Download WS Getting Warmed Up Key

Getting Warmed Up Worksheet KEY
Name ______________________
Worksheet Getting Warmed Up
Practice with Powers
Directions: Compute and simplify each expression completely. Show work using the correct mathematical or
algebraic methods.) DO NOT USE A CALCULATOR!!!! You must get rid of negative exponents in
your final answers!!
2
1


4
1. (81)  = (3)2


4.
 5x
3
7. 32

3
5
3
= 2 
1
8
 72a3b3 

5 3 
 2a b 

1
1
2
3
3
=
5 x 5 y 9
5.
15 x 6 y
5
 15x 5 z 3 
10. 
4
4 
 25x yz 
12.
=9
y 2 z 2  2 x 1 y 2 z 7  =
10z
x4

 4b 4 

2. 
 8a 
3
1
2
=
2x 3
y
2
=  36a b
 =
13.
=


2 4
3
1
81x 4 y 8
=
2
125x5 y15
1
ab 3
6
 3xy 
2 5
6. 2 x 5 x y
 3x 9 z 7 
5y
5 x9 yz 7
  9 7 
= 
3
3x z
 5y 
1
6  2
3.
y8
= 3x
 4x y 
8. 
2 5 
 x y 
8
8a 3
 2a 

 4
b12
b 
 2a
9. 256

3
4
3
= 4 
1
64
2
6 9 12
11.  64x y z  3 =
16 x 4 y 6 z 8 
  6a
4 2 3
b
=
2 3 2
b
16 y 6
x4 z8
a12 36b6
g
8b6 a 4
=

9a 8
2
VIP, “Very Important Problem(s)” Extending some ideas:
14.
e x ge x
=
e2x
x
x
15. 3 g3
=
30  1
16.
ex 1 gex 1 = e2 x 1
2x2
or e
anything to the 0 power = 1
17. Since  ab   a 2b 2 , some people believe that the property extends to  a  b   a 2  b 2 . Show that
this assumption is incorrect.
2
 ab    ab  ab   a 2b2 using the definition of squaring and the product of powers property while
2
a  b
2
2
  a  b  a  b   a 2  2ab  b 2 There is no power of a sum property! Also, no distributive
property for exponents!! Simply expand.
S. Stirling 2015
Page 1 of 10
Getting Warmed Up Worksheet KEY
Name ______________________
Radicals On #1 – 15, compute and completely simplify each expression completely.
1.
43
12


75


2. 3 8  5 24 
3
6.
3
53 3
2
3  375 3
3
25  3 30
3
3
3
4
5 5 5 6 5 6
3

25
7.
4

4

5 3
3
3
8.
2
10.
8 2  3  15 3  5 2
3 2  12 2
62 6 2 6 4
3 2  14 3
3 2 2 6
10  4 6
13.
VIP On #16 – 18, multiply
by the conjugate so that you
may evaluate each
expression. Follow the
example below.

34

32

3 2 3  4 3 8
5  2 3
16. Evaluate at x = 4.
Ex:
Evaluate
x 9
at x = 9.
3 x
99 0
 You can’t.
3 9 0
So multiply by the conjugate
3  x 
3  x  3  x 
 x  9 3  x 
=
 x  9

9 x
=  x  9 3  x 
  x  9
= 3  x
Now evaluate. 3  9  6
Neat trick huh?!
S. Stirling 2015
1
x5 3
At x = 4:
1
1

45 3 6

2
12.
6
6  18
2

2
15.
18
2
 3 2 3
x
x2  x
17. Evaluate at x = 0.
18. Evaluate at x = 0.
x2  2
x

x 1
12  3 2  4 2  2
10  2
x2  2
x
x22
x 5 3 x 5 3
x4
x5 3
x 59
x5 3
62
14.  3  2  4  2 
x5 3
x4
 x  4 

4
6  18
3
6 3 2

 2 2
3
3
32
32 4
 16  2
2
11.
3
16 3
9.  4 8  3   5 27  5 2 
2 3  12
48
4.
6
15 8 8 3  120 3
3 25  30
5.  5 3 
50
3.
x

x2 2
x
x4 2


x  x  4  2
x2  2
x2 2
x

x4 2


x  4  2
x42
x44
1
x2  2
x42
At x = 0:
1
1

02  2 2 2
No need to rationalize.
At x = 0:
04 2  4
Page 2 of 10
Getting Warmed Up Worksheet KEY
Name ______________________
Polynomials, Factoring and Zeros of Functions
No decimal answers!! Use the factoring
flowchart on page 5 of “General Concepts” for help.
1. Write in standard form  x  4 
2
= (x – 4) (x – 4)
= x2 – 8x + 16
8. Write in standard form  x  3
3
= (x + 3) (x2 + 6x + 9)
= x3 + 6x2 + 9x + 3x2 +18x + 27
= x3 + 9x2 + 27x + 27
2
2. Factor 40 x  80  5 x
= –5(x2 + 8x + 16) always take out GCF 1st!
= –5(x + 4)2
3
2
4
3. Factor 13x  2 x  15 x
= –x2(15x2 – 13x – 2)
= –x2(x – 1)(15x + 2)
3
2
9. Factor 8 x  18 x  4 x  9
= 2x(4x2 – 9) + 1(4x2 – 9)
= (2x + 1)(4x2 – 9)
= (2x + 1)(2x – 3)(2x + 3)
2
2
10. Factor x y  25 y  3x  75
 x y  25 y    3x  75
y  x  25   3  x  25 
2
3
4. Find zeros of f ( x)  x  27
= (x + 3)(x2 – 3x + 9)
x
x = – 3 or
2
2
2
 y  3 x  5 x  5
3  (3)  4(1)(9)
2(1)
2
3  27 So only 1 real solution.
x
2
4
3
2
11. Factor f ( x)  2 x  11x  x  23x  15
Must use the Rational Root Theorem.
2 11
1
(2x – 5) (4x2 + 10x + 25)
2 9
2
3
5. Factor completely 8 x  125
= –2(4x2 – 20x + 25) recognize perfect squares!
= –2(2x – 5) 2
9
23 15
8
15
8 15
0
( x  1)(2 x3  9 x 2  8x  15)
2
2
6. Factor 8 x  40 x  50
1
1
2
9
8
15
2
7
15
7
15
0
( x  1)2 (2 x 2  7 x  15)
( x  1)2 (2 x  3)( x  5)
3
2
7. Factor 16 x  56 x  49 x
= –x(16x2 – 56x + 49)
KNOW long and synthetic division WELL!
recognize perfect squares!
= –x(4x – 7)2
S. Stirling 2015
Page 3 of 10
Getting Warmed Up Worksheet KEY
Long Division and Synthetic Division
1.
Divide  x3  3x 2  5    x  3
Name ______________________
Practice both techniques on each of the following.
2. Divide  18 x  3x3  9  6 x 2    3  3x 
x 2  3x  3
3 x  3 3 x 3  6 x 2  18 x  9
x2
x  3 x3  3x 2  5
3 x 3  3 x 2
 x3  3x 2
9x 2  18 x
5
x2 
Write the answer:
5
x3
1
1
3
0
5
3
0
0
0
0
5
 9x  9
9x  9
0
x  3  0 , x  3 now
For synthetic, find the root:
3
 9 x2  9 x
For synthetic, find root:
1
3
3
5
Answer: x 
x3
3x  3  0 , x  1 now
6
18
9
3
9
9
9
9
0
2
So:
3 x 2  9 x  9 , factored 3  x 2  3 x  3
But the threes would divide out, so answer
3. Divide  x3  20    x  3
4. Rewrite
x 2  3x  9
x  3 x 3  20
 x  3x
3
x2
in an equivalent form.
x 1
x 1
x 1 x2
2
3 x 2  20
 x2  x
 3x2  9 x
x
9 x  20
 x 1
 9 x  27
1
7
Answer:
x 2  3x  9 
Answer:
7
x 3
3
1
1
1
0
0
20
3
9
27
3
9
7
S. Stirling 2015
x 1
1
x 1
With synthetic:
With synthetic:
1 0 0
1 1
same answer!
x 2  3x  3 .
same answer!
1 1 1
Page 4 of 10
Getting Warmed Up Worksheet KEY
Name ______________________
Simplify Rational Expressions
I. Simplify completely, if possible.
x 2  3x
x
1.
x  x  3
x
x3
or
2.
x
x  x  1
x 2 3x

x
x
x3
1
x 1
This IS the distributive property of
division over subtraction.
 5x
x 2  8 x  15 x  1

x 5
x3
5.
 7 x   x  5 x  7 x
3
x2  2 x  3
x2  6 x  9
x 2  2 x  3 ( x  3)( x  1)
=
=
x 2  6 x  9 ( x  3)( x  3)
( x  1)
( x  3)
3.
This is NOT the distributive property!
4 x  5 x 2  3x
4.
1
x
2
x
2
x x
=
2
( x  3)( x  5) x  1

= x 1
x5
x3
or in factored form
x3  x 2  5 x  3
6.
x3  3x  2
3
2
Num. x  x  5 x  3
1
 x  5x  7 
2
3 x
2
x  2x  8
( x  3)
=
( x  2)( x  4)
3
1
2
3
1 2 3
0
 x  1  x 2  2 x  3
 x 1 x 1 x  3
This IS the distributive property of
division over subtraction.
7.
1 1 5
x2  2 x  1
=
x3  1
( x  1)( x  1)
=
( x  1)( x 2  x  1)
( x  1)
2
( x  x  1)
8.
3
Denom. x  3 x  2
1 0 3
1
2
1
2
1 1 2
0
1
 x  1  x 2  x  2 
 x 1 x 1 x  2
So
3x 2 3x 2

9.
x 1 x 1
=
3x3  3x 2  3x3  3x 2
=
( x  1)( x  1)
6 x2
x2 1
S. Stirling 2015
10.
x3  x
x 1
x  x  1 x  1
 x  1
x3  x 2  5 x  3
x3  3x  2
x3
( x  1)( x  3)( x  1)
=
=
( x  1)( x  2)( x  1) x  2
x  x  1
Page 5 of 10
Getting Warmed Up Worksheet KEY
4a 2b 8b3t 2

11.
3st 15a
=
12.
Name ______________________
k
k2

x2  4 x  2
13.
k
x2
=
=
( x  2)( x  2) k 2
4a 2 b 15a
5a 3

=
3st 8b3t 2
2b 2 st 3
=
=
7
2
 3
2
5x
3x
21x  10
15 x 3
=
x
x  10
5
x
2  x  10 
2
16.
=
=
6x  3
3
2
3(2
x

1)
6
x
x
4
1

x  4x x  4
15.
2
4
x

=
x( x  4) x( x  4)
x4
x( x  4)
17.
2x  3
3

2
x  25 5  x
2x  3
3( x  5)

=
( x  5)( x  5) ( x  5)( x  5)
2 x  3  3x  15
=  x  12
( x  5)( x  5)
( x  5)( x  5)
19.
=
4
  x  5
6
1

3  x  5 3  x  5
(4n  1)(n  5) (n  5)(2n  4)
=

(n  5)(n  5)
(n  5)(n  5)
4n 2  21n  5  2n 2  6n  20
=
(n  5)(n  5)
2n 2  27 n  25
(n  5)(n  5)
4 x
( x  4)
8
10

=
2
x  16
10
 x  4 x  4 
8
S. Stirling 2015
=
20.
4
  x  5
3  x  5  12
4



5
  x  5
5
5
3  x  5
x  x  10 
6(2 x  1)
4n  1 2n  4

n5
n5
4
5 x
2
1

5  x 3x  15
=

18.
2x  3
3

=
=
( x  5)( x  5) ( x  5)
( x  6)( x  5)
=
4( x  5)( x  5)
( x  6)
4( x  5)
1
k ( x  2)
14.
x 2  11x  30
100  4 x 2
=
1
4
4


5 ( x  4) 5 x  20
Page 6 of 10
Getting Warmed Up Worksheet KEY
Name ______________________
II. Simplify completely, if possible.
4  2( x  3)
4
2

x 2  9 x  3  x  3 x  3
1.
1
1 = x 3 x 3

 x  3 x  3
x 3 x 3
=
 x  3 x  3
4  2x  6

2x
 x  3 x  3
2 x  10 2  x  5 x  5



2x
2x
x
VIP!
3a.
“Very Important Procedure”
3b.
5.
1
2
 4  x2   4x 2  x
1

5
2  x  1
 x  1
x
 1
2

2
2
 x2  2x
x
2
 1
2
x2  2 x  2
x
 1
2
2
4. Simplify by factoring: 2 x

4
1  12
x
4  5x 2
2
1
4  5x2 
1 
2x 2
4  5x2

1
1
1
 x 3 16  x 2  2  x 16  x 2  2
2
1
1
1
x 16  x 2  2   x 2  2 16  x 2  


2
x
 3 x 2  32 
1 
2 16  x 2  2
S. Stirling 2015
 x  1
2x2  4x  2
2
Simplify by factoring
1
1
1 21

x  16  x 2  2  2 x    x 16  x 2  2
2 2

2 16  x 2
2x2  2x  x2

2
3
1  12
x
4 x 2  6x2
2
3
3
4
1  12
1 1
x
4 x 2  x 2 6 x 2  2 x  3x 2
2
2
 x  3 x 2  32 
2
See page 8 of the “General Concepts” notes.
Multiply (use fractional exponents)
x  4  x2 
x
 x  1 2 x  x 2
2
 x  1
2
2.
2x
6.
1
2

2
Simplify by factoring
1
 1 1 
x 1  x  x 2 
2


1
2
1


 x 1

x 1
1

2


 2  x 12 x 12  1 



x 1
2
 2  x  x 12 

 2  x  x

3
3
2 x 1
2 x 1




Page 7 of 10
5 3
 x 2
2
Getting Warmed Up Worksheet KEY
Solving Equations
Name ______________________
No decimal answers!! Simplify all answers completely! Exclude imaginary roots!
1. Solve 5 x  3   x
5. Solve by completing the square
2
x 2  10 x  13
5x2  x  3  0
x 2  10 x  25  13  25
1  12  4(5)(3)
x
2(5)
x
 x  5
2
 12
x  5   12
1  61
10
x  5 2 3
2. Solve x3  5 x  3x 2  15
x
3
 3x 2    5 x  15   0
x 2  x  3  5  x  3  0
x
2
 5   x  3  0
x   5 or x  3
WRONG!
x  x 2  5  3  x 2  5
x3
You lost two roots!
3. Solve 0  x 5  x 3  20 x
0  x  x 4  x 2  20 
0  x  x 2  5  x 2  4 
x0
0  x2  4
0  x2  5
x  2
x  i 5
5x
10

x  3 2x  6
5x
10
6( x  3) 
( x  3) 
( x  3)
( x  3)
2( x  3)
6. Solve 6 
6 x  18  5x  5
11x  23
23
x
 2.09
11
Eliminate the imaginary. Calculus only deals with
real numbers.
4. Solve
2
1

x2  x x 1
x2 – x = 2x – 2 Used cross multiplication.
x2 – 3x + 2 = 0
(x – 2)(x – 1) = 0
x = 2 or x = 1
only 2, since 1 is not in the domain.
S. Stirling 2015
Page 8 of 10
Getting Warmed Up Worksheet KEY
2 x  6  x  3
7. Solve
2 x  6   x  3 
Name ______________________
9.
2
2
Don’t divide out the (x + 3)! But can divide out
the 2.
x  x 2  6 x  9   2( x  3)
2 x  6  x 2  6 x  9
0  x 2  8 x  15
x3  6 x 2  9 x  2 x  6
0   x  3 x  5 
x3  6 x 2  7 x  6  0
x = –3 or –5
Check because could gain a root
2  3  6   3  3
Solve 2 x  x  3  4( x  3)
Now rational root theorem:
2  5  6   5  3
00
3
1
2  2
So x = –3 is the only solution.
1
7
6
3 9
6
2
0
6
3
( x  3)( x 2  3x  2)  0
8. Solve
0  a bc
7  4a  2b  c
10  4a  b
3  32  4(1)(2)
x  3 or x 
2(1)
x
3  17
2
Used the elimination method, but could have
used substitution.
7  4a  2b  c
0  a  b  c
7  3a  b
10  4a  b
7  3a  b
3a
10  4  3  b
2  b
0  3 2 c
1  c
3
2
10. Solve 6 x  60  36 x  10 x
Again, need to get = 0.
6x3 – 36x2 + 10x – 60 = 0
2 [3x3 – 18x2 + 5x – 30] = 0
2 [3x2(x – 6) + 5(x – 6)] = 0
2 [(3x2 + 5)(x – 6)] = 0
2(3x2 + 5)(x – 6) = 0
So x   
5
and x = 6
3
Square root of negative extraneous. Eliminate
the imaginary. Calculus only deals with real
numbers.
Only solution is x = 6.
a = 3, b = –2, c = –1
S. Stirling 2015
Page 9 of 10
Getting Warmed Up Worksheet KEY
Name ______________________
Some Miscellaneous Things to Think About
1. Find
f ( x  x)  f ( x)
, if f ( x)  3x  2 .
x
2. Find
Use composition
3( x  x)  2   3 x  2 
x
3 x  3x  2  3 x  2
x
3x
3
x
f ( x  h)  f ( x )
, if f ( x)  x 2  4 x .
h
Use composition
( x  h) 2  4( x  h)   x 2  4 x 
h
x  2 xh  h  4 x  4h  x 2  4 x
h
2
2 xh  h  4h
h
h  2x  h  4
 2x  h  4
h
2x  h  4
2
2
On # 3 – 6, you may want to simplify the expression before evaluating!
3. Evaluate for a = 0.
b  a 
2
b
4.
2
a
b 2  2ab  a 2  b 2
a
a  2b  a 
a
2b  a

Evaluate for x = 4.
x 2
x4

x 2
x 2


x 2
1
x 2
Now if x = 4
1
1

42 4
Now if a = 0
2b  0  2b

5. Evaluate for x = 0
1
1

x4 4
x
4   x  4
4  x  4
x
x
1
4  x  4 x
1
4  x  4
So
1
1

4  0  4
16
6. Evaluate for x = 0. Hint: multiply numerator and denominator by the conjugate.
x  x  1  x  1
1
x
x  x  1  x  1
x  x  1  x  1
x  x  1  x  1

x


x  x  1   x  1
x
x


x  x  1  x  1
x
x  x  1  x  1
S. Stirling 2015
x  x  1 



x 1
For
x=0
1
x  0 1  x 1
1
2 x 1
Page 10 of 10