Dividing Monomials and Simplifying Expressions with Negative Exponents
The objective of this lesson targets simplifying expressions with negative exponents. The student
should already understand how to use the quotient rule for exponents (
ar
 a r  s ). In this lesson, the
s
a
student will use the quotient rule for exponents to simplify a rational expression then simplify further by
writing the expression with positive exponents only.
Negative exponents have nothing to do with the sign of the base. Instead, negative exponents
indicate the position of the base. By position, we refer to the two positions of a ratio, the numerator and the
denominator. To simplify an expression with a negative exponent, simply move the base of the negative
exponent. For example, the expression 4-2 is equivalent to
1
1
, which, in turn, equals
:
2
16
4
4 2 
1
1

2
4
16
Negative exponents originate with the quotient rule for simplifying rational expressions. The
quotient rule states:
ar
 a r s
as
We can prove the quotient rule by expanding a rational expression and canceling common factors:
a 5 a  a  a  a  a a  a  a  a  a


 a  a  a  a3
2
a
aa
a  a
Using the quotient rule with the same expression:
a5
 a 5 2  a 3
a2
To illustrate how the quotient rule defines the meaning of a negative exponent, we will repeat the exercise
above using a rational expression where the exponent in the denominator is greater than the exponent in the
numerator:
a2
aa
a  a
1


 3
5
a
a  a  a  a  a a  a  a  a  a a
Using the quotient rule for the same expression:
a2
1
 a 25  a 3  3 .
5
a
a
We see from the preceding example that a negative exponent indicates that the base should change position
from the numerator to the denominator or visa versa. The following examples will further demonstrate.
#1
32 
1 1

32 9
#2
(ab 2 ) 1 
1
1
 2
2 1
(ab )
ab
#3
(3 xy  3 )  2  3  2  x  2  y 32  3  2  x  2  y 6 
1 1
y6
6


y

32 x 2
9x 2
#4
2 x  2 x 
4
1
2
3

1
1
1 1 6
1 x6
x6
x2
2
 3 2
2
6





2
x


2
x



x




1
2 x 4 
2 x 4 
2x 4 22
2 x 4 4 8x 4
8
#5
a 1
b2
2
3
2
3
2
2
b  a 1  b  a  b a  b 2  4 a 31  b  2 a 2  1  a 2  a
a 3
b 4 a 1 b 4 a 1b 4
b2
b2
b 4
a3
#5 (again)
a 1
1
4
1  4
2
2
b  a  b  a b  a 1 3 b  4  2  a 1 3 b  4  2  a 2 b  2  a 2  1  a
a 3 b  2 a 3 b  2 a 3
b2 b2
b 4
#6 (remember that a negative exponent indicates that the base should switch positions)
1
 
2
2 2 1
2
1 2 1
1
 2   2  
 2   OR
2
4 4 2
2
4 2
2
1
 
2
2
2
2