Here are the steps to evaluate (8)2 in either order: 1) Evaluate the exponent first: (8)2 = 8 * 8 = 64 2) Evaluate the base first: 8 = 23 (23)2 = (2 * 2 * 2 * 2 * 2 * 2 * 2 * 2)2 = 25 * 25 = 625 So evaluating the exponent first yields 64, while evaluating the base first yields 625. In general, the order of operations matters for expressions involving exponents. Exponents should be evaluated before other operations like multiplication or division.

1. Radicals and
Rational
Exponents
19

2. RADICALS
A perfect square is the square of a natural number. 1, 4,
9, 16, 25, and 36 are the first six perfect squares.
A perfect cube is the cube of a natural number. 1, 8, 27,
64, 125, and 216 are the first six perfect cubes.
Perfect Powers
This idea can be expanded to perfect powers of a variable
for any radicand.
The radicand xn is a perfect power when n is a multiple of
the index of the radicand.
A quick way to determine if a radicand xn is a perfect
power for an index is to determine if the exponent n is
divisible by the index of the radical.
Since the exponent, 20, is divisible by the index, 5,
Example: 5
x 20
x20 is a perfect fifth power.

3. For all real numbers a and b and all rational numbers m For all real numbers a and b and all rational numbers m
and n, and n,
Product rule: am • an = am + n Zero exponent rule: a0 = 1, a 0
a m
Raising a power to a power:
a   a
m n m n
 a mn , a  0
an
Quotient rule:
Raising a product to a power : abm  a m b m
m
m 1 a am
a  m, a0    m , b0
Negative exponent rule: a Raising a quotient to a power : b b
Rational

4. Product Rule for Radicals
For nonnegative real numbers a and b,
n
a  n b  n ab
Example 1:
3
32  3 8 3 4  23 4
3
 250  3  125 3 2  53 2
4
48  4
16 4 3  24 3

5. Quotient Rule for Radicals
For nonnegative real numbers a and b,
n
a n a
n
 , b0
b b
Example 2:
81 81 9
 
100 100 10
75 25
 Simplify radicand, if possible.  25  5
3 1

6. Quotient Rule for Radicals
Example 3:
64 x 6 3
64 x 6 4x2
3
12
  4
y 3
y12 y
64 x 5 64 x 5 32 x 2 16 x 2 2
 3
   4x 2
2x 3 2x 1 1
3a 6b5 3a 8 4
3a 8 4
a8 4 3 a2 4 3
4
 2 13
4 8
  
16a b 16b 4
16b8 4
16b8 2b 2

7. Rules of Exponents
Example 4:
1.) Simplify x-1/2x-2/5.
1 2 2 5 -1 2   2 5  5 10   4 10  9 10 1
x x x x x 
x 9 10
2.) Simplify (y-4/5)1/3.
y  4 5 1 3
 y  4 5 1 3  y  4 15 
1
y 4 15
3.) Multiply –3a-4/9(2a1/9 – a2).
6
 6a 4 9 1 9   3a 4 9  2  6a 3 9  3a14 9    3a14 9
a1 3

8. Factoring Expressions
Example 5:
1.) Factor x1/4 – x5/4. The smallest of the two exponents is 1/4.
x1/4 – x5/4 = x1/4 (1 – x5/4-(1/4)) = x1/4 (1 – x4/4) = x1/4 (1 – x)
Original Exponent
exponent factored out
2.) Factor x-1/2 + x1/2. The smallest of the two exponents is -1/2.
-1/2 + x1/2 = x -1/2 (1– x1/2-(-1/2) ) = x -1/2 (1– x) =
1 x
x
x1 2
Original Exponent
exponent factored out

9. SEATWORK HOMEWORK
1 1
2 2
3 3
4 4
5 5

10. SEATWORK #1 :
Like radicals are radicals having the same
radicands. They are added the same way
like terms are added.
5 2 4 2 9 2
4 4 4
Answer:
54 2  44 2  94 2
3 xyz  10 xyz  5 xyz  8 xyz
2 2 2 2
65 7  75 6 Cannot be simplified further.

11. SEATWORK #2 : Adding & Subtracting
To Add or Subtract Radicals
1. Simplify each radical expression.
2. Combine like radicals (if there are any).
3 250  5 160 3 y 4 48x5  x4 3x5 y 4
Answer:
3 250  5 160  3  25 10  5  16 10 
3  5 10  5  4 10  15 10  20 10  35 10
3 y 4 48x5  x4 3x5 y 4  3 y 4 16x 4 4 3x  x4 x 4 y 4 4 3x 
3 y  2 x 4 3 x  x  xy 4 3 x  6 xy 4 3 x  x 2 y 4 3x 
 4 3 x (6 xy  x 2 y )

12. SEATWORK #3 : Multiplying Radicals
Multiply: 3( 5  x )
Answer:
3 5 3 x
(8  5 )( 6  2 )  Use the FOIL method.
48  8 2  6 5  10
( 3  6)( 3  6)   3  6
2
3  6 3  36 
3  36  33 Notice that the inner and outer terms cancel.

13. SEATWORK #4 : Rationalizing Denominators
To Rationalize a Denominator
Multiply both the numerator and the denominator of the fraction
by a radical that will result in the radicand in the denominator
becoming a perfect power.
x2 5 pq 4
y3 2r
Answer :
x2 x2 y3 x 2 y 3 xy y x y
3
   3
 3
 2
y y 3
y 3 y y y
5 pq 4 5 pq 4 2r 10 pq 4 r 10 pq 4 r q 2 10 pr
    
2r 2r 2r 2r 2r 2r

14. SEATWORK #5 : Simplifying Radicals
Simplify by rationalizing the denominator:
5 c  2d
2 1 c d
Answer :
5 5 2  1 5( 2  1)
  
2 1 2  1 2 1 2 1
c  2d c  2d c d
  
c d c d c d
( c  2d )( c  d ) c  cd  2cd  d 2

( c  d )( c  d ) cd

15. HOMEWORK #1 : Simplifying Radicals
1 3 8 100 6
( r  3) 5
Simplify:  2 4
3 3 3 6 3
( r  3)5
Answer:
1 3 1 3 3 3 3 2 3
      
3 3 3 3 3 3 3 3
8 100 8 3 100 6 24 600 4 6 100 6
2 4 2  4   2 4  4 4 
3 6 3 3 6 6 3 6 6 6
2 6 10 6 8 6 40 6 32 6 16 6
4 4    
6 6 6 6 6 3
6
( r  3)5 ( r  3)5 6 ( 5 6 ) ( 5 3 )
  ( r  3)  ( r  3)( 5 6 )(10 6 )  ( r  3) 5 6 
( r  3)5 ( r  3)
53
3
1
(r  3) 5 6

16. HOMEWORK #2 :
8
2
Evaluating in Either Order
3
 8
Answer:
8  2   4
2 2
 3 2
3
or
8
2
3  8  64  4
3 2 3
16

17. HOMEWORK #3 :
8
2
Evaluating a-m/n 
3
Answer:
8 1 1 1 1
2

   2 
 8 2 4
3
8
2 2
3
3

18. HOMEWORK #4 :
y   
1 1 1

Simplify: 6 6  a b ab2 3
 
 
Answer:
y 
1
6 6  6
y6  y
 1 1   1 1  1 1
 a 2b 3

ab  

 a 2b 3
 
 a b
 
   
1 1
1  1
a 2
b 3
3 2
a b 2 3 18

19. HOMEWORK #5 :
Exponent 1/n When n Is Odd
1
1
 1 
 27
1 5
27 3 3  
 32 
Answer: 1
27 3
 3
27  3
 27 
1
3  3
 27  3
1
 1  5 1 1
   5 
 32  32 2 19

20. Submitted
by:
John Marion G. De Guzman
2nd year BTTE AUTOMOTIVE
ALGEBRA 111