Download 5-2 Dividing Monomials

5-2 Dividing Monomials
Doesn’t it seem logical, that if you add exponents when multiplying monomials, you
would subtract exponents when dividing monomials.
For instance,
Simplify
x 11
.
x5
Subtract the exponent on bottom from the
one on top.
x6
The property of dividing powers is:
For any real number a, except a=0, and intergers m and n,
am
 a mn .
an
Another property of exponents is that of negative exponents. This states:
For any real number a, except, a=0, and any integer n, a  n 
1
1
and  n  a n .
n
a
a
This rule can be used to write exceptionally small numbers in scientific notation.
.0064 in scientific notation would be 6.4x10-3.
Whenever you are asked to simplify an expression, check to make sure all of the
following are true:
Are all exponents in positive form? (i.e. if you have a negative
exponent express it as its inverse).
Have you cleared all parentheses?
Is each base reduced to only one appearance?
Are all fractions in their simplest forms?
Below are some examples:
Simplify 3 5  3 2
3 5  3 2
Reduce the exponents as
much as possible
Clear the negative exponent
 5  2  3
3 3
3 3 
1
1

3
27
3
Solve for all constants.
(rs 2 ) 3 (r 3 s ) 2
r 2 s 4 t 2
(rs 2 ) 3 (r 3 s ) 2
r 2 s 4 t 2
(r 3 s 6 )( r 6 s 2 )
r 2 s 4 t 2
There are many ways to start, I would start
by distributing the exponents.
(r 3 s 6 )( r 6 s 2 )
r 2 s 4 t 2
r3  r 6  s6  s2
r 2 s 4 t 2
r 9s8
r 2 s 4 t 2
r 9s8
r 2 s 4 t 2
r 9  2 s 8 4
t 2
r7s4
t 2
r 7 s 4t 2
Now work the top and bottom of the
fraction.
Simplify
Simplify
Then combine the exponents across the top.
Finally clear any negative exponents by
using their inverse.
You have now simplified this as much as
possible.
5 ( 2n 4)
5 ( 2 n2)
(2n  4)  (2n  2)
Notice that we have the same base, i.e. 5.
As such, we can subtract the divided
exponents.
Apply this to our original base.
2n  4  2n  2
6
56 or15,625
2
Simplify  
5
2
 
5
2
5
 
2
2
Because it is such a large number, it can be
left in the 56 form.
2
In this case, remember that a negative
exponent is equal to the inverse. You
should start by clearing the negative.
Notice that the fraction has been inverted,
and the negative has been removed.
However, you still need to distribute that
exponent.
5
 
2
52
22
25
4
There you have your solution.
2
This leads to our final property for this lesson:
For any nonzero real numbers a and b, and integer n,
n
an
a
a
   n and  
b
b
b
n
n
bn
b
   or n .
a
a