Precalculus Summer Algebra Review Page 1 of 10

The Laws and Properties of Exponents

Operation Action Law/Property Example(s)

Multiplication Add exponents of the a m a n = a m+n x 3x 4 = x 7
same base
Division Subtract exponents of am x7
the same base = a m−n = x4
an x3
Power to a Power Multiply exponents
(a ) = a
m n mn
(x ) = x 4 3 12

Power to a Power of a Multiply exponents (ab) = a nb n (3x ) = 3 x = 27 x

2 3
n 3 6 6

Product
⎛ a ⎞ an ⎛ y ⎞ y3
n 3
Power to a Power of a Multiply exponents
Quotient ⎜ ⎟ = n ⎜ ⎟ = 3
⎝b⎠ b ⎝x⎠ x
Zero Exponent a 0 = 1, a ≠ 0 5 0 = 1, (2 x 3 y 2 ) = 1
0

Negative Exponents Take reciprocal of the 1 1

base and change the a −n = x −3 =
an x3
sign of exponent to
positive
a −n b n x −3 y 4
= =
b −n a n y−4 x 3

−n
⎛a⎞ ⎛b⎞
n −3
⎛x⎞ ⎛ y ⎞ y3
3
⎜ ⎟ =⎜ ⎟ ⎜ ⎟ =⎜ ⎟ = 3
⎝b⎠ ⎝a⎠ ⎝y⎠ ⎝x⎠ x

Simplify with only positive exponents.

1. (3x )(4 x )
3 5
2.
x5 y3 3. (x y )
3 2 4

x2 y

4. (2 x 3 y 2 ) ⎛ 2x 3 ⎞ 6. (5x 0 y 3 )
4 3 2

5. ⎜ 2 ⎟
⎝ 3y ⎠

7.
x −5 y 4 ⎛ 3x 2 ⎞
8. ⎜ 3 ⎟
−3
9. (x )(3x )(4x )
5 2 7

w −3 ⎝ y ⎠

10. (2x 4 y−3 ) (3y )(2y )

−2 −2
3 2 3 ⎛ 3x 5 y 4 ⎞
11. 12. ⎜ 0 −3 ⎟
(y ) 4 3 ⎝x y ⎠
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The Laws and Properties of Radicals

n
In a , n is the index.

Operation Law Example Law Example

Product n
ab = n a n a 50 = 25 ⋅ 2 = 5 2 n
a n a = n ab 2⋅ 3= 6
Same index
Quotient a na 5 35 35 n
a na 200 200
Same index n = 3 = = = = = 100 = 10
b nb 8 38 2 n
b b 2 2
Radical m n
a = mn a 3
64 = 6 64 = 2
within a
Radical

n
Properties of a , where n is a positive integer.

Property Examples when the index equals the exponent

( a ) = a if ( a ) is a real number ( 5) = 5 ( −8 ) = −8
n 2 3
n n 3
(1)

(2) n
a n = a if a ≥ 0 5 2 =5 3
23 = 2
a n = a if a < 0 and n is odd (−2) = − 2 (−2) = −2
n 3 5
(3) 3 5

a n = a if a < 0 and n is even (−3) = −3 =3 (−2) = −2 = −2

n 2 4
(4) 4

Additional Examples
Simplify each radical. Assume all variables represent positive real numbers.

Example 1a: Using Perfect Squares 75

75 = 25 ⋅ 3 = 5 3

Example 1b: Using Prime Factorization 75

75 = 5 2 ⋅ 3 = 5 3

3
Example 2a: Using Perfect Cubes 320
3
320 = 3 64 ⋅ 5 = 3 64 ⋅ 3 5 = 4 3 5

3
Example 2b: Using Prime Factorization 320
3
320 = 3 2 6 ⋅ 5 = 3 2 6 ⋅ 3 5 = 4 3 5

Example 3: 3x 2 y 3 ⋅ 6 x 5 y
3x 2 y 3 ⋅ 6 x 5 y = 18 x 7 y 4 = 3x 3 y 2 2 x
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Simplify, remove radicals if at all possible. Assume all variables represent positive real numbers.

1. 98 3 27
2. 3.
4 36

4. 16x 8 5. 3 125 6. 3 108

7. 5 −64 8. 4
81x 5 y 8 9. 3 16xy 2 ⋅ 3 3x 2 y 2

In problems 10-15 determine the exponent needed to remove the radical symbol

10. 5 ⋅ 5? = 5 11. 7
4 2 ⋅ 7 4? = 4 12. 5
32 ⋅ 5 3? = 3

13. 8
x 3 ⋅ 8 x? = x 14. 3
y ⋅ 3 y? = y 15. 5
z 4 ⋅ 5 z? = z
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Multiplication of Polynomials

Example 1: Distributive Property 5x (2x − 3)

5x (2x − 3) = 10x 2 −15x

Example 2: Multiplying binomials (4x + 5)(3x − 2)

(4 x + 5)(3x − 2) = 12x 2 − 8x +15x −10 = 12x 2 + 7x − 10
Product Formulas

Formula Example
(x + y)(x − y) = x 2 − y2 (x + 5)(x − 5) = x 2 − 25
or or
(x − y)(x + y) = x 2 − y2 (x − 5 )(x + 5) = x 2 − 25
(x + y) = x 2 + 2xy + y 2 (2x + 3) = (2x ) + 2 (2x )(3) + (3) = 4x 2 +12x + 9
2 2 2 2

(x − y) = x 2 − 2xy + y 2 (2x − 3) = (2x ) − 2 (2x )(3) + (3) = 4x 2 −12x + 9

2 2 2 2

(x + y) = x 3 + 3x 2 y + 3xy 2 + y 3 (2x + 3) = (2x ) + 3 (2x ) (3) + 3 (2x )(3) + (3)

3 3 3 2 2 3

= 8x 3 + 36x 2 + 54x + 27
(x − y) = x 3 − 3x 2 y + 3xy 2 − y 3 (2x − 3) = (2x ) − 3 (2x ) (3) + 3 (2x )(3) − (3)
3 3 3 2 2 3

= 8x 3 − 36x 2 + 54x − 27

Multiply, use product formulas when they apply.

1. 3x (2x 2 + 5x − 3) 2. (x + 5)(x + 7)

3. (3x + 4 )(2x − 7) 4. (x − 3)
2

(x + 5) (3x − 2)
2 2
5. 6.

(x + 3) 8. (2x + 5 )
3 3
7.

(x − 4 ) (3x −1)
3 3
9. 10.
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Factoring Polynomials

Example 1: Greatest Common Factor 15x 3 − 25x

15x 3 − 25x = 5x (3x 2 − 5 )

Examples 2 through 4 are Trinomial Factoring of ax 2 + bx + c where a = 1

Example 2: Factors of Same Signs x 2 + 7x +12

x 2 + 7x + 12 = (x + )(x + )
x 2 + 7x + 12 = (x + 3 )(x + 4 )

Example 3: Factors of Same Signs x 2 − 7x +12

x 2 + 7 x + 12 = (x − )(x − )
x 2
+ 7 x + 12 = (x − 3 )(x − 4 )

Example 4: Factors of Different Signs x 2 − 2x −15

x 2 − 2 x − 15 = (x − )(x + )
x 2 − 2 x − 15 = (x − 5 )(x + 3 )

Examples 5: Trinomial Factoring of ax 2 + bx + c where a ≠ 1, a ≠ 0

Factors of Same Signs 6x 2 +19x +10

Factors of a ⋅ c with sum of 19 6 ⋅10 = 60 , 4 +15 = 19
Rewrite trinomial as 6x 2 +19 x +10 = 6 x 2 + 4 x +15 x +10
Group as two binomials 6 x 2 +19 x +10 = (6 x 2 + 4 x )+ (15 x +10 )
Factor Greatest Common Factor 6x 2 +19x +10 = 2x (3x + 2 ) + 5 (3x + 2 )
Factor Greatest Common Factor 6x 2 +19x +10 = (3x + 2 )(2x + 5 )

Factoring Formulas

Name Formula Example

Difference x − y = (x + y )(x − y )
2 2
9 x − 16 = (3x ) − (4 ) = (3x + 4 )(3x − 4 )
2 2 2

of Two
Squares
Difference x 3 − y 3 = (x − y )(x 2 + xy + y 2 ) 8x 3 − 27 = (2x ) − (3) = (2x − 3)⎡⎣(2x ) + (2x )(3) + (3) ⎤⎦
3 3 2 2

of Two
Cubes = (2 x − 3)(4 x 2 + 6 x + 9 )
Sum x 3 + y 3 = (x + y )(x 2 − xy + y 2 ) 8x 3 + 27 = (2x ) + (3) = (2x + 3)⎡⎣(2x ) − (2x )(3) + (3) ⎤⎦
3 3 2 2

of Two
Cubes = (2 x + 3)(4 x 2 − 6 x + 9 )
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Problems for Factoring Polynomials

Factor each completely.

1. 21x + 14 2. x 2 +12 x + 35

3. x 2 −10 x + 24 4. x 2 − 4 x − 21

5. x 2 − 3x − 40 6. x 2 +13x + 30

7. x 2 −1 8. 25 x 2 − 49 y 2

9. 2 x 2 − 98 10. 6 x 2 +13x + 5

11. 3x 2 −11x + 10 12. 12 x 2 + x − 6

13. x 3 − 8 14. x 3 −125

15. x 3 + 27 16. 8 x 3 +125

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Fractional Expressions

A fractional expression is a quotient of two algebraic expressions. As in any fraction, the denominator
may not equal zero, division by zero is not permitted. A special case of fractional expressions is the
quotient of two polynomials, which is referred to as a rational expression.

Follow these three steps in simplifying fractional expressions and rational expressions.

Step 1 Factor the numerator and the denominator if possible

Step 2 Determine any values that would make the denominator equal to zero.
Step 3 Remove any factors of one, factors that are the same in the numerator and
denominator.

Example 1: Simplify the fractional expression and determine any restricted values.

x 2 + 3x − 10
x 2 − 25

x 2 + 3x −10 (x + 5 )(x − 2 )
Step 1 Factor =
x 2 − 25 (x + 5)(x − 5)
Step 2 Determine Restricted Values x ≠ ±5

x 2 + 3x −10 (x + 5 )(x − 2 ) (x − 2 )
Step 3 Remove Factors of One = =
x 2 − 25 (x + 5)(x − 5) (x − 5)

Example 2: Simplify the fractional expression and determine any restricted values.

3x 2 − 5 x − 2
x 2 + 3x − 10

3x 2 − 5 x − 2 (3x +1)(x − 2 )
Step 1 Factor =
x 2 + 3x −10 (x + 5 )(x − 2 )

Step 2 Determine Restricted Values x ≠ −5, x ≠ 2

3x 2 − 5 x − 2 (3x +1)(x − 2 ) (3x +1)

Step 3 Remove Factors of One = =
x 2 + 3x −10 (x + 5 )(x − 2 ) (x + 5 )
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Problems for Fractional Expressions

Simplify each if possible and state any restrictions.

1.
(x + 2)(x − 4 ) 2.
5x + 20
(x + 2)(x + 5) x + 10x + 24
2

x 2 + 13x + 30 x 2 + 12 x + 35
3. 4.
x2 − 9 x 2 − 3x − 40

x 2 − 49 x 3 −125
5. 6.
x 2 + 14x + 49 x 2 − 25