25

2 Algebraic Expressions 2.1 Evaluating Algebraic Expressions Chapter 1 was a review of the real numbers and the rules governing numeric expressions – these are expressions that consist of numbers and the opera-tions of addition, subtraction, multiplication, division, exponentiation, and possibly symbols of grouping. Now we begin our study of algebra by consid-ering algebraic expressions. The only difference is we now introduce letters or variables. Examples The following are examples of algebraic expressions; on the right you will see how to read them:

( )

2

3

2

4 2 four x minus two y2 1 x squared plus two x plus one

2 the quantity x minus two, cubedthe quantity x plus three, over3

th9

x yx x

x

xx

−

+ +

−

+−

2 4

e quantity x squared minus nine

4 the square root of four x squared y z to the fourthx yz

The letters x and y are the most popular letters to represent the variables or the unknowns, but we can also use other letters like a, b, c, etc. When there is no operation written in between a number and a variable or in between variables, the operation is multiplication. Thus, 4x means “4 times x”. It will be useful to be able to identify the numeric part and the variable part of an algebraic expression. The numeric part is also called the numerical co-efficient. Before we can do so, we sometimes would need to split the expression into terms. Terms are separated by plus (+) or minus (-) signs.

3 2 7 numeric3 algebraicx y⋅ −−

26

Example 1

( )

2 2

3

2

. 4 2 2 terms 4 and 2

. 2 1 3 terms , 2 , and 1

. 2 the cube of 2 terms and 2the ratio of 2 quantities, ea3.

9

x y x yx x x x

x x

xx

− −

+ +

− −

+−

ab

c

d 2

2 4 2 4

ch having

two terms: and 3; and 9

. 4 the square root of 1 term: 4

x x

x yz x yz

−

e

The last example can also be thought of a one term consisting of 2 44x yz .

The numerical coefficient of this term is 1 and the variable part is 2 44x yz . The radicand 2 44x yz has numerical coefficient 4 and variable part 2 4.x yz Note that a term can be thought of as the product of letters and a number. The numerical coefficient is the numerical factor; the variable part consists of the variable factors. Evaluating Algebraic Expressions In order to find the value of an algebraic expression for given values of the variables, we simply replace the variables by the given values and evaluate the resulting numeric expression following the rules governing the operations. Example 2. Evaluate 4 2x y− for 3 and 1.x y= = −

Solution: ( )4 2 4 3 2 1 12 2 14x y− = ⋅ − − = + = Example 3. Find the value of 2 2 1x x+ + when 1 2.x =

Solution: ( ) ( )22 142 1 1 2 2 1 2 1 1 4 1 1 2x x+ + = + + = + + =

Example 4. What is the value of ( )32x − when x is replaced by 0? By 2? By – 1? Solution: Replace x by 0: ( ) ( ) ( )3 3 32 0 2 2 8x − = − = − = −

Replace x by 2: ( ) ( )3 3 32 2 2 0 0x − = − = =

Replace x by – 1: ( ) ( )3 31 2 3 27− − = − = −

27

Example 5. For the expression 2

39

xx+−

, determine if x can be replaced by 0.

Can it be replaced by – 3? Can it be replaced by 3?

Solution: Replace x by 0: 2 2

3 0 3 3 19 0 9 9 3

xx+ +

= = = −− − −

Yes, 0 works!

Replace x by – 3: ( )22

3 3 3 0 0 UND9 9 9 03 9

xx+ − +

= = = =− −− −

---No

Replace x by 3: 2 2

3 3 3 6 UND9 3 9 0

xx+ +

= = =− −

----No

Recall that UND (page 5) is shortcut for “undefined”. We see that both 3 and

3− will not work because they make the denominator of 2

39

xx+−

equal to 0

which makes the expression 2

39

xx+−

undefined.

Example 6. Evaluate 2 44x yz for 2, 1, and 1.x y z= = = −

Solution: ( ) ( )( )2 42 44 4 2 1 1 4 4 1 1 16 4x yz = − = ⋅ ⋅ ⋅ = = Example 7. Evaluate 2 44x yz for 2, 1, and 1.x y z= = − = −

Solution: ( ) ( )( ) ( )2 42 44 4 2 1 1 4 4 1 1 16 not realx yz = − − = ⋅ ⋅ − ⋅ = − We end up with the square root of 16− which is undefined in the set of real numbers. We will see later that we can define this over a larger set of numbers, namely, the set of complex numbers.

To evaluate an algebraic expression, simply replace each variable by its given numerical value and follow the rules governing the operations. The denominator can never be 0 and we cannot have the square root of a negative number.

28

2.2 Combining Like Terms

We now look at how we can perform the operations of addition or subtraction on algebraic expressions. The key is to look for like terms. These are simply terms that have the same variable part. Example 1. The following show 4 groups of like terms.

a. 23 , 8 , , , 1.3 3

x x x x x− c. , 21 , 9ab ab ab−

b. 2 2 2 21,0.9 , ,104

y y y y− d. 2 5, , ,3 2 4

x x x xy y y y

−

The nice thing about being able to identify which terms are like terms is we can now just add/subtract the numerical coefficients to combine the like terms.

Example 2. Simplify: 23 8 1.33

x x x x x− + − +

Solution: Since all the terms are like terms, we simply perform 23 8 1 1.3 3

− + − + either by hand or using a calculator to get -3.366…, giving

the final answer 3.3666... or 3.36 .x x− −

Example 3. Simplify: 2 2 2 210.9 104

y y y y− − + +

Solution: The four terms are all like terms, so we perform the calculation 11 0.9 104

− − + + to get 8.35 and the simplified form of the algebraic expres-

sion is 2 .8.35y Example 4. Simplify: 21 9ab ab ab+ −

Solution: The three terms are all like terms, so we simply perform 1 21 9+ − to get 13, giving the final answer 13 .ab

29

Example 5. Simplify: 2 53 2 4

x x x xy y y y− + −

Solution: We can rewrite the expression as

1 1 523 2 4

x x x xy y y y

⋅ − ⋅ + ⋅ − ⋅

to see that the variable part is xy

and that we just need to perform

1 1 523 2 4

− + − for which we see that the LCD is 12 and we can proceed as

follows:

1 1 5 2 1 1 5 24 4 6 15 112 =3 2 4 1 3 2 4 12 12

− + −− + − = ⋅ − ⋅ + ⋅ − ⋅ =

12 4 6 312 4 6 3

We then conclude that the simplified form of the expression is 11 .12

xy

Example 6. Simplify: ( ) ( )2 23 12 2 2 2 22 2

x x x xx x

+ − + − − − − + −− −

Solution: ( ) ( )2 23 12 2 2 2 22 2

x x x xx x∗ ∗

+ − + − − − − + −− −

We note that there are 2 ∗ terms that can be combined, 2 terms that can be combined, and 2 terms that can be combined:

terms:∗ 3 1 2

2 2 2x x x− =

− − −

terms:

( ) ( )2 22 2 0x x− − − = terms:

2 2 2 3 2x x x− + − = −

We then get the final simplified form 2 3 2

2x

x+ −

−.

So, the variable part can be anything for as long as it’s the same for the terms, I can combine them!

3 3 3

2 2 2

3 25 5 5x y x y x y

z z z− = Neat!!

30

2.3 Removing Symbols of Grouping To simplify an algebraic expression that has symbols of grouping like paren-theses, brackets, or braces, we must first remove the symbol(s) of grouping before combining the like terms.

To remove the symbols of grouping, we use the distributive property of multiplication over addition: ( )a b c ab ac+ = +

Example 1. ( ) ( )Simplify 2 3 .x y z x y z+ + + − + Solution: ( ) ( )x y z x y z+ + + − +2 3

"distribute" 2 and 3

combine like terms

2 2 2 3 3 3

5 5

x y z x y z

x y z

⊕∗ ∗ ⊕⊗ ⊗+ + + − +

− +

=

=

Note: Distributing a negative number will change the signs of all the terms inside the parentheses when the symbol of grouping is removed.

Example 2. ( ) ( ) ( )Simplify 4 2 2 3 2 4 .a b a b a b− − + − − Solution: ( ) ( ) ( )4 2 2 3 2 4a b a b a b− − + − −

( )remove

combine like terms

4 8 6 2 2 4

4 6

a b a b a b

a b

− − − − +

− −

=

=

Example 3. Simplify ( ) ( )3 22 1 2

4 3.x x+ − +

Solution: ( ) ( )( ) ( )

remove

3 2 3 4 4 3

3 2 3 3 2 22 1 2 2 1 2

4 3 4 4 3 3 x x x x+ − + = ⋅ + ⋅ − ⋅ − ⋅

1

( ) ( )

LCD 12LCD 6

multiply coefficients put like terms together3 3 2 4 3 2 3 4

2 4 3 3 2 3 4 3 x x x x

==⊗ ∗ ⊗ ∗

+ − − − + −= =2 3

( )

( ) equivalent fractions combine like terms

9 6 4 6 9 12 16 12

3 2 3 4 5 72 3 4 3 6 12

x x x− + − = −= ⋅ ⋅ ⋅ ⋅4 53 2 3 4

3 2 3 4

31

Example 4. ( ) ( )Simplify 1.2 0.1 3 0.5 2 6 .y y− − −

Solution: ( ) ( )( ) ( ) remove

0.12 3.6 3 1.2 0.1 3 0.5 2 6 y yy y = − − +− − −1

( ) ( )

-3.6 -1.003.00.12

-0.6 -0.88

put like terms together combine like terms0.12 3.6 3 0.88 0.6 y y y

++

− − + = − −=2 3

We can repeatedly apply the distributive property of multiplication over ad-dition to remove nested symbols of grouping. An example of an expression with nested grouping symbols is ( )( )( )5 3 2 2 1x x x− − + − . Let us use dif-

ferent grouping symbols to clarify which terms or expressions belong together. So, instead of using only parentheses ( ), we can also use curly braces { } and brackets [ ]. We can then rewrite the example as

( ){ }5 3 2 2 1x x x − − + − . We will use parentheses for the innermost

grouping symbol, followed by brackets, and lastly by curly braces. We will remove ( ) first, followed by [ ], and lastly { }. Example 5. Simplify ( ){ }5 3 2 2 1x x x − − + − .

Solution: ( ){ } 5 3 2 2 1x x x− − + ⋅ − 1 ( ) { }( )

[ ]{ }( ) { }( )

{ }( )

"distribute" 1

combine

"distribute" -2

combine

5 3 2 2 1

5 3 3 1

5 3 6 2

5 5 2

x x x

x x

x x

x

∗ ∗

∗

= − − + −

= − − −

= − − +

= − − +

=

1

2

3

4

5

2

3

( )

"distribute" -3

combine

5 15 6

15 1

x

x

+ −

= −6

Start from the innermost grouping symbol.

32

Example 6. Simplify ( ) ( )5 3 2 2 1x x x− − − + −

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

( )

( )[ ]

( )

( )

##

"distribute" 5 and 2

combine

"distribute" -1

combine like terms

15 10 2 2

15 10 3 2

15 10 3 2

18 8

x x x

x x

x x

x

= − − − + −

= − − − ⋅ −

= − − − +

= − −

1

2

3

4

1

2.4 Integer Exponents I Next let us learn the principles governing multiplication of algebraic expres-sions. The key is to identify the like factors. These are factors that have the same base. Examples The following can be multiplied because they have like factors.

( )2 5 27

x x x −

common base is x

( )( )( )3 42 3y y y− common base is y So how do we perform the multiplication? Let us develop some exponent rules, where the exponents are positive integers. We know that for a positive integer n, the exponential expression nb means

factors

n

b b b⋅ ⋅ ⋅

.

So, if we have two (or more) exponential expressions with the same base b, say and m nb b , where m and n are positive integers and we want the product

m nb b⋅ , we have the following:

factors factors factors

m n m n

m n m n

b b b b b b b b b b b b +

+

⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ =

33

Thus, we have the first rule called the product rule for exponents:

m n m nb b b +⋅ = (1) The rule says that we simply add the exponents to multiply like factors.

Example 1. Simplify ( )2 5 27

x x x −

.

Solution: ( ) ( )

2 5 2 5 8

2 72 5 1

product rulecollect like factors2 2 2 1 1 7 7 7x

x x x x x x x+ +

−

− = ⋅ − ⋅ ⋅ ⋅ ⋅ = −

Example 2. Simplify ( )( )( )3 42 3 .y y y−

Solution: ( )( )( ) ( )( )( )

3 4 3 4 8

3 4 1

product rulecollect like factors

6

2 3 2 1 3 6y

y y y y y y y+ +−

− = − = −

Next, suppose we want to write ( )532y− without parentheses. We can do it by

multiplying ( )( )( )( )( )3 3 3 3 32 2 2 2 2y y y y y− − − − − but this will be very inef-ficient. We need to develop more rules to do this one.

Suppose we want to find ( )mnb for positive integers m and n. We again use the definition of an exponential expression to come up with a rule.

( )

factors

factors factors factors factors factors

factors of

,

n

mmn

n n n n mn

m b b b

b b b b b b b b b b b b b b b b

⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

( )

factors

factors factors factors factors factors

factors of

OR

n

mn n n n

m n n n mn

m b b b

b b b b b b b b b b b b b b b b

⋅ ⋅ ⋅

= ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ ⋅ ⋅

When multiplying like factors, just ADD the exponents.

34

Either way we do it, we get the second rule called the power rule:

( ) mn nmb b= (2)

The rule says that to raise an exponential expression to a positive integer, we simply multiply the exponents.

With this rule, we still can’t simplify ( )532y− since there are two factors inside

the quantity. We need to develop a rule for ( )nab .

( ) ( )( ) ( ) ( )( ) factors factors factors

n n n

n n n

ab ab ab ab aa a bb b a b= = =

The third rule called the distributive property of exponentiation over mul-tiplication is then:

( ) n n nab a b= (3)

This rule tells us that we may distribute the exponent to each factor in order to remove the parentheses. Note that this rule applies to 3 or more factors.

Example 3. Simplify ( )532 .y−

Solution: ( ) ( )

( )

5 553 3 15

3 5

power rule"distribute" 5

32

2 2 32y

y y y⋅−

− = − = −

( )54 20x x=

( )32 3 3 63 3ab a b=

35

Example 4. Simplify ( )32 43 .x yz

Solution: ( ) ( )

( )

( )

( )

3 3 33 32 4 2 4 6 3 12

32 3 4 3

power rule"distribute" 3

27

3 3 27yx z

x yz x y z x y z⋅ ⋅

= =

Let us summarize the exponent rules governing multiplication: Product Rule m n m nb b b +⋅ =

Power Rule ( )mn nmb b=

Distributivity over Multiplication ( )n n nab a b= 2.5 Integer Exponents II

Next, we want to establish rules governing division of algebraic expressions. The key is again identifying like factors of the form nb , where b is any nonzero real number and n is a positive integer (we will extend this later to 0 and negative integers). Let us first look at the expression ( )na b . For real numbers a and b and b is

nonzero, and for a positive integer n, ( )na b can be written as:

factors

factors factors

nn n

n

nn

a a a a aa a ab b b b bb b b

= = =

Thus, exponentiation is also distributive over division and we have the first rule called the distributive property of exponentiation over division:

n n

na ab b

= (1)

Add Multiply Distribute

36

Example 1. Simplify. a. 32a

b

b. 34

2

xy

Solution: a. ( )

33 22 6

3 3

2 3

power rule"distribute" 3

a

aa ab b b

⋅

= =

b. ( )

( )

33 44 12

32 62

4 3

2 3

power rule"distribute" 3

x

y

xx xy yy

⋅

⋅

= =

Next, let us consider the expression m

n

bb

, where b is nonzero, and m and n are

positive integers with m > n. Using the definition of an exponential expression, we have

factors factors

factors

m nm

n

n

b bb b bb bb bb b

= =

factors

factors

m n

n

bb bbb b

−

⋅

factors

.m n

m nbb b b−

−= =

Thus, we have the first quotient rule for exponents:

, if m

m nn

b b m nb

−= > (2)

Example 2. Simplify: 16

10. xx

a 7

4

8. 2

yy

b 4 7

5

5. 10

a bab

−c

Solution:16

16 10 610. x x x

x−= =a

7 77 4 3

4 4

quotient rule #1collect like factors8 8. 4 42 2

y y y yy y

−= ⋅ = =b

4 7 4 74 1 7 5 3 2

5 5

quotient rule #1collect like factors5 5 1 1. 10 10 2 2

a b a b a b a bab a b

− −− −= ⋅ ⋅ = − = −c

37

Now, what happens when m < n? factors factors

factors

m mm

n

n

b bb b bb bb bb b

= =

factorsm

bb b

factors factors

1 1n m

n mn m

bb b bbb b −

−−

= =⋅

We then have for the case m < n the second quotient rule:

1 , if m

n n m

b m nb b −= < (3)

Example 3. Simplify: 6

13. cc

a b. 2

9

1520

nn−

c. 3

4 6

2xyx y

Solution: 6

13 13 6 7

1 1. cc c c−= =a

2 2

9 9

quotient rule #2collect like factors15 15 5. 20 20

n nn n

= − ⋅ = −−

b 35⋅

9 2 7 7

1 3 1 34 44 n n n−⋅ = − ⋅ = −

⋅

3 3

4 6 4 6 4 1 6 3 3 3

quotient rule #2collect like factors2 1 1 2. 2 2xy x yx y x y x y x y− −= ⋅ ⋅ = ⋅ ⋅ =c

Let us summarize the exponent rules we learned in this section.

Distributivity over Division n n

n

a ab b

=

Quotient Rule I , if m

m nn

b b m nb

−= >

Quotient Rule II 1 , if m

n n m

b n mb b −= >

3 3 5 43

2 6 2 10 6

1; ; x x a yay y a y y

= = =

38

2.6 Zero and Negative Exponents

When dividing m

n

bb

and the exponents m and n are equal, i.e. m n= , we get

m n

n

b bb

=nb

1.=

But using Quotient Rule 1, we also get

0.m n

n nn n

b b b bb b

−= = =

Since the two answers should be the same, we get the zero exponent rule:

0 1, 0 b b= ≠ (1) Example 1 Simplify. a. 08 b. 08w c. ( )0 0 08 8 8w w w− + d. ( )00 08 8 8w w w− +

Solution:

0. 8 1=a

b. 0 08 8 8 1 8w w= ⋅ = ⋅ =

c. ( ) ( )0 00 0 0 08 8 8 8 8 8w w w w w w− + = − ⋅ + ⋅

1 1 8 1 1 8 9 w w w− ⋅ + ⋅ = − + = −=

d. ( )00 08 8 8 1w w w− + = since the entire quantity is raised to zero. Let us next look at the relationship of Quotient Rule I and Quotient Rule II. Consider the following:

1212 18 6

186

612

18 18 12 6

quotient rule #1

quotient rule #2

1

1 1

b b bb

bbb

b b b

− −

−

−

= =⟩ ∴ =

= =

39

Likewise, consider:

2828 15 13

131513 13

1328

15 15 28 13

quotient rule #1

quotient rule #2

1 1 or

1 1

b b bb b b

b bbb b b

−−

−

− −

= = ⟩ = = = =

We can now state the negative exponent rule for nonzero b and positive in-teger n.

1 1 or , 0

nn

nb bb b

− = ≠

. (2)

The rule says that b raised to a negative integer exponent is equal to the reciprocal of the base b raised to the positive integer exponent. Example 2. Simplify: a. 32− b. 32x− c. ( ) 32x −

Solution: 33

1 1. 22 8

− = =a 3 33 3

1 2. 2 2 2x xx x

− −= ⋅ = ⋅ =b

( )( )

33 3 3 3

1 1 1. 22 82

xx xx

− = = =c

Example 3. Simplify: a. 2

14−

b. 2

14y−

c. ( ) 2

14y −

Solution: 2

22

1 1. 4 164 4

−

−

= = =

a

2 22

2

1 1 1 1 . 4 4 4 4

yyy y

−

−

= ⋅ = ⋅ =

b

( )( )

22 2 2 2

2

1 1. 4 4 1644

y y yyy

−

−

= = = =

c

( )( )

22

2 2

22

1 14 ; ;4 4 4

1 44

yy

yy

− −

−

= =

=

40

Example 4. Simplify: 2

3. xy

−

−a 3

2

2. xy

−

b 1 2

3

2. xy

−

−c

Solution: 2 3

2 33 3 2 2

1 1. x yx yy y x x

−−

− −= ⋅ = ⋅ =a

33

2 2 3 2 3 2

2 1 1 1 2. 2 2x xy y x y x y

−−= ⋅ ⋅ = ⋅ ⋅ =b

1 2 2 31 2 2 3

3 3 1

2 1 1. 22 2

x x yx x yy y

−−

− −= ⋅ ⋅ = ⋅ =c

Note that if we skipped all the middle steps (just do the boxed steps – problem and final answer), the factors with negative exponents in the numerator moved to the denominator and the exponents became positive; similarly, those fac-tors with negative exponents in the denominator moved to the numerator and the exponents became positive. Also note that we do NOT move any factor that is raised to a positive exponent.

2.7 Summary of Integer Exponent Rules For the any nonzero real base and the any integer exponent:

(1) Product Rule n m n mb b b +⋅ = (2) Power Rule ( )nm mnb b=

(3) Distributivity over Multiplication ( )n n nab a b=

(4) Distributivity over Division n n

n

b bc c

=

(5) Quotient Rule m

m nn

b bb

−=

(6) Zero Exponent Rule 0 1b =

(7) Negative Exponent Rule 1 1 or

nn

nbb b

− =

2 3 3 1 2 2 3

3 2 2 2 3 3

2 2 2; ;

2

x y x x x y

y x y y x y

− − −

− −= = =

41

Note that we compressed the quotient rule into one rule: subtract the expo-nents.

Example 1. Simplify 33 2

1 0

2x y zxy z−

Solution: We can remove the parentheses right away.

( )

( )

( )

( )

( )

( )

3 33 3 3 2 33 2 9 6 3

3 31 0 3 3 03 1 0

9 6 3

3 3 0

2 33 3

1 3 0 3

39 3 6 3

( ) "distribute" 3 ( ) power rule

collect like factors

22 8

8

yx

y z

zxy

x y zx y z x y zxy z x y zx y z

x y zx y z

− −−

−

⋅⋅

− ⋅ ⋅

− − −

= =

= ⋅ ⋅ ⋅3

1 2

6 9 3

0

( ) quotient rule 8 zx y

−

=4

Or, we can simplify the quantity inside the parentheses first before remov-ing the parentheses.

( )

( )

( )

( )

33

3 2 3 2 32 311 0

3 33 2 3 3

3 1 2 1

2 3 3 3

( ) collect like factors ( ) quotient rule

( ) "distribute" 3 ( ) power rule

1

2 2 21

2 z

zxy

x y

x y z x y z x y zx yxy z

x y

−−

− − −

⋅ ⋅

=

= ⋅ ⋅ ⋅ =

= =

1 2

3 46 9 3 8 y z x

Multiply…ADD Divide…SUBTRACT Ok to distribute to products and quo-tients “Anything” to 0 gives 1 Base to negative…flip the base, change exponent to positive

42

Example 2. Simplify: ( )42 0

3

4x xyx x

−−

⋅

Solution:

42 0

2 4 2

3 3 4

2 4

1 3

( ) zero exponent rule ( ) product rule1

44 4

x

x

x x yx x x

x x x x x

−−

− +

+

=

− − ⋅ − = =

⋅ ⋅

1 2

2

2

2

( ) quotient rule ( ) negative exponent rule

1

4 4

x

xx

−= − = −3 4

Example 3. Simplify: ( )( )

43 1

13

2

5 2

n m

nm

−

−

−

−

Solution: ( )( )

( )( )

4 43 1 12 4

1 1 1 33

( ) "distribute" 4 and -12 2

5 25 2

n m n mn mnm

− −

− − − −

− −=

−−

1

( ) ( ) ( ) ( )

( )

4 1 12 1 4 3

13 5 13 1

32

( ) quotient rule

( ) do the arithmetic ( ) negative exponent rule

1 25

1 32 2 5 5

n m

nn mm

− − − − − − −

−

−

= ⋅ −

−= ⋅ − =

2

3 4

One application of integer exponents is in scientific notation. A number in standard notation can be written in scientific notation and vice-versa. Sci-entific notation uses the powers of 10:

3 3

2 2

1 1

0

1

2

3

10 1 10 1 1000 0.001

10 1 10 1 100 0.01

10 1 10 1 10 0.1

10 1

10 10

10 100

10 1000

−

−

−

= = =

= = =

= = =

=

=

=

=

43

The scientific notation for a number is 10nN × , where the N is a number be-tween 1 and 10 (or 1 and 10− − , if negative), and n is an integer.

How to Write a Number from Standard Notation to Scientific Notation

1. Move the decimal point immediately to the right of the first nonzero digit of the number. 2. Multiply by a power of 10 determined by counting the number of places the decimal point has been moved. a. If the original number is greater than 10, the exponent is positive. b. If the original number is between 1 and 10, the exponent is 0. c. If the original number is between 0 and 1, the exponent is negative.

Example 4. Write the following in scientific notation.

a. 9,000,000 . 0.0007−b

Solution: a. 6 for the 6 zeros that disappeared

move left put after 9

9,000,000 9, 000,000 00 9 10.↓

←

= = ⋅

4 for the 4 decimal places that disappeared

right. 0 0007 7 10. −

↓→↑

− = − ⋅b

Conversely, we would want to be able to write a number in scientific notation to a number in standard notation. How to Write a Number from Scientific Notation to Standard Notation

1. Move the decimal point as many spaces as the exponent indicates. a. If the exponent is positive, move the decimal point to the right. b. If the exponent is negative, move the decimal point to the left. 2. Put the appropriate number of zeros.

Example 5. Write the following in standard notation.

a. 122 10× b. 74.51 10× c. 56.3 10−× Solution:

44

a. 12

12 places2 . 10 2,000,000,000,000 . 2,000,000,000,000↓ ↑→

× = =

b. 7

7 places4 . 51 10 45,100,000 . 45,100,000↓ ↑→

× = =

c. 5

5 places6 . 3 10 0 . 00006 3 0.000063−

↓ ←× = =

Performing Numerical Calculations Using Scientific Notation

An application of scientific notation is in performing calculations involving very large or very small numbers containing a lot of zeros. In the examples, we will use scientific notation to perform the calculations and we will ex-press our final answer in both scientific notation and standard notation. Example 6. Find: ( )( )( )4,000,000 30,000,000,000 0.00 000 02 Solution:

( )( )( )

( ) ( ) ( )6 10 7write in scientific notation

collect like factors

24

4,000,000 30,000,000,000 0.00 000 02

4 10 3 10 2 10

4 3 2

−= ⋅ ⋅ ⋅

= ⋅ ⋅

6 10 7

9

10

6 10 7

product rule10

10 10 10

= 24 10 24,000,000,000 standard notation

2.4 10

−

+ −

⋅ ⋅ ⋅

⋅ =

= ×

scientific notation

Example 7. Use scientific notation to calculate:

( )( )4,500,000

0.0000003 5,000

Solution:

( )( ) ( )( )( )

( )

6

7 3

6 6 7 3 10

7 3

( ) write in scientific notation

collect like factors ( ) product and quotient rules

4,500,000 4.5 10 0.0000003 5,000 3 10 5 10

4.5 10 4.5 10 0.3 103 5 10 10 15

3,000,000,000 stan

−

− − −−

⋅=

⋅ ⋅

= ⋅ = × = ×⋅ ⋅

=

2

1

3

9

dard notation

3 10 scientific notation= ⋅

45

Example 8. Use scientific notation to calculate: ( ) ( )3 22,000,000 0.0004 − Solution:

( ) ( )( ) ( )

3 2

3 26 4

3 18 2 8

18 8 102

10

11

2,000,000 0.0004

2 10 4 10

2 10 4 101 88 10 104 16

0.5 10

0.00 000 000 005 standard notation

5 10 scientific notation

−

− −−

− −

− + −

−

−

= ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅ = ⋅

= ×

=

= ⋅

2.8 English to Algebra Now that we know the rules governing manipulations of algebraic expres-sions, we next learn how to produce those algebraic expressions from given English phrases or sentences. This skill will help us solve word problems which is how most problems in the real world are stated: in words. We usually use letters to denote unknowns – the most popular are x, y¸ and z; other letters can be used, of course. The following shows what English words correspond to some popular math symbols Symbol English Words + sum, plus, added to, more than, increased by − difference, minus, subtracted from, less, less than decreased by or ⋅ × product, times, multiplied by, of

or ÷ quotient, divided by, ratio of

( ) quantity

( ) ( )2 3, quantity squared, quantity cubed Let us use x to represent the number and y if there is another number.

Use scientific notation to handle the zeros.

46

Example 1 English Algebra a. the sum of a number and 10 10x + b. a number increased by 2 2x + c. 3 more than a number added to another number

3 or 3x y x y+ + + +

d. a number plus 8 8x + Example 2

English Algebra a. the difference of a number and 10 10x − b. a number decreased by 2 2x − c. 3 less than a number 3x − d. 3 less a number 3 x− e. 5 subtracted from a number 5x −

Example 3 English Algebra a. The product of 7 and a number 7x b. 21 times a number 21x c. A number multiplied by 100 100x d. The product of two numbers xy

Example 4

English Algebra a. The quotient of a number and 5 5 or 5 or 5x x x÷

b. The ratio of 20 and a number 20 or 20 or 20x x x÷

c. A number divided by another number

or or x y x y y x÷

Example 5

English Algebra a. A number squared 2x b. The cube of a number 3x c. Twice a number 2x d. Half a number ( )2 or 1 2x x

47

Example 6 English Algebra a.

( )2 1

twice the quantity 1 more than a numberx+⋅

( )2 1x +

( )2 4. the square of the quantity 4 less than a number

x−b

( )24x −

1. the product of the quantity 1 more than a number

x+c

4and the quantity 4 less the number

x−

( )( )1 4x x+ −

2. the ratio of twice a number and

xd

3

the sum of the number and 3x+

23

xx +

Next, let us translate and simplify the resulting algebraic expression. Example 7. Translate and simplify: the sum of twice a number and three times the quantity 4 more than the number Translation: Let us use x to denote the number

( )

2

4

3 4

the sum of twice a number and

three times the quantity 4 more than the numberx

x

x

+

+

+

Thus, the translation is ( )2 3 4x x+ + and we simplify the expression as fol-

lows:

( )"distribute" 3 like terms

2 3 4 2 3 12 5 12x x x x x+ + = + + = +

48

Example 8. Translate and simplify: the ratio of 4 times a number and 16 times the square of the number Translation:

24 16

the ratio of 4 times a number and 16 times the square of the numberx x

Thus, the translation is 2

416

xx

and we simplify this expression as follows:

2

4

like factors

1

4 4 16 16

xx

⋅

=

44⋅

1 2 12

quotient rule 1 1 1 4 4 4

x x xx x

− −⋅ = ⋅ = =

Example 9. Translate and simplify: the cube of the product of a number and the square of another number, divided by the square of the product of the two numbers Translation: Let us start from the word “number” and work our way out.

( )

2

2

32

the cube of the product of a number and the square of another number ,

divided by the square of the prod

x y

y

xy

xy

( )2

uct of the two numbersxy

xy

Thus, the translation is ( )( )

32

2

xy

xy and we simplify as follows:

( )( )

32 3 63 2 6 2 4

2 2 2

quotient rule"distribute" 3 and 2

xy x y x y xyx yxy

− −= = =

49

Now that we know how to translate English phrases to algebraic expressions, we now translate English sentences to algebraic equations. An equation con-sists of two expressions separated by the equal sign. Following are words that translate to the equal sign:

is, is the same as, is equal to, equals, gives, yields The strategy is to first look for the word or words that translate to the equal sign. This word(s) divides the sentence into a left phrase and a right phrase that respectively translate to the left side and right side of the equation. Example 10. A number plus 3 equals twice the number less 4 Translation: In this example the equal sign corresponds to the word “equals”.

3 22 4

A number plus 3 equals twice the number less 4x x

x xx

+

−

=

Thus, the equation is 3 2 4 .x x+ = − Example 11. The square of a number is the same as the sum of twice the number and 3. Translation: The words “is the same as” translate to the equal sign.

2 22 3

The square of a number is the same as the sum of twice the number and 3.x x

xxx+

=

Thus, the equation is 2 2 3x x= + .

50

Better review each section and do more problems.

( )

0

2 2 2

4 5 1 2 2

4 33 2 3 6

2 3 5 2 2

2 3 6 2 2

6 2 12 2 2

6 3 2 22

x x x x

x x x x x

x x x x x

x x x xx

−

−

+ = =

⋅ = =

⋅ = =

= =

If I know what he knows, I’ll pass??