics
Rational Numbers (1 of 6)
Let us look at integer multiplication. For example, if:
then we say 2 and 3 are integer factors of 6. Now, we
introduce a new operation called integer division. If we
have the product 2 ⋅ 3 = 6, then we define the quotients:
Let us extend the concept to quotients, such as:
Since there are no integers 𝑚 and 𝑛, such that:
Then the above quotients are not integers.
2 ⋅ 3 = 6
6
3
= 2
6
2
= 3and
5
3
6
4
and
3 ⋅ 𝑛 = 5 4 ⋅ 𝑚 = 6and

3. Lehman College, Department of Mathematics
Rational Numbers (2 of 6)
Let 𝑝 and 𝑞 be integers with 𝑞 ≠ 0, then the quotient:
Is called a rational number. The adjective rational
comes from the word ratio, meaning quotient.
Why is division by zero not allowed? Suppose we have:
Where 𝑟 is a rational number. Then 𝑟 ⋅ 0 = 2, but the
product of any number with zero is zero, so no such
number 𝑟 can exist, and the qotient is thus undefined.
How about the quotient
0
0
? Suppose
0
0
= 𝑟, so 𝑟 ⋅ 0 = 0.
In this case, 𝑟 could be any real, and is thus undefined.
𝑝
𝑞
2
0
= 𝑟

4. Lehman College, Department of Mathematics
Rational Numbers (3 of 6)
How do we add or subtract rational numbers?
Example 1. Perform the following operation:
Solution.
2
3
+
1
2
Step 1: Find the least common multiple
(LCM) of the denominators:
LCM = 2 ⋅ 3 = 6
Step 2: Bring each fraction to a common denominator:
2
3
+
1
2
=
2
3
2
2
+
1
2
3
3
=
4 + 3
6
=
7
6

5. Lehman College, Department of Mathematics
Rational Numbers (3 of 6)
Example 2. Perform the following operation:
Solution.
3
10
−
1
15
Step 1: Factor the denominators to find
their least common multiple (LCM).
LCM = 2 ⋅ 3 ⋅ 5 = 30
Step 2: Bring each fraction to a common denominator:
3
10
3
3
−
1
15
2
2
=
9 − 2
30
=
7
30
3
10
−
1
15
=
3
10
−
1
15
=
3
2 ⋅ 5
−
1
3 ⋅ 5

6. Lehman College, Department of Mathematics
Rational Numbers (3 of 6)
Example 3. Perform the following operation. Here 𝑎, 𝑏,
𝑐, and 𝑑 are integers, with 𝑏 ≠ 0 and 𝑑 ≠ 0.
Solution.
𝑎
𝑏
+
𝑐
𝑑
Step 1: Find the product of the denominators:
𝑏𝑑
Step 2: Bring each fraction to a common denominator:
𝑎
𝑏
+
𝑐
𝑑
=
𝑎
𝑏
𝑑
𝑑
+
𝑐
𝑑
𝑏
𝑏
=
𝑎𝑑 + 𝑏𝑐
𝑏𝑑
Above is the general formula for the sum of two rational
numbers. Note the cross-multiplication.

7. Lehman College, Department of Mathematics
Rational Function Definition (1 of 1)
Let 𝑝(𝑥) and 𝑞(𝑥) be polynomials with 𝑞(𝑥) ≠ 0, then
the quotient:
is called a rational function. For example, the following
are rational functions:
How do we add or subtract rational functions?
Answer: Similar to adding and subtracting rational
numbers. Here the role of prime factors will be played
by linear polynomial factors (and irreducible quadratics).
𝑝(𝑥)
𝑞(𝑥)
1
𝑥
𝑥 + 1
𝑥2 + 3𝑥 − 4
5𝑥3 + 2𝑥 + 3
3𝑥2 + 2𝑥 − 1
(a) (b) (c)

8. Lehman College, Department of Mathematics
Sum of Rational Functions (1 of 2)
Example 3. Perform the following algebraic operation:
Solution.
1
𝑥
+
1
𝑥 + 1
Step 1: Factor the denominators to find
their least common multiple (LCM).
LCM = 𝑥(𝑥 + 1)
Step 2: Bring each fraction to a common denominator:
1
𝑥
𝑥 + 1
𝑥 + 1
+
1
𝑥 + 1
𝑥
𝑥
=
𝑥 + 1 + 𝑥
𝑥(𝑥 + 1)
=
2𝑥 + 1
𝑥(𝑥 + 1)
Since the denominators are linear polynomials, they are
already simplified. The LCM is given by their product:
1
𝑥
+
1
𝑥 + 1
=

9. Lehman College, Department of Mathematics
Difference of Rational Functions (1 of 2)
Example 4. Combine into a single fraction:
Solution.
𝑥 + 6
𝑥2 + 𝑥 − 20
−
3
𝑥 − 4
Step 1: Factor the denominators to find
their least common multiple (LCM).
LCM = (𝑥 − 4)(𝑥 + 5)
Step 2: Bring each fraction to a common denominator:
3
𝑥 − 4
𝑥 + 5
𝑥 + 5
=
𝑥 + 6 − 3(𝑥 + 5)
𝑥 − 4 𝑥 + 5
𝑥 + 6
𝑥2 + 𝑥 − 20
−
3
𝑥 − 4
=
𝑥 + 6
𝑥 − 4 𝑥 + 5
−
3
𝑥 − 4
𝑥 + 6
𝑥 − 4 𝑥 + 5
−

10. Lehman College, Department of Mathematics
Difference of Rational Functions (2 of 2)
Solution (cont’d). Step 2. Bring each fraction to a
common denominator :
3
𝑥 − 4
𝑥 + 5
𝑥 + 5
=
𝑥 + 6 − 3(𝑥 + 5)
𝑥 − 4 𝑥 + 5
𝑥 + 6
𝑥 − 4 𝑥 + 5
−
=
𝑥 + 6 − 3𝑥 − 15
𝑥 − 4 𝑥 + 5
=
−2𝑥 − 9
𝑥 − 4 𝑥 + 5
= −
2𝑥 + 9
𝑥2 + 𝑥 − 20

11. Lehman College, Department of Mathematics
Difference of Rational Functions (1 of 2)
Example 5. Combine into a single fraction:
Solution.
𝑥
𝑥2 − 𝑥 − 12
−
5
12𝑥 − 48
Step 1: Factor the denominators to find
their least common multiple (LCM).
LCM = 12 𝑥 − 4 𝑥 + 3
Step 2: Bring each fraction to a common denominator:
5
12 𝑥 − 4
𝑥 + 3
𝑥 + 3
=
12𝑥 − 5(𝑥 + 3)
12 𝑥 − 4 𝑥 + 3
𝑥
𝑥2 − 𝑥 − 12
−
5
12𝑥 − 48
=
𝑥
𝑥 − 4 𝑥 + 3
−
5
12(𝑥 − 4)
𝑥
𝑥 − 4 𝑥 + 3
12
12
−

12. Lehman College, Department of Mathematics
Difference of Rational Functions (2 of 2)
Solution (cont’d). Step 2. Bring each fraction to a
common denominator :
=
12𝑥 − 5𝑥 − 15
12 𝑥 − 4 𝑥 + 3
=
7𝑥 − 15
12 𝑥 − 4 𝑥 + 3
=
7𝑥 − 15
12 𝑥2 − 𝑥 − 12
5
12 𝑥 − 4
𝑥 + 3
𝑥 + 3
=
12𝑥 − 5(𝑥 + 3)
12 𝑥 − 4 𝑥 + 3
𝑥
𝑥 − 4 𝑥 + 3
12
12
−

13. Lehman College, Department of Mathematics
Product of Rational Functions (1 of 1)
Example 6. Find the product of the following rational
functions:
Step 1. Factor all polynomials in the product:
Step 2. Identify and cancel common factors:
Step 3. Distribute the numerator and denominator
𝑥2
+ 3𝑥 − 10
𝑥2 − 𝑥 − 6
⋅
𝑥2
− 2𝑥 − 3
𝑥2 + 4𝑥 − 5
𝑥2
+ 3𝑥 − 10
𝑥2 − 𝑥 − 6
⋅
𝑥2
− 2𝑥 − 3
𝑥2 + 4𝑥 − 5
=
=
𝑥 − 2 𝑥 + 5
𝑥 − 3 𝑥 + 2
⋅
𝑥 − 3 𝑥 + 1
𝑥 + 5 𝑥 − 1
=
𝑥 − 2 𝑥 + 5
𝑥 − 3 𝑥 + 2
⋅
𝑥 − 3 𝑥 + 1
𝑥 + 5 𝑥 − 1
=
𝑥 − 2 𝑥 + 1
𝑥 + 2 𝑥 − 1

14. Lehman College, Department of Mathematics
Quotient of Rational Functions (1 of 1)
Example 7. Find the quotient of the following rational
functions:
Step 1. Change division to multiplication by reciprocal:
Step 2. Factor all polynomials in the product :
Step 3. Identify and cancel common factors:
Step 4. Distribute the numerator and denominator. In
this particular case, no simplification is achieved.
𝑥2
+ 𝑥 − 6
10𝑥2
÷
𝑥2
− 9
2𝑥8
𝑥 − 2 𝑥 + 3
10𝑥2
⋅
2𝑥8
𝑥 − 3 𝑥 + 3
=
𝑥6
𝑥 − 2
5 𝑥 − 3
𝑥2
+ 𝑥 − 6
10𝑥2
÷
𝑥2
− 9
2𝑥8
=
𝑥2
+ 𝑥 − 6
10𝑥2
⋅
2𝑥8
𝑥2 − 9

15. Lehman College, Department of Mathematics
Simplifying Complex Fractions (1 of 8)
Example 8. Simplify the following complex fraction:
Solution 1. Evaluate the numerator and denominator
separately (determine LCM and combine fractions):
2
3
−
3
5
3
4
+
5
12
2
3
−
3
5
3
4
+
5
12
=
2
3
5
5
−
3
5
3
3
3
4
3
3
+
5
12
=
10 − 9
15
9 + 5
12
=
1
15
÷
14
12
=
1
15
⋅
12
14
=
2 ⋅ 2 ⋅ 3
3 ⋅ 5 ⋅ 2 ⋅ 7
=
2
35

16. Lehman College, Department of Mathematics
Simplifying Complex Fractions (2 of 8)
Solution 2. Evaluate the numerator and denominator
together (determine LCM of all denominators):
2
3
−
3
5
3
4
+
5
12
=
2 ⋅ 4 ⋅ 5 − 3 ⋅ 3 ⋅ 4
3 ⋅ 3 ⋅ 5 + 5 ⋅ 5
=
4 10 − 9
5 9 + 5
=
4
5 ⋅ 14
=
LCM = 3 ⋅ 4 ⋅ 5 = 60
2
3
−
3
5
3
4
+
5
12
⋅
3 ⋅ 4 ⋅ 5
3 ⋅ 4 ⋅ 5
=
2
35

17. Lehman College, Department of Mathematics
Simplifying Complex Fractions (3 of 8)
Example 8. Simplify the following complex fraction:
Solution 1. Evaluate the numerator and denominator
separately (determine LCM and combine fractions):
1 −
1
𝑥2
3
𝑥 + 1
+
5
𝑥
1 −
1
𝑥2
3
𝑥 + 1
+
5
𝑥
=
1
1
𝑥2
𝑥2 −
1
𝑥2
3
𝑥 + 1
𝑥
𝑥
+
5
𝑥
𝑥 + 1
𝑥 + 1
=
𝑥2
− 1
𝑥2
3𝑥 + 5 𝑥 + 1
𝑥 𝑥 + 1
=
𝑥2 − 1
𝑥2
÷
3𝑥 + 5𝑥 + 5
𝑥 𝑥 + 1
=
𝑥2 − 1
𝑥2
⋅
𝑥 𝑥 + 1
8𝑥 + 5

18. Lehman College, Department of Mathematics
Simplifying Complex Fractions (4 of 8)
Solution 1 (cont’d). From the previous slide:
1 −
4
𝑥2
3
𝑥 + 1
+
5
𝑥
=
𝑥2 − 1
𝑥2
÷
3𝑥 + 5𝑥 + 5
𝑥 𝑥 + 1
=
𝑥2 − 1
𝑥2
⋅
𝑥 𝑥 + 1
8𝑥 + 5
=
𝑥 − 1 𝑥 + 1
𝑥
⋅
𝑥 + 1
8𝑥 + 5
=
𝑥 − 1 𝑥 + 1 2
𝑥 8𝑥 + 5

19. Lehman College, Department of Mathematics
Simplifying Complex Fractions (5 of 8)
Example 9. Simplify the following complex fraction:
Solution 2. Evaluate the numerator and denominator
together (determine LCM of all denominators):
1
𝑥2 +
5
𝑥
+ 6
2𝑥 + 1
1
𝑥2 +
5
𝑥
+ 6
2𝑥 + 1
=
1
𝑥2 +
5
𝑥
+ 6
2𝑥 + 1
⋅
𝑥2
𝑥2
=
1
𝑥2 +
5
𝑥
+ 6 ⋅ 𝑥2
2𝑥 + 1 ⋅ 𝑥2
=
1 + 5𝑥 + 6𝑥2
𝑥2 2𝑥 + 1
=
2𝑥 + 1 3𝑥 + 1
𝑥2 2𝑥 + 1
=
3𝑥 + 1
𝑥2

20. Lehman College, Department of Mathematics
Simplifying Complex Fractions (6 of 8)
Example 10. In the previous slide, we factored:
Solution. The factoring number is:
Splitting the linear term yields:
1 + 5𝑥 + 6𝑥2
= 6𝑥2 + 5𝑥 + 1
6 ⋅ 1 = 6
Factor 1 Factor 2 Sum
1 6 7
2 3 5
6𝑥2 + 5𝑥 + 1 = 6𝑥2
+ 2𝑥 + 3𝑥 + 1
= 2𝑥 3𝑥 + 1 + 1 ⋅ 3𝑥 + 1
= 2𝑥 + 1 3𝑥 + 1

21. Lehman College, Department of Mathematics
Simplifying Complex Fractions (7 of 8)
Example 11. Simplify the following complex fraction:
Solution 2. Evaluate the numerator and denominator
together (determine LCM of all denominators):
4
𝑥 − 6
−
4
𝑥 + 6
8
𝑥2 − 36
4
𝑥 − 6
−
4
𝑥 + 6
8
𝑥2 − 36
=
4
𝑥 − 6
−
4
𝑥 + 6
8
𝑥 − 6 𝑥 + 6
=
4
𝑥 − 6
−
4
𝑥 + 6
⋅ 𝑥 − 6 𝑥 + 6
8
𝑥 − 6 𝑥 + 6
⋅ 𝑥 − 6 𝑥 + 6
=
4 𝑥 + 6 − 4 𝑥 − 6
8
4 𝑥 + 6 − 4 𝑥 − 6
8
4 𝑥 + 6 − 4 𝑥 − 6
8
4 𝑥 + 6 − 4 𝑥 − 6
8

22. Lehman College, Department of Mathematics
Simplifying Complex Fractions (8 of 8)
Solution 2 (cont’d). From the previous slide:
4 𝑥 + 6 − 4 𝑥 − 6
8
4
𝑥 − 6
−
4
𝑥 + 6
8
𝑥2 − 36
=
=
4𝑥 + 24 − 4𝑥 + 24
8
=
4𝑥 + 24 − 4𝑥 + 24
8
=
4𝑥 + 24 − 4𝑥 + 24
8
=
4𝑥 + 24 − 4𝑥 + 24
8
=
48
8
= 6