Download Algebra Review

Review
Algebra
1- 1
Section R.1
Polynomials
1- 2
Your Turn 1
Perform the operation 3( x 2  4 x  5)  4(3x 2  5x  7).
Solution: Multiply each polynomial by the coefficient in front
of the polynomial and then combine like terms.
 3x 2  12 x  15  12 x 2  20 x  28
 9 x 2  8 x  13
Your Turn 2
Perform the operation (3 y  2)(4 y 2  2 y  5).
Solution : Using the Distributive property yields
 3 y(4 y 2  2 y  5)  2(4 y 2  2 y  5)
 3 y(4 y 2 )  3 y(2 y)  3 y(5)  2(4 y 2 )  2(2 y)  2(5)
 12 y3  6 y 2  15 y  8 y 2  4 y  10
 12 y3  2 y 2  19 y  10.
Section R.2
Factoring
1- 6
Your Turn 1
Factor out the greatest common factor in 4 z 4  4 z 3  18z 2 .
Solution: Each of these terms is divisible by 2z2.
4 z 4  4 z 3  18z 2  2 z 2 (2 z 2 )  2 z 2 (2 z )  2 z 2 (9)
 2 z 2 (2 z 2  2 z  9).
Example 4
Section R.3
Rational
Expressions
1 - 10
Your Turn 1
z 2  5z  6
Write in lowest terms
.
2
2z  7z  3
Solution: Factor both numerator and denominator in order to
identify any common factors.
( z  2)( z  3)

(2 z  1)( z  3)
( z  2)

.
(2 z  1)
The answer can not be further simplified.
Your Turn 2
z 2  5z  6 2 z 2  z 1
Perform the following operations 2
 2
.
2 z  5z  3 z  2 z  3
Solution: Factor where possible.
( z  2)( z  3) (2 z  1)( z  1)


(2 z  1)( z  3) ( z  3)( z  1)
( z  2)

( z  3)
Section R.4
Equations
1 - 14
Your Turn 1
Solve 3x  7  4(5 x  2)  7 x.
Solution: 3x  7  20 x  8  7 x
3x  7  13x  8
10 x  7  8
Distributive Property
Combine the like terms.
Add  13x to both sides.
10 x  15
Add 7 to both sides.
10 x 15

10 10
1
Multiply both sides by
.
10
3
x
2
Your Turn 2
Solve 2m2  7m  15.
Solution: First write the equation in standard form.
2m2  7m  15  0
Now factor 2m2  7m  15 to get
(2m  3)(m  5)  0.
By the zero-factor property, the product (2m  3)(m  5)
can equal 0 if and only if
2m  3  0
or
m  5  0.
Solve each of these equations separately to find that the
3
solutions are
and  5.
2
Your Turn 3
Solve z 2  6  8 z.
Solution: First, add  8 z on both sides of the equal sign in order
to get the equation in standard form.
z 2  8 z  6  0.
Now identify the letters a, b, and c.
Here, a  1,
b  8, and c  6.
Substitute these numbers into the quadratic formula.
(8)  (8) 2  4(1)(6)
x
2(1)
Continued
Your Turn 3 continued
8  64  24
x
2
8  40 8  2 10
x

2
2
2(4  10)

2
Factor 8  2 10
 4  10 Reduce to lowest terms.
The two solutions are 4+ 10 and 4  10.
Your Turn 4
1
2
1
Solve 2

 .
x 4 x2 x
Solution: Factor x 2  4 as ( x  2)( x  2).
1
2
1


( x  2)( x  2) x  2 x
The least common denominator for all the fractions
is x( x  2)( x  2). Multiplying both sides by x( x  2)( x  2)
gives the following:
1
2 
1

x( x  2)( x  2) 

 x( x  2)( x  2)

x
 ( x  2)( x  2) x  2 
Continued
Your Turn 4 continued
x  2 x( x  2)  ( x  2)( x  2)
x  2 x2  4 x  x2  2x  2x  4
Distributive property
2 x2  5x  x2  4
Add  x 2 and 4, Rearrange terms.
x2  5x  4  0
( x  1)( x  4)  0
x 1  0
or
x40
x  1
x  4
Verify that the solutions are  1 and  4.
Section R.5
Inequalities
1 - 24
Your Turn 1
Solve 3z  2  5 z  7.
Solution: 3 z  2  2  5 z  7  2
3z  5 z  9
Add 2 to both sides.
3z  (5 z )  5 z  9  (5 z )
 2z  9
1
1
 2 z   9
2
2
Add  5 z to both sides.
Multiplying by a negative changes
the direction of the inequality.
9
z
2
Section R.6
Exponents
1 - 28
Your Turn 1
2 4
2
y z 
Simplify
 y 3 z 4 


2 2
4 2
(
y
)
(
z
)
Solution:
 3 2 4 2
( y ) (z )
Property 4 and 5
y 4 z 8
 6 8
y z
Property 3
z 8( 8)
 6( 4)
y
Property 2
z16
 10 .
y
Section R.7
Radicals
1 - 34
Example 2
Your Turn 2
Rationalize the denominator in
5
.
x y
Solution: The best approach here is to multiply both numerator
and denominator by √x + √y . The expressions √x + √y
and √x − √y are conjugates, and their product is
( x ) 2  ( y ) 2  x  y.
Thus,
5 x 5 y
5( x  y )


.
x y
( x  y )( x  y )