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M098
Carson Elementary and Intermediate Algebra 3e
Section 5.6
Objectives
1.
2.
3.
4.
Divide exponential forms with the same base.
Divide monomials
Divide a polynomial by a monomial (Omit long division 5.6.5)
Simplify expressions using rules of exponents.
Vocabulary
Prior Knowledge
Exponent Rules
0
If a is a real number except 0, then a = 1.
If a is a real number except 0 and n is a natural number, then a  n 
If a is a real number except 0 and n is a natural number, then
1
a n
1
an
.
 an .
a
If a and b are real numbers except 0 and n is a natural number, then  
b
If a is a real number and m and n are integers, then am  an  am n .
n
n
b
  .
a
n
If a is a real number and m and n are integers, then  am   amn .


n n
n
If a and b are real numbers and n is an integer, then ab  a b .
New Concepts
Extra Credit: Scientific Notation
Skip 5.6.5 – Use long division to divide polynomials
am
If m and n are integers and a is any real number except 0, then am  an 
 am  n .
n
a
Always leave answers with positive exponents.
Example 1:
a.
u5  u3  u 5  3  u 2
b.
b.
1
j 6  j14  j 6 14  j  8 
j8
c.
V. Zabrocki 2011
m 6
1
 m 6 3  m9 
3
m
m9
d.
x5y3
 x3y
2
2
x y
e.
4 5 xy 4
 4 3 xy  64xy
2
3
4 y
f.
28p 5 q 4r
2p 3r

42p 2 q7
3q3
page 1
M098
Carson Elementary and Intermediate Algebra 3e
If a, b, and c are real numbers, variables, or expressions with c ≠ 0, then
Section 5.6
ab
a b
  .
c
c c
Example 2:
a.
16hk 4  28hk  4h 2k 2
16hk 4
28hk
4h 2k 2
7



 4k 2   h
k
4hk 2
4hk 2
4hk 2
4hk 2
b.
30x 3 yz 5  15xyz 2
 6x 2 z 3  3
2
 5xyz
Combination of exponent rules.
1.
2.
3.
4.
Use the power rule to remove parentheses.
Move powers so all exponents are positive.
Simplify numerator and denominator using the product rule.
Use the quotient rule to obtain final result. Always leave the answer with positive exponents.
Example 3:
a.
b.
 2x  2 y 


 1 
 z

c.
d.
e.
V. Zabrocki 2011
3
 p 5

 3
 p





p15
 p15p 9  p 24

9
p
4

24 x 8 y 4
24 y 4z 4
16y 4 z 4


z4
x8
x8
5
 y 3 
y 15
y8
y8
1
 




y 6 y 8
y 6 y 8
y15 y 6
y 21
y13
3m 6n 1p 2
3m 6m1n3
3m7n3
3m7n 2



m 1n  3p
n1p 2p
n1p 3
p3
 3y 4 


2
 6y 7 


3

32 y8
9y 8
1


3
21
21
6 y
216y
24y13
page 2
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