# Algebra 2 CP1 Review Packet by C92iV8

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```									                                       Algebra 2 Summer Review Packet

Summer Review Packet
For Algebra 2 CP1
Name _________________________________________________________

Current Course _____________________ Math Teacher________________________

Introduction
Algebra 2 builds on topics studied from both Algebra 1 and Geometry. Certain topics are
sufficiently involved that they call for some review during the year in Algebra 2. However, some topics and
basic skills are so fundamental to any Algebra 1 course that MASTERY of these topics is expected and
required PRIOR to beginning Algebra 2. This packet covers these topics, which are:
Solving simple equations in one variable,
Factoring polynomials – GCF, General trinomials, perfect square trinomials
and difference of squares.
Graphing lines in y  mx  b form, and slopes of lines,
Computation with fractions without the use of a calculator,
(Computations using a calculator – NOT in this packet, but it will be done in class)

We will be using a TI-83 or TI-84 in our classes

Note about calculators: The school will issue each student who needs/wants a TI-83 or TI-84 graphing
calculator, just as we issue a textbook to each student. If it is lost or broken the student must pay us the
replacement value of about \$90. A student may prefer to purchase his/her own. If so, we recommend the
TI-84 Plus (Silver Edition optional). We then highly recommend the calculator be etched with the student’s
name clearly on the front face, and on the back for security purposes. A sticker or a name written with a
marker is not enough.

During this summer you are to complete this review packet, following the directions for each
section. Although you will have some opportunity to get some help, if you find you need a lot of help, you
should consider getting it before school begins. You will be tested on this material shortly after the
beginning of school. The topics on computation with and without a calculator will require true mastery – a
100% result, and will be formative up to that point.
We recommend you do this work close to the onset of school. Do the work neatly on loose-leaf
paper or graph paper in case your teacher decides to collect it.

DO NOT LOSE THIS PACKET!!
It will also be available from the school website via the math department.

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Algebra 2 Summer Review Packet

COMPUTATIONS WITHOUT THE USE OF A CALCULATOR
SOME NOTES…..
You are expected to know how to do operations involving fractions, mixed numbers and signed numbers
without the use of a calculator.

Here are a few basic principles:

A mixed number like     42
3    really means    4  2 , which becomes 4  3   3  12  3  14
3                     3
2
3
2
3

Adding and subtracting fractions requires that all the fractions have the same denominator.
Some examples:

7 1 7 3 1 8 21 8 29
1)                DO NOT change this to a mixed number.
8 3 8 3 3 8 24 24 24

2)   5 1  2 5  (5  1 )  (2  5 ) 
4
3
4
3       21
4     13 
5
21
4    ( 5 )  13 ( 4 )  105  52 
5
5 4        20   20
53
20

a c a c
Multiplying fractions: The rule is              , and you MUST reduce as possible.
b d bd
Some examples:

7 12 7 12 1 4 4
3)                   It is always better to reduce BEFORE you multiply
9 21 9  21 3  3 9

4)

(7 1 )(3 5 )  (7  1 )(3  5 )  ( 15 )( 5 )  (15)( 17)  (3)(2 17)   51
2
2
2
2
2
17
(2)(5)                   2

Dividing fractions: The rule is to change the division problem into a multiplication problem by
multiplying the numerator by the reciprocal of the denominator, and then multiply as above.
a
a c a d ad
That is: b                   , and then reduce as possible.
d
c
b d b c bc
Some examples:

11
11 7 77
5)
4
2
      
7         4 2 8

2 2 2  3  2        8
8 5 (8)(5) (1)(5) 5      5
6)
3
               3
                       or 
31
5     3 1
5
16
5       3 16 (3)(16) (3)(2)   6      6

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Algebra 2 Summer Review Packet

COMPUTATIONS WITHOUT THE USE OF A CALCULATOR
WORKSHEET…

For the worksheet on this topic, you will be given an algebraic expression using variables, and the values for
these variables. You will then substitute the values for the variables and find the final simplified answer as
a reduced fraction or an integer. NO MIXED NUMBERS as answers.

Instructions: Given the following values or the six variables, find the value of each expression below in
simplest, reduced form.

1       4                                                                       5
a  , b   , c  2 3 , x  1,
4                                          y  2, z 
3       5                                                                       6
1) a  b  y                                                       2) bz

c                                                                  a y
3)                                                                 4)
b                                                                  cb

z                                                                z b
5)                                                                 6)
a b                                                                 c

7) 3 y  12 z  3a                                                 8) y  c

5b  8c                                                             18az
9)                                                                 10)
6a  3 x                                                              x

a b
11)                                                               12) a 2  b 2
z y

(3 x  y ) 2                                                       y x
13)                                                                14)     
4a                                                             2 7

a b                                                                x z y
15)                                                               16)      
b a                                                                y x z

23                  2                   55               140
And the final answers are:          1)                  2)                 3)                4)
15                  3                   16               117

25                     2                                3                 18
5)                   6)           7) 17               8)                 9)                 10) 5
34                    165                               4                  5

4                169              3                   8                     169                56
11)               12)              13)               14)                   15)               16)         Solved
5                225              4                   7                      60                15
Examples

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Algebra 2 Summer Review Packet
Solving Equations

1
Example 1:     Solve        x2  9
3

1
x  2  9
3
2 2                    Eliminate the +2 by subtracting 2 from both sides.
1
Solution:              x  7
3
1 
3  x   3  7           Eliminate the    1
3
by multiplying by its reciprocal, 3, on both sides.
3 
x  21

Example 2:     Solve     3y  8  7  11

3y  8   7  11
3y  24  7  11              Use the distributive property and combine to simplify the left side
3y  17  11             of the equation.
Solution:                - 17  - 17          Eliminate the +17 by subtracting 17 from both sides.
3y   -6
                Eliminate the 3 by dividing both sides by 3.
3    3
y  -2

Example 3:     Solve 7a  2  6  2a  8  a

7 a  2   6  2a  8  a
7 a  14  6  2a  8  a           Use the distributive property and combine to simplify both
7 a  20  3a  8               sides of the equation.

 3a             3a                Eliminate the variable term on the right by subtracting 3a.
Solution:           4a  20  8
 20         20            Eliminate the –20 by adding 20 to both sides.
4a        28
                   Eliminate the 4 by dividing both sides by 4.
4          4
a       7

Example 4:     Solve     3(1  n)  5n  2(n  1)

3(1  n)  5n  2(n  1)            Continue as in the previous example.
3  3n  5n  2n  2
Solution:              3  2n  2n  2
 2n        2n
Correct steps have resulted in a false statement.
32
Conclude that the equation has no solution.
no solution or Ø or {}
Solved Examples

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Algebra 2 Summer Review Packet
Factoring (Common Monomial Factor)

Example 5:      Factor 15x 3 y 2 z  20x 2 yz 2

Solution:       Find the greatest common factor on the terms in the polynomial.
GCF is 5x 2 yz . Divide each term in the polynomial by the GCF to find the other factor.

15x 3 y 2 z                                     20 x 2 yz 2
 3xy        and                                 4z
5x 2 yz                                         5 x 2 yz

15x 3 y 2 z  20x 2 yz 2  5x 2 yz3xy  4 z 

Factoring (Difference of Squares)

Example 6:      Factor 16x 2  25

Solution:       A term is a square if the exponents on all variables in it are even and the coefficient
is the square of an integer. Both 16x2 and 25 are squares. This will only factor if the two squares are
subtracted – thus a difference of squares. In particular, 16x2 = (4x)2 and 25 = (5)2. The binomial factors
as the sum and the difference of the squares roots:
16x2 – 25 = (4x + 5)(4x – 5)

Factoring (Perfect Square Trinomials)

Example 7:      Factor 4x2 – 20xy + 25y2

Solution:       A perfect square trinomial is a polynomial with three terms that results from
squaring a binomial. In other words, perfect square trinomial = (some binomial) 2.
To factor, we must first check that the trinomial is a perfect square by answering
the following three questions:
1.       Is the first term a square?        Answer: Yes, 4x2 = (2x)2
2.       Is the last term a square?         Answer: Yes, 25y2 = (5y)2
3.       Ignoring the sign at this time,
is the middle term twice the
product of 2x and 5y?              Answer: Yes, 20xy = 2 (2x5y)
Having established that the trinomial is a perfect square, it is easy to find the
binomial of which it is a square; its terms are in the parenthesis above.
4x2 – 20xy + 25y2 = (2x - 5y)2

Since the middle term in the trinomial is negative, the factored result is a subtraction.

Factoring (Product – Sum)

Example 8:      Factor y2 + 14y + 40

Solution:       If, as above, the trinomial is in standard form, the product number is the numerical
term, 40, and the sum number is the coefficient of the linear term, 14. The unique
pair of number whose sum is 14 and product is 40 is 4 and 10. Therefore, the
trinomial factors as
y2 + 14y + 40 = (y + 4)(y + 10)
(Note: all factoring problems can be checked by multiplying out)
Solved Examples

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Algebra 2 Summer Review Packet
Graphing

Example 9:    Find the slope and the y-intercept of the line whose equation is 3y = 2x + 9 and use
them to graph the equation.

3y  2 x  9
3y 2 x 9
Solution:                       Divide by 3 to solve for y.
3     3 3
2
y  x3
3              Now the equation is in the familiar form y = mx + b.

2
The slope is the coefficient of the x-term,    , and the y-intercept is the constant
3
2
term, 3. To graph the line, plot the y-intercept, (0 , 3). Then, since the slope is , move 2
3
units up(rise) and 3 units right (run) to locate a second point. (See below.)
y

+3

+2

x
Example 10:   a)      Find the slope of the line passing through the points (-2 , 5) and (4 , 8).
b)      Given that the y-intercept is 6, write an equation for the line.

y 2  y1 85 3 1
Solution:     a)       slope                    
x 2  x1   42 6 2
b)      In a linear equation of the form y = mx + b, the slope is the coefficient of the x-term or linear
term, namely m. The numerical term or constant term is the y-intercept, namely b. So, starting
with y = mx + b, substitute ½ for m and 6 for b. Therefore, the equation is
1
y  x6
2

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Algebra 2 Summer Review Packet
Exercises

Solve each equation.

1.      2x – 1 = 11                                2.       -4n + 9 = -3

1                                                  2
3.         x7 6                                  4.         x8
2                                                  3

2x  1                                             1 x
5.              5                                 6.            7
3                                                 2

7.      7x – 4x = 56                               8.       -3n – 5n = 240

9.      y – 7 + 4y = 13                            10.      0 = n – 13 – 4n

11.     35 = 5 (n + 2)                             12.      2 (4 – x) = 22

13.     y + 5 – 4y = -10                           14.      15 = 8x – 5 + 2x

1
15.           (x  4)  16                        16.      3 (a – 5) + 19 = -1
2

17.     -3 = 4 (k + 7) - 15                        18.      4c + 3 (c – 2) = -34

8  2x                                            3
19.      0                                        20.      1     (v  2)  5
5                                               4

21.     -9 – 3 (2q – 1) = -18                      22.      -2 = 4 (s + 8) – 3s

23.     c – (1 – 2c) + (c – 3) = -4                24.      3x = 27 – 15x

25.     51a – 56 = 44a                             26.      -7a = -12a - 65

27.     3p – 8 = 13 – 4p                           28.      5n + 1 = 5n -1

2
29.     8 (5 – n) = 2n                             30.        x7  x
3

9  2y                                             23  11c
31.             y                                 32.                5c
7                                                   7

33.     3 (30 + s) = 4 (s + 19)                    34.      2 (g – 2) – 4 = 2 (g – 3)

35.     4 (3y – 1) + 13 = 5y + 2                   36.      4 (a + 2) = 14 – 2 (3 – 2a)

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Algebra 2 Summer Review Packet
Exercises

Factor.

1.    15a – 25b + 20                     2.       6x2 + 10x

3.    6p2q – 9pq                         4.       7y3 – 21y2 – 14y

5.    6ab2 – 8a2b                        6.       -15x2y2 – 6xy2

7.    5ax2 + 10a2x – 15a3                8.       48a3b2 + 72a2b3

9.    c2 - 36                            10.      9a2 - 100

11.   169m2 - 225                        12.      144 – y2

13.   m6 – 25                            14.      25m4 – n2

15.   81p8 – 121q6                       16.      49a2 – 9b10

17.   x2 + 5x + 4                        18.      c2 – 10c + 16

19.   q2 + 16q + 15                      20.      a2 + 10ab + 24b2

21.   u2 – 50uv + 49v2                   22.      y2 + 5y - 6

23.   x2 + 2x - 8                        24.      x2 – 25x - 54

25.   x2 – 10xy – 75y2                   26.      a2 + 5ab – 84b2

27.   4w2 + 12w + 9                      28.      25m2 – 60mn + 36n2

29.   49a2 + 28ab + 4b2                  30.      4s2 – 36st + 81t2

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Algebra 2 Summer Review Packet
Exercises

Find the slope and the y-intercept of the line determined by each equation. Use them to graph the equation on the graph
paper provided. You may put two problems on a graph.

2
1.       y     x8                  2.    y = 8 – 2x                      3.      y = -x
3

4.      y=4                          5.    3x + y = 9                      6.      4x – 3y = 9

7.      x + 5y = 5                   8.    6x + 4y = 8

Find the slope of the line passing through each pair of points.

9.      (0 , 7) and (1 , 9)          10.   (5 , 2) and (7 , 0)             11.     (8 , 1) and (-8 , 1)

12.     (3 , -1) and (6 , 7)         13.   (-2 , 0) and (2 , -3)           14.     (-2 , -6) and (-2 , 4)

Find the equation of the line passing through each given point with the given slope and y-intercept.

2
15.     slope = 3, y-intercept = 7                  16.          slope =     , y-intercept = -5
3

17.     slope = 0, y-intercept = -6

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Algebra 2 Summer Review Packet

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Algebra 2 Summer Review Packet
Key to Exercises

Solve Equations

1.       x=6                        2.        n=3                      3.         x = -2
4.       x = 12                     5.        x=8                      6.         x = -13
56
7.       x                         8.        n = -30                  9.         y=4
3
13
10.      n                        11.       n=5                      12.        x = -7
3
13.       y=5                       14.       x=2                      15.        x = -36
5
16.      a                        17.       k = -4                   18.        c = -4
3
19.      x=4                        20.       v=6                      21.        q=2
3
22.      s = - 34                   23.       c=0                      24.        x
2
25.      a=8                        26.       a = -13                  27.        p=3
28.      No solution                29.       n=4                      30.        x = -21
1
31.      y=1                        32.       c                       33.        s = 14
2
34.      No solution                35.       y = -1                   36.        Any Real Number

Factor

1.       5 (3a – 5b + 4)                                2.    2x (3x + 5)
3.       3pq (2p – 3)                                   4.    7y (y2 – 3y – 2)
5.       2ab (3b – 4a)                                  6.    -3xy2 (5x + 2)
7.       5a (x2 + 2ax – 3a2)                            8.    24a2b2 (2a + 3b)
9.       (c + 6)(c – 6)                                 10.   (3a + 10)(3a – 10)
11.      (13m + 15)(13m – 15)                           12.   (12 + y)(12 – y)
3          3
13.      (m + 5)(m – 5)                                 14.   (5m2 + n)(5m2 – n)
15.      (9p4 + 11q3)(9p4 – 11q3)                       16.   (7a + 3b5)(7a – 3b5)
17.      (x + 1)(x + 4)                                 18.   (c – 2)(c – 8)
19.      (q + 1)(q + 15)                                20.   (a + 6b)(a + 4b)
21.      (u – 49v)(u – v)                               22.   (y + 6)(y – 1)
23.      (x + 4)(x – 2)                                 24.   (x – 27)(x + 2)
25.      (x – 15y)(x + 5y)                              26.   (a – 7b)(a + 12b)
27.      (2w + 3)2                                      28.   (5m – 6n)2
29.      (7a + 2b)2                                     30.   (2s – 9t)2

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Algebra 2 Summer Review Packet
Key to Exercises

Graph

2
1.      slope =     , y-intercept = 8                                  5.         slope = -3, y-intercept = 9
3

4
2.      slope = -2, y-intercept = 8                                    6.         slope =      , y-intercept = -3
3

1
3.      slope = -1, y-intercept = 0                                    7.         slope =      , y-intercept = 1
5

3
4.      slope = 0 , y-intercept = 4                                    8.         slope =      , y-intercept = 2
2

#3                        #2     y            #1                                  #8                 #5
y              #6

#4
#7

x

9.      2                          10.   -1                            11.        0

8                                     3
12.                                13.                                14.        no slope
3                                     4

2
15.     y = 3x + 7                 16.   y         x5                17.        y = -6
3

LHS                                                      12

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