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Algebra 1 Second Semester Final Exam Study Guide 2009-2010
Chapter 8 - Exponents and Radicals

Simplifying Expressions using Exponent Properties (Positive Exponents)
Simplify each expression.
Example 1)
Example 2)
Example 3)
 4x y 
 4 x   y
 4 x  y
m8
m5
x3  x 2  x  x 7
 x 3  2 1  7
m
 x13
Example 4)
2
3
2
8 5
2
 m3
3 2
3 2
 4c 
 5
d 
2

2
 16 x 6 y 2
2
 4c  2
d 
5 2

42 c 2
d5 2

16c 2
d 10
Simplify.
m5  m3  m
1)

2)
3
x11
x4
3)
(6xy5 )2
 x5 
 7 
 2y 
4)
Simplifying Expressions using Exponent Properties (Zero and Negative Exponents)
Evaluate.
Example 1)
(6)
0
1
Example 2)
4 2 3
3
 4  8 1
 32
Simplify each expression. Write answers using only positive exponents.
Example 3)
Example 4)
Example 5)
m 5
0
1
 
m
1
 5
m
5
  5 n  2
3 x 4
4
2
1
 3 
x
 1 
 3 4 
x 
3
 4
x
 1 


 5n 
1

 5n  2


5) Evaluate.
1
 5  2 n 2
1
25n 2
50  6  32
Simplify. Write all answers using only positive exponents.
6)
c7
Algebra 1 2nd Semester Review
5/15/2008
7)
 2x 3
8)
 2n4
Page 1 out of 13
N. Cavalieri

Simplifying Radical Expressions
Simplify each expression
Example 1)
5 18
 5 2 33
Example 2)
Example 3)
20 x3 y
3
8
 2 25 x  x  x  y
 5 2  32
 5 2  3
 22  5  x 2  x  y
 5  32  2
 2 x 5 xy
2

3
8

3
2 2

3
2

2 2 2

6
2 4
 53 2
 15 2
6
22
6

4

Simplify.
9)

10)
7 12
50x 2 y3
11)
5
18
Apply Pythagorean Theorem and its Converse
Let a and b represent the lengths of the legs of a right triangle, and
let c represent the length of its hypotenuse. Find the unknown length.
Example 1)
12)
a  5, b  10
Example 2)
b  3, c  7
Tell whether the triangle with the given
side lengths is a right triangle.
Example 3)
2, 6, 7
Replace c with 7 since it is the
length of the longest side.
a 2  b2  c2
a 2  b2  c2
52  102  c 2
a 2  32  7 2
a 2  b2
c2
25  100  c 2
a 2  9  49
22  62
72
125  c 2
a 2  40
c  125
a  40
c5 5
a  2 10
4  36
49
40
49
Since 40  49 , the triangle is NOT a
right triangle
Let a and b represent the lengths of the legs of a right triangle, and let c represent the length of its hypotenuse
a  4, b  8 . Find c.
13) Tell whether the triangle with the side lengths 8, 15, 17 is a right triangle.
Algebra 1 2nd Semester Review
Page 2 out of 13
N. Cavalieri

Cube roots and Fractional Exponents
Evaluate.
Example 1)
Example 2)
 27  3
since
3
82
since
3
23  8
( 3)  27
Example 4)
Example 5)
25
1 2
Example 3)
 3  64  (4)
4
3
 25
8
1 3
 5
Example 6)
27 2
 8
3
 2
3

 
 27 1
3
27
  3

3 2

2
2
 9
Evaluate.
14)  3 125
15) (27) 1
16) 16 3
3
2
Chapter 9 Polynomials and Factoring

Adding and Subtracting Polynomials
Example 1) Given 3x2  5x3  2  x
(a) Write the polynomial so that the terms are in descending order. Answer : 5x3  3x2  x  2
(b) Find the degree of the polynomial.
Answer : The degree is 3
(c) Identify the leading coefficient.
Answer : The leading coefficient is –5
Example 2) Add.
(2 x3  5 x 2  1)  (2 x 2  9 x  4 x3  7)
Example 3) Subtract.
(7 x 2  3x  8)  ( x 2  6 x  9)
 (2 x3  5 x 2  1)  (4 x3  2 x 2  9 x  7)
 (7 x 2  3x  8)  ( x 2  6 x  9)
 2 x3  4 x3  5 x 2  2 x 2  9 x  7  1
 7 x 2  x 2  3x  6 x  8  9
 6 x3  3x 2  9 x  8
 6 x 2  3 x  1
Rewrite as
addition: 1st
polynomial plus
the opposite of
the 2nd
17) Identify the leading coefficient and the degree of the polynomial. 4 x3  x  9 x4  2
18) Add. (3x3  4x2  7)  (2x2  9x  2x3  12)

Multiplying polynomials
Example 1)
Multiply.
(3x  5)( x  2)
 3x( x)  3 x( 2)  5( x)  5( 2)
 3x 2  6 x  5 x  10
 3x 2  x  10
Multiply.
20) 3x2 (4x3  2x2  7 x)
23)
19) Subtract. (7 x3  2x2  8x)  (4 x3  4 x2  3x)
( x  2)( x  6 x  7)
2
Algebra 1 2nd Semester Review
Special Products:
(a  b)2  a2  2ab  b2
(a  b)(a  b)  a2  b2
(a  b)2  a2  2ab  b2
Example 2) Multiply.
Example 3) Multiply.
(4 x  3) 2
(2 x  5)(2 x  5)
 (4 x)2  2(4 x)(3)  (3) 2
 (2 x) 2  (5) 2
 16 x 2  24 x  8
 4 x 2  25
21) (2 x  7)( x  3)
24)
(3x  8)
22) ( x  8)( x  8)
2
Page 3 out of 13
N. Cavalieri

Dividing polynomials
Example 1)
Example 2) Divide.
(4 x 2  8 x  7)  (2 x  1)
Divide.
Example 3) Divide.
( x 2  6)  ( x  2)
(3x3  6 x 2  15 x)  (3x)

2x  5
3x3 6 x 2 15 x


3x
3x
3x
x2
2x  1 4 x  8x  7
x  2 x2  0x  6
2
 x2  2x  5
 (4 x 2  2 x)
 ( x 2  2 x)
 10 x  7
2x  6
  10 x  5 
  2x  4
12
2
Answer :
2x  5 
Divide.
25) (4x4  20x3  12 x2 )  (4 x)

Answer :
12
2x 1
x2
2
x2
26) ( x2  3x  5)  ( x  2)
Factoring
Steps in factoring:
Step 1: Factor out the GCF.
Step 2: Look at the number of terms
Binomial: If the binomial is a difference of squares, factor:
a 2  b 2  (a  b)(a  b)
Trinomial:
(a) If the trinomial is a perfect square trinomial, factor:
a 2  2ab  b 2  (a  b)( a  b)
a 2  2ab  b 2  (a  b)(a  b)
(b) If the trinomial is not a perfect square trinomial, try to unFOIL
Four or more terms: try to factor by grouping.
Example 1) Factor.
2 x  6 x  36 x
Example 2) Factor.
3 x 4  48
 2 x( x 2  3x  18)
 3( x 4  16)
 2 x( x  6)( x  3)
 3( x 2  4)( x 2  4)
3
Note 9 and 25 are
perfect squares, and
2
 9  15   30
2
 3( x 2  4)( x  2)( x  2)
so this is a perfect
square trinomial.
Example 3) Factor.
9 x 2  30 xy  25 y 2
 (3 x  5 y )(3 x  5 y )
 (3 x  5 y )
2
Example 4)
Factor.
4 x3  8 x 2  x  2
 (4 x 3  8 x 2 )  ( x  2)
 4 x 2 ( x  2)  (1)( x  2)
 ( x  2)(4 x 2  1)
 ( x  2)(2 x  1)(2 x  1)
Algebra 1 2nd Semester Review
Page 4 out of 13
N. Cavalieri
Missing a term of
degree 1. Put in
0x as a place
holder
Factor each expression completely.
27) x  1
28) 2 x  32 x
4
3
29) a b  49c
2
6
5 xy  40 xy 2
2
32) 3x  4 x  15
2
34) 25 x  70 x  49
3
2
36) x  3 x  16 x  48
4
30)
31) m  9m  36
2
33) 2x3  6x2  36x
35) x3  8x2  5x  40
Chapter 10 Quadratic Equations and Functions
Finding the Axis of Symmetry and the Vertex of Quadratic Equations
Example: Find the axis of symmetry and the vertex of y  2 x  4 x  1 .
1st) Identify a, b, and c: a = 2, b = 4, and c = –1.
3rd) Find the vertex. The equation for the axis of
symmetry tells you the x coordinate, so x = – 1.
2nd) Find the Axis of Symmetry:
Now replace x in your equation with –1 to find the y coordinate.
b
x
y  2x2  4x 1
2a
2
y  2(1) 2  4(1)  1
4
2( 2)
4
x
4
x  1
x
y  2(1)  4  1
y  2  4 1
y  3
The vertex is (–1, –3)
Since a  0 , the vertex is a minimum point.
Find the equation for the axis of symmetry and the coordinates of the vertex. Determine if the vertex is a minimum or a
maximum point.
37) y  3x 2  12 x  20
Graphing Quadratic Equations in the form y  ax2  bx  c
Example: Graph y 
1st) Find the vertex
b
x
2a
4
x
2(1)
4
x
2
x  2
x 2  4x  3 .
y  x 2  4x  3
y   ( 2 )  4( 2 )  3
2nd) Make a table.
x
y  x 2  4x  3
-4
-3
-2
-1
0
y  (4)  4(4)  3
y  (3) 2  4(3)  3
y  (2) 2  4(2)  3
y  (1) 2  4(1)  3
y  (0) 2  4(0)  3
2
y
(x, y)
-3
(–4, –3)
0
(–3, 0)
1
(–2, 1)
0
(–1, 0)
-3
(–2, –3)
3rd) Graph the points and draw the parabola
2
y   ( 4)  8  3
y  4  8  3
y 1
The vertex is (–2, 1)
38)
Graph.
y  x 2  2x  3
Algebra 1 2nd Semester Review
Page 5 out of 13
N. Cavalieri
Graphing Quadratic Equations in the form y  a( x  p)( x  q)
Example: Graph y  2( x  3)( x  1) .
1st) Find and graph the x-intercepts.
4th) Draw a parabola through the vertex and the x-intercepts.
Let y equal 0.
y  2( x  3)( x  1)
0  2( x  3)( x  1)
x  3  0 or x  1  0
x3
x  1
The x-intercepts are (3, 0) and (–1, 0).
2nd) Find and graph the vertex.
pq
2
3  1
x
2
x 1
y  2( x  3)( x  1)
x
y  2(1  3)(1  1)
y  2(2)(2)
y 8
The vertex is (1, 8)
3rd) Graph the axis of symmetry.
pq
2
x 1
x
Graph.
39) y  ( x  2)( x  4)
Solving Quadratic Equations

Solving equations using the zero-product property.
Example 1): Solve.
3x( x  5)( 2 x  7)  0
Example 2): Solve.
2 x 3  6 x 2  56 x  0
3 x  0 or x  5  0 or
2x  7  0
x  5  5  0  5 2x  7  7  0  7
3x 0

x5
2x  7
3
3

2
2
x0
2 x( x 2  3x  28)  0
2 x( x  7)( x  4)  0
2x  0 x  7  0 x  4  0
x0
x7
x  4
2x  7
Solution Set: 0, 5,  7 

x
7
2
Solution Set: 0, 7,  4
2
Solve each equation.
40) 9 x(7 x  5)  0
41) ( x  7)( 4 x  9)  0
Algebra 1 2nd Semester Review
42) x  8 x  33  0
2
43) 2 x  162 x
3
Page 6 out of 13
N. Cavalieri
 Solving by Graphing
When you solve by graphing, you are looking for the x intercepts of the graph.
Example 1) What values appear to be the zeros of the
Example 2) Solve by graphing. x 2  2 x  8  0
equation whose graph is shown?
Graph the related function: y  x 2  2 x  8
Vertex:
y  (1) 2  2(1)  8
2
y  1 2  8
2(1)
y  9
x  1
The vertex is (–1, –9).
x
Table: x
–3
–2
–1
0
1
y  x 2  2x  8
y  (3)  2(3)  8
2
y  (2) 2  2(2)  8
y  (1) 2  2(1)  8
y  (0) 2  2(0)  8
y  (1) 2  2(1)  8
y
–5
–8
–9
–8
–5
(x, y)
(–3, –5)
(–2, –8)
(–1, –9)
(0, –8)
(1, –5)
Graph
:
Answer: The graph intersects the x axis at –3 and –1,
so the zeros are –3 and –1.
Solutions: The zeros are –4 and 2.
Solve by graphing.
44)  x2  4 x  5  0

Solving by Taking the Square Root of Both Sides
Example 1) 3 x  12  0
2
Example 2)
3x 2  12
( x  5) 2  9
Example 3)
( x  7) 2   12
( x  5) 2   9
x2  4
x  5  3
x  4
2
( x  7) 2  12
x 7  2 2
x  5  3
x  2
x  7 2 2
x  5  3 or x  5  3
x  2
x  8
Solve.
45)  6 x  54  0
2

46)
( x  3) 2  4
47)
( x  2) 2  18
Solving by Completing the Square
Algebra 1 2nd Semester Review
Page 7 out of 13
N. Cavalieri
Example 1) x  8 x  12  0
2
4 x 2  24 x  12  0
Example 2)
x 2  8 x  12
4x
24 x 12 0



4
4
4 4
x 2  6x  3  0
x  8 x  16  12  16
2
1

 2 (8)


2
 (4)
( x  4) 2  4
2
x 2  6x  3
x4  4
x 2  6x  9  3  9
x  4  2
( x  3) 2  12
x  42
x  4  2 or x  4  2
x6
x2
 16
Divide by 4
to make the
quadratic
coefficient 1.
2
x  3   12
x  3  12
x  3 2 3
Solve by completing the square.
48) x  4 x  3  0
49) 3 x  30 x  21  0
2

2
Solving Using the Quadratic Formula
 b  b 2  4ac
x
2a
Example 1)  3 x  x  10
x 2  5x  2  0
a = 1 , b = 5 , c = –2
2
Example 2)
 3x 2  x  10  0
a = –3 , b = –1 , c = 10
Put in standard form:
x
 5  (5) 2  4(1)( 2)
2(1)
x
(1)  (1) 2  4(3)(10)
2(3)
 5  25  8
2
 5  33
x
2
x
1  1  120
x
6
1  121
6
1  11
x
6
x
1  11
1  11
or x 
6
6
12
 10
x
x
6
6
5
x  2
x
3
x
Solve using the quadratic formula.
50) 3 x  5 x  2
2
51)  x  4 x  2  0
2
Determining the number of real roots using the discriminant b  4ac
2
If b  4ac  0 (the discriminant is negative), then there are no real solutions.
2
Algebra 1 2nd Semester
Review
2
If b  4ac  0 (the discriminant is zero), then there is only one real solution.
Page 8 out of 13
N. Cavalieri
Example 1) 3 x  2 x  4  0
2
b 2  4ac  (2) 2  4(3)(4)
 4  48
Example 2) 4 x  12 x  9  0
2
Example 3)  x  3 x  2  0
b 2  4ac  (12) 2  4(4)(9)
 144  144
b 2  4ac  (3) 2  4(1)(2)
 98
2
0
 44
Since the discriminant is negative,
there are no real solutions.
 17
Since the discriminant is zero,
there is one real solution.
Since the discriminant is positive,
there are two real solutions.
Use the discriminant to determine the number of real roots for each equation.
52) 2 x  5 x  2  0
2
53)
3x 2  x  2  0
54)  x  6 x  9  0
2
Chapter 11 Rational Expressions and Equations
Simplifying , Multiplying, and Dividing Rational Expressions
Example 1) Simplify.
Example 2)
2 x 2  18
x  7 x  12
2
2( x  9)
x  7 x  12
2( x  3)( x  3)

( x  4)( x  3)
2( x  3)( x  3)

( x  4)( x  3)
2( x  3)

x4
2

2
55) Simplify.
x  x  30
x 2  4x  5
2
Multiply.
3x  9
x 4
 2
x  3x  10 x  5 x  6
2
2
3( x  3)
( x  2)( x  2)

( x  5)( x  2) ( x  2)( x  3)
3( x  3)( x  2)( x  2)

( x  3)( x  2)( x  2)( x  5)
3

x5

56) Multiply.
x  x  12 7 x  21

x 2  3x  18 x 2  16
2
Example 3)
Divide.
5x  25
4x  8

2
3x  3x  60 x
6x
3
5 x  25
6x

3 x 3  3 x 2  60 x 4 x  8
5( x  5)
2  3x


3 x( x 2  x  20) 4( x  2)
5( x  5)
2  3x


3 x( x  5)( x  4) 4( x  2)
2  3x  5( x  5)

3 x  4( x  5)( x  4)( x  2)
25

4( x  4)( x  2)
10

4( x  4)( x  2)

57) Divide.
2x
4 x 2  12 x

x 2  8 x  7 x 2  4 x  21
Adding and Subtracting Rational Expressions (NOTE: Must have common denominators.)
Example 1) Add.
Algebra 1 2nd Semester Review
Example 2)
Factor to find
theLCD. The
LCD is
Subtract.
Page 9 out of 13
N. Cavalieri
5
3

x4 x
5
x 3 ( x  4)

  
x  4 x x ( x  4)
5x
3( x  4)


x( x  4) x( x  4)
5 x  3( x  4)

x( x  4)
5 x  3 x  12

x( x  4)
2 x  12

x( x  4)
2( x  6)

x( x  4)
12
2

m  4m  5 m  5
2







12
2

(m  5)( m  1) m  5
(m  1)
12
2


(m  5)( m  1) (m  5) (m  1)
Multiply by a
form of 1 to get
12  2(m  1)
fractions with
(m  5)( m  1)
the same
12  2m  2
denominator.
(m  5)( m  1)
2m  10
(m  5)( m  1)
2(m  5)
Add and
simplify.
(m  5)( m  1)
2
m 1
Add or subtract. Simplify answers.
58) Add.
59) Subtract.
4
2

2
x  6x  8 x  2
5
3

n2 n
Solving Rational Equations
Example 1)
Factor the
denominators.
Clear the
fractions by
multiplying
both sides of
the equation
by the LCD.
The LCD is
2y  3
4
5
 2

y2 y 4 y2
2y  3
4
5


y  2 ( y  2)( y  2) y  2
2y  3
4
5
( y  2)( y  2)
 ( y  2)( y  2)
 ( y  2)( y  2)
y2
( y  2)( y  2)
y2
4( y  2)  (2 y  3)  5( y  2)
20
x
3
x
5
 20 
 x
5 x    5 x  3  5 x  
 x 
5
100  15 x  x 2
x 2  15 x  100  0
( x  20)( x  5)  0
x  20  0 or x  5  0
4 y  8  2 y  3  5 y  10
( y  2)( y  2)
Example 2)
x  20 or
2 y  11  5 y  10
x  5
 3 y  21
y7
60)
Solve.
61)
1
2
2


x  16 x  4 x  4
5
x
2
x
4
2
Work Problems
Solve.
(work rate)(time together )
= part done
Example1:
It takes a painter 15 hours to paint a house. A second
Painter 1can
the same
house in 12 hours. How
Algebra
2ndpaint
Semester
Review
long will it take to paint the house if they work together?
Example 2:
Together, you and your cousin can wash and dry six
cars in 3 hours. It would take your
cousin
Page 10
out oftwice
13 as
long as you take to wash the six cars
alone.
How long
N. Cavalieri
Time Work
alone rate
Time
Part
together done
1st painter
15
1
15
t
t
15
2nd painter
12
1
12
t
t
12
t
t

1
15 12
t 
 t
60    601
15
12


4t  5t  60
This total is always
one. The parts add up
to 1 whole job done.
9t  60
60
9
2
t6
3
t
It will take 6 2 hours
3
working together.
you
your cousin
Time Work
alone rate
1
x
x
2x
1
2x
Time
Part
together done
3
3
x
3
3
2x
3 3

1
x 2x
3 3 
2 x    2 x1
 x 2x 
6  3  2x
9
2 x  2 
 2
9
2x  9
It will take my cousin 9
hours alone.
9
x
2
Solve.
62) You and a co-worker at a grocery store have to place sales tags under each item that is on sale. You are able to
complete the job in 40 minutes, and your co-worker takes 60 minutes to complete the same job. If you work
together, how long will it take?
63) Maria can clean the garage in 4 hours when working alone. With the help of her brother Julio, the job takes
only 3 hours to complete. How long would Julio take to clean the garage by himself?
Algebra 1 2nd Semester Review
Page 11 out of 13
N. Cavalieri
Castro Valley Unified School District
Castro Valley High School
Algebra 1 Second Semester Final Exam Study Guide
2008-2009
ANSWERS
17)
The leading coefficient is 9 and
the degree is 4.
1)
m9
18) 5 x 3  2 x 2  9 x  5
2)
x7
19) 11x 3  6 x 2  5 x
3)
36x 2 y10
x15
4)
8 y 21
20) 12 x 5  6 x 4  21x 3
21) 2 x 2  x  21
22) x 2  64
5)
54
23) x 3  8 x 2  19 x  14
6)
1
c7
24) 9 x 2  48 x  64
7)
1
8x 3
2
n4
8)
9)
10)
14 3
5xy 2 y
25) x3  5 x 2  3x
26) x  5 
27)
15
x2
( x 2  1)( x  1)( x  1)
28) 2 x( x  4)( x  4)
29)
(ab3  7c 2 )(ab3  7c 2 )
30) 5 xy (1  8 y )
11)
10
6
12)
4 5
32) (3x  5)( x  3)
13)
Yes, since 82  152  17 2
33) 2 x( x  6)( x  3)
14)
5
34)
(5 x  7) 2
15)
3
35)
( x  8)( x 2  5)
16)
64
31) (m  12)(m  3)
36) ( x  3)( x  4)( x  4)
37)
Algebra 1 2nd Semester Review
Axis of symmetry: x  2
Vertex: (2,8)
The vertex is a minimum
Page 12 out of 13
N. Cavalieri
45) x   3
y
38)
5
46) x  1 or x  5
4
3
47) x  2  3 2
2
48) x  1 or x  3
1
–5
–4
–3
–2
–1
–1
1
2
3
4
5
x
49) x  5  3 2
–2
50) x 
–3
–4
–5
39)
51) x  2 
y
10
9
8
7
6
5
4
3
2
1
–10–9 –8 –7 –6 –5 –4 –3 –2–1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
40)
41)
2 real roots (D = 9)
53)
0 real roots (D = –23)
54)
1 real root (D = 0)
x6
x 1
7( x  3)
56)
 x  6 x  4
55)
1 2 3 4 5 6 7 8 9 10
x
57)
1
2  x  1
2
x4
x  0 or x  
5
7
58)
x  7 or x 
9
4
59) 2  n  3 
n  n  2
43) x  0 or x  9 or x  9
60) x 
y
9
8
7
6
5
4
3
2
1
–9 –8 –7 –6 –5 –4 –3 –2 –1
–2
–3
–4
–5
–6
–7
–8
–9
6
52)
42) x  11 or x  3
44)
1
or x  2
3
1
4
61) x  10 or x  2
1 2 3 4 5 6 7 8 9
Algebra 1 2nd Semester Review
62)
24 minutes
63)
12 hours
x
Page 13 out of 13
N. Cavalieri