Worksheet 8
RATIONAL FUNCTIONS
Name:____________________________
Section: _______________________
Warm up Exercises
1.
2.
3
5
𝑎
𝑏
1
×6
1
2
𝑎
𝑐
3. 8 ÷ 3
𝑐
×𝑑
4. 𝑏 ÷ 𝑑
Rational Expressions
A rational expression is an expression that is the ratio of two polynomials. The general
𝑎 𝑥 𝑛 +𝑎
𝑥 𝑛−1 +⋯+𝑎 𝑥+𝑎0
form of these expressions is 𝑦 = 𝑏 𝑛𝑥 𝑚+𝑏𝑛−1 𝑥 𝑚−1 +⋯+𝑏1
𝑚
𝑚−1
1 𝑥+𝑏0
2𝑥 3 +8𝑥 2
. But you don’t have to worry
𝑥+8
𝑥 2 −5𝑥+6
about this form. Examples of rational expressions: 6𝑥 2 +42𝑥, 𝑥 2 −64, 𝑥 2 −7𝑥+12. As we shall
soon see, working with rational expressions is extremely similar to working with
fractions.
Simplifying Rational Expressions
Recall that one method of simplifying fractions is to prime factor the top and bottom and
cancel common factors:
12 2 × 2 × 3 2
=
=
42 2 × 3 × 7 7
This carries over exactly the same way to rational expressions:
2𝑥 3 + 8𝑥 2 2𝑥 2 (𝑥 + 4) 𝑥(𝑥 + 4)
=
=
6𝑥 2 + 42𝑥
6𝑥(𝑥 + 7)
3(𝑥 + 7)
Remark: It is often easier to leave it in factored form as opposed to multiplying out the
top and bottom.
Steps to simplifying rational expressions:
1. Factor both the numerator and the denominator.
2. Cancel out common factors
Exercise 1: Simplify the following expressions.
a)
𝑥−𝑦
𝑦−𝑥
𝑎+3
b) 3+𝑎
𝑥+8
c) 𝑥 2 −64
𝑥 2 −5𝑥+6
d) 𝑥 2 −7𝑥+12
Adding and Subtracting Rational Expressions
Recall Example 6 from Activity 1.
1 1
Compute
+ =_____.
4 7
Solution: The LCM for 4 and 7 is 28.
7 ´1 1´ 4 7 4 7 + 4 11
+
= + =
=
7 ´ 4 7´ 4 28 28 28 28
Just as with numerical fractions, the key concept to adding and subtracting rational
expressions is finding a common denominator. Moreover, we want the lowest common
denominator which will be the LCM of the denominators of the two rational expressions.
The following example will show you how the LCD is found for rational expressions, but
in general, the LCD is the product of all the factors of the denominators without repeating
common factors in the product.
3
5
Example 1: Find the LCD of 𝑥 2 −4 and 2𝑥−4.
3
3
5
5
Solution: Note that 𝑥 2 −4 = (𝑥−2)(𝑥+2) and 2𝑥−4 = 2(𝑥−2) which means that the
LCD of these two rational expressions is 2(𝑥 − 2)(𝑥 + 2)
Regardless, once a common denominator is found, we add the numerators just like
before.
3
5
Example 2: Compute 𝑥 2 −4 + 2𝑥−4.
Solution: In Example 1, we found that the LCD of these two rational expressions
is 2(𝑥 − 2)(𝑥 + 2). Therefore,
3
5
3
5
3
2
5
𝑥+2
+
=
+
=
(
)
+
(
)
2
(𝑥−2)(𝑥+2) 2(𝑥−2)
(𝑥−2)(𝑥+2) 2
𝑥 −4
2𝑥−4
2(𝑥−2) 𝑥+2
6
5(𝑥+2)
6
5𝑥+10
= 2(𝑥−2)(𝑥+2) + 2(𝑥−2)(𝑥+2)
= 2(𝑥−2)(𝑥+2) + 2(𝑥−2)(𝑥+2)
6+(5𝑥+10)
= 2(𝑥−2)(𝑥+2)
5x+16
= 2(x−2)(x+2)
Steps to adding and subtracting rational expressions:
1. Factor the denominators
2. Find LCD of the two rational expressions
3. Add/subtract
Exercise 2: Perform the following operators and simplify
a)
c)
6
𝑥
7
𝑥−1
+𝑥
3𝑡−2
3𝑡+2
𝑥+1
b) 𝑥+1 − 𝑥−1
3𝑡+2
− 3𝑡−2
𝑦
d)
𝑦
𝑦𝑥
𝑦−𝑥
+ (𝑦−𝑥)2
3
e) 𝑦 2 +5𝑦+4 + 𝑦 2 +11𝑦+10
Multiplying and Dividing Rational Expressions
Multiplication
Recall that when multiplying two numerical fractions, we simply multiply the
numerators and the denominators to find our new fraction. Recall Example 8 from
Activity 1:
3
5
Example 3: Compute 4 × 6.
Solution: As mentioned above, all we have to do is multiply the two numerators
3
5
3×5
15
15 ÷3
5
and denominators, 4 × 6 = 4×6 = 24 = 24÷3 = 8.
Therefore, this same technique holds for rational expressions. For rational expressions, it
helps to factor the top and bottom of both fractions before multiplying.
8𝑥 2 −4𝑥
Example 9: Evaluate 2𝑥 2 +5𝑥−3 ×
Solution:
8𝑥 2 −4𝑥
×
2𝑥 2 +5𝑥−3
𝑥 2 −9
4𝑥 2
𝑥 2 −9
4𝑥 2
4𝑥(2𝑥−1)
= (2𝑥−1)(𝑥+3) ×
(𝑥+3)(𝑥−3)
4𝑥 2
=
𝑥−3
𝑥
Steps to multiplying rational expressions
1. Factor the numerator and denominator of both expressions in the product.
2. Cancel common factors on top and bottom.
3. Multiply the numerators and denominators.
Division
Once again, division of rational expressions is identical to division of numerical
fractions. Recall Example 10 from Activity 1:
5
3
Example 4: Compute 6 ÷ 4.
Solution:
5
3
5
4
5×4
20
20÷2
÷ 4 = 6 × 3 = 6×3 = 18 = 18÷2 =
6
10
9
.
Then the following example is almost identical
Example 3: Evaluate
Solution:
𝑥 2 +4𝑥+3
𝑥−5
÷
𝑥 2 +4𝑥+3
𝑥−5
2𝑥 2 −𝑥−3
𝑥 2 −25
÷
=
=
2𝑥 2 −𝑥−3
𝑥 2 −25
𝑥 2 +4𝑥+3
𝑥 2 −25
× 2𝑥 2 −𝑥−3
𝑥−5
(𝑥+3)(𝑥+1)
=
=
(𝑥+5)(𝑥−5)
× (𝑥+1)(2𝑥−3)
𝑥−5
(𝑥+3)(𝑥+5)
2𝑥−3
𝑥 2 +8𝑥+15
2𝑥−3
Steps to dividing rational expressions:
1. Find the reciprocal of the divisor.
2. Change the operation from division to multiplication.
3. Multiply and simplify.
Exercise Set 3: Perform the following operations and simplify
a)
c)
e)
𝑥 2 −25
𝑥−2
𝑥 2 −4
× 𝑥 2 −7𝑥+10
3𝑥2 𝑦
𝑧4
2𝑥3
𝑦3 𝑧
x 2 + 4x + 4
reduced to lowest term
x 2 + 3x + 2
b)
d)
4𝑥 3 𝑦 2
𝑧3
×
9𝑥 2 −16
2𝑥+1
𝑦3𝑧4
2𝑥 5
÷
6𝑥 2 −11𝑥+4
4𝑥 2 +4𝑥+1
f)
x 2 - 8x +16 2x - 8
¸ 2
x -3
x -9
g) What is
h)
x 2 - 3x + 2
reduced to lowest term?
x 2 - 5x + 6
3
2
- 2
x - 6x x + 6x
2