Algebra Honors (Period 6 & 7)

Properties of Rational Numbers 11-1

A real number that can be expressed as the quotient of two integers is called a rational number
A rational number can be written as a quotient of integers in an unlimited number of ways.
3 = 3/1= 6/2 = 12/4 = -15/-5
To determine which of two rational numbers is greater, you can write them with the same positive denominator and compare the numerators
Which is greater 8/3 or 17/7?
the LCD is 21
8/3 = 56/21
17/7 = 51/21
so 8/3 > 17/7

For all integers a and b and all positive integers c and d
a/c > b/d if an only if ad > bc
a/c < b/d if and only if ad < bc This method compares the product of the extremes with the product of the means Thus 4/7 > 3/8 because (4)(8) > (3)(7)

The Density Property for Rational Numbers
Between every pair of different rational numbers there is another rational number
The density property implies that it is possible to find an unlimited or endless umber of rational numbers between two given rational numbers.

If a and b are rational numbers and a< b then the number halfway from a to b is
a + (1/2)(b-a);
the number one third of the way from a to b would be
a + (1/3)(b-a) and so on

Math 6 Honors ( Periods 1, 2, & 3)

Fractions 6-1

The symbol 1/4 can mean several things:
1) It means one divided by four
2) It represents one out of four equal parts
3) It is a number that has a position on a number line.



1/8 means 1 divided by 8 or 1 ÷ 8
A fraction consists of two numbers
The denominator tells the number of equal parts into which the whole has been divided.
The numerator tells how many of these parts are being considered.
we noted that we could abbreviate ...

denominator as denom with a line above it

and numerator as numer

we found that you could add

1/3 + 1/3 + 1/3 = 3/3 = 1
or 1/4 + 1/4 + 1/4 + 1/4 = 4/4 = 1
we also noted that 8 X 1/8 = 8/8 = 1

We also noticed that 2/7 X 3 = 6/7


So we discussed the properties
For any whole numbers a, b,and c with b not equal to zero

1/b + 1/b + 1/b ... + 1/b = b/b = 1 for b numbers added together

and we noticed that b X 1/b = b/b = 1
we also noticed that
(a/b) X c = ac/b

We talked about the parking lot problem on Page 180

A count of cars and trucks was taken at a parking lot on several different days. For each count, give the fraction of the total vehicles represented by
(a) cars

(b) trucks

Given: 8 cars and 7 trucks
We noticed that you needed to find the total vehicles or 8 + 7 = 15 vehicles
so

(a) fraction represented by cars is 8/15
(b) fraction represented by trucks is 7/15


What if the given was: 12 trucks and 15 cars


(a) fraction represented by cars is 15/27
(b) fraction represented by trucks is 12/27

What about
GIVEN:
9cars
35 vehicles
This time we need to find out how many trucks there are
35 -9 = 26
so
(a) 9/35
(b) 26/35

We aren't simplifying YET

Math 6 Honors ( Periods 1, 2, & 3)

Least Common Multiple 5-6
Here is a review of that lesson...
Let’s look at the nonzero multiples of 8 and 12—listed in order
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72…

Multiples of 12: 12, 24, 36, 48, 60, 72, ….

The numbers 24, 48, and 72, ... are called common multiples of 8 and 12. The least of these multiples is 24 and is therefore called the least common multiple.

LCM(8, 12) = 24

To find the LCM of two whole numbers, we can write out lists of multiples of the two numbers.

Or, we can use prime factorization

Lets find LCM(12, 15)

12 = 2²∙3
15 = 3∙5

The LCM will be made up of the greatest power of each factor

LCM will be 2²∙3∙5 = 60
The book has a third option or method
you can check out, if you’d like

Let’s find LCM (54, 60)


54= 2∙3∙3∙3 = 2∙3³
60 = 2∙2∙3∙5 = 2²∙3∙5

The greatest power of 2 that occurs in either prime factorization is 2²
The greatest power of 3 that occurs in either prime factorization is 3³
The greatest power of 5 that occurs in either prime factorization is 5
Therefore, LCM(54,60) is 2²∙3³∙5 = 540


REMEMBER:
The GCF (greatest common factor) is a factor. The GCF of two numbers will be either the smaller of the two or smaller than both

The LCM (least common multiple) is a multiple. The LCM of the two numbers will be the largest of the two or larger than both.



To find the LCM of two whole numbers you could write out the lists of multiples-- and that works relatively easily with small numbers... but there are more efficient ways to find the least common multiple of two whole numbers.

1. Write out the first few multiples of the larger of the two numbers and test each multiple for divisibility by the smaller number. The first multiple of the larger number that is divisible by the smaller number is the LCM

2. You can use prime factorization to find the LCM. The LCM is EVERY factor to its GREATEST power!!

LCM(54, 60)
54 = 2⋅ 3⋅ 3⋅ 3 = 2⋅ 3³
60 = 2⋅ 2⋅ 3⋅ 5 = 2²⋅ 3⋅ 5
So the greatest power of 2 is 2²
The greatest power of 3 is just 3
and the greatest pwoer of 5 is just 5
so the product of 2²⋅ 3⋅ 5 will be the LCM
LCM(54, 60) = 540

3. You may use the BOX method as shown in class... unfortunately it does not show well here. Remember you need to create a L. The numbers on the side of the box represent the GCF!! You need to multiple them with the last row of factors.
See me before or after class if you want any review!!

We reviewed the concept of relatively prime and noticed that any two prime numbers are relatively prime. We also noticed that if two numbers are relatively prime-- neither of them must be prime....

We also found out that if one number is a factor of a second number, the GCF of the two numbers is the first number AND... if one whole number is a factor of a second whole number the LCM of the two numbers is the second number!!
GCF(12,24) = 12
LCM(12,24) = 24

WOW!!

If two whole numbers are relatively prime---
their GCF = 1
and their LCM is their product!!
GCF(8,9) =1
GCF(8,9) = 72

WOW!!

LCM & GCF Story PRoblems
1) Read the problem
2) Re-read the problem!!
3) Figure out what is being asked for!!
4) find the "magic " word... to help you determine if you are finding GCF or LCM
5) When in doubt... draw it out!!

Math 6 Honors ( Periods 1, 2, & 3)

Greatest Common Factor 5-5

If we list the factors of 30 and 42, we notice
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42

We notice that 1, 2, 3, and 6 are all COMMON factors of these two numbers. The number 6 is the greatest of these and therefore is called the
GREATEST COMMON FACTOR of the two numbers. We write
GCF(30,42) = 6

Although listing the factors of two numbers and then comparing their common factors is one way to determine the greatest common factor, using prime factorization is another easy way to find the GCF

Find GCF(54, 72)
54 = 2 ⋅ 3 ⋅ 3 ⋅ 3
72 = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3
Find the greatest power of 2 that occurs IN BOTH prime factorization. The greatest power of 2 that occurs in both is just 2 ¹
Find the greatest power of 3 that occurs IN BOTH prime factorizations. The greatest power of 3 that occurs in both is 3²
Therefore
GCF(54, 72) = 2 ⋅ 3² = 18

In class we circled the common factors and realized that
GCF(54, 72) = 2 ⋅ 3 ⋅ 3 = 18


Fin the GCF( 45, 60)
45 = 3 ⋅ 3⋅ 5
60 = 2⋅ 2⋅ 3⋅ 5
Since 2 is NOT a factor of 45-- there is NO greatest power of 2 that occurs in both prime factorizations.
The greatest power of 3 is just 3¹
and the greatest power of 5 is just 5¹
Therefore,
GCF(45,60) = 3⋅ 5 = 15

The number 1 is a common factor of any two whole numbers!! If 1 is the GCF , then the two numbers are said to be RELATIVELY PRIME. Two numbers can be relatively prime even if one or both of them are composite.

Show that 15 and 16 are relatively prime
List the factors of each number
FACTORS of 15: 1, 3, 5, 15
FACTORS of 16: 1, 2, 4, 8, 16

Since the GCF(15,16) = 1. The two numbers are relatively prime!!

Algebra Honors (Period 6 & 7)

Multiplying Fractions 6-2
You know from previous years that
ac/bd = a/b ⋅ c/d
and you know the converse is also true
a/b ⋅ c/d = ac/bd
That means you could solve
8/9⋅3/10 by either multiplying first and then simplify or you could simplify first and then multiply.
I find it works so much better to simplify first
8/9⋅3/10 = 4/15

6x/y³⋅y²/15 = 2x/5y where y ≠0

Which simplifies to

This textbook wants us to keep the factored form as our answers--> so let's continue to do that. In addition it states, " ...from now on, assume that the domains of the variables do not include values for which any denominator is ZERO. Therefore it will NOT be necessary to show the excluded [or restrictions] values of the variables."
I know everyone is jumping for joy!!

Rule of Exponents for a Power of a Quotient
(a/b)^m = a^m/b^m
(x/3)³ = x³/27

(-c/2)²⋅4/3c
you must do the exponent portion first!!
c²/4⋅(4/3c)
c/3

Find the volume of a cube if each edge has length 6n/7 in
You just need to cube each factor
(6n/7)³ = 216n³/343 inches cubed

If you traveled for 7t/60 hours at 80r/9 mi/h, how far have you gone?
Just multiply
7t/60⋅80r/9
but simplify first and you get
28rt/27 miles

Math 6 Honors ( Periods 1, 2, & 3)

Prime Numbers & Composite Numbers 5-4

A prime number is one that has only two factors: 1 and the number itself, such as 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31...
A counting number that has more than two factors is called a composite number, such as 4, 6, 8, 9, 10...

Since one has exactly ONE factor, it is NEITHER PRIME NOR COMPOSITE!!
Zero is also NEITHER PRIME NOR COMPOSITE!!
Sieve of Eratosthenes - We did it!! :)

Every counting number greater than 1 has at least one prime factor -- which may be the number itself.
You can factor a number into PRIME FACTORS by using a factor tree or the inverted division, as shown in class.

Using the inverted division, you also start with the smallest prime number that is a factor... and work down
give the prime factors of 42
2⎣42
3⎣21
7

When we write 42 as 2⋅3⋅7 this product of prime factors is called the prime factorization of 42.

Two is the only even prime number because all the other even numbers have two as a factor.

Explain how you know that each of the following numbers must be composite...
111; 111,111; 111,111,111; and so on....
Using your divisibility rules you notice that the sums of the digits are multiples of 3.

List all the possible digits that can be the last digit of a prime number that is greater than 10.
1, 3, 7, 9.

Choose any six digit number such that the last three digits are a repeat of the first three digits. For example
652,652. You will find that 7, 11, and 13 are all factors of that number... no matter what number you choose... why is that???? email me your response.

Math 6 Honors ( Periods 1, 2, & 3)

Prime Numbers and Composite Numbers 5-4

A prime number is a positive integer greater than 1 with exactly two factors, 1 and the number itself. The numbers 2, 3, 5, 7 are examples of prime numbers

A composite number is a positive integer greater than 1 with more than two factors. The numbers 4, 6, 8, 9, and 10 are examples of composite numbers.

Since 1 has exactly 1 factor, it is neither prime nor composite.

About 230 BCE Erathosthenes, a Greek Mathematician suggested a way to find prime numbers—up to a specific number. The method is called the Sieve of Eratosthenes because it picks out the prime numbers as a strainer, or sieve, picks out solid particles from a liquid.

You may factor a number into prime factors by using either of the following methods
➢ Inverted short division
➢ Factor tree

Both were shown in class.

Could you start the factor tree differently? If so, would you end up with the same answer?


The prime factors of 42 are the same in either factor tree, except for their order.

Every composite number greater than 1 can be written as a product of prime factors in exactly one way, except for the order of the factors.

When we write 42 as 2 ∙ 3 ∙ 7 this product is called the prime factorization of 42

Notice the order in which prime factorization is written.

Let’s try finding the prime factorization of 60

The prime factorization of 60 = 2 ∙ 2∙ 3 ∙ 5 or 2²∙ 3∙ 5

Math 6 Honors ( Periods 1, 2, & 3)

Square Numbers and Square Roots 5-3

Numbers such as 1, 4, 9, 16, 25, 36, 49... are called square numbers or PERFECT SQUARES.

One of two EQUAL factors of a square is called the square root of the number. To denote a square root of a number we use a radical sign (looks like a check mark with an extension) See our textbook page 157.

Although we use a radical sign to denote cube roots, fourth roots and more, without a small number on the radical sign, we have come to call that the square root.
SQRT = stands for square root, since this blog will not let me use the proper symbol) √ is the closest to the symbol

so the SQRT of 25 is 5. Actually 5 is the principal square root. Since 5 X 5 = 25
There is another root because
(-5)(-5) = 25 but in this class we are primarily interested in the principal square root or the positive square root.

Evaluate the following:
SQRT 36 + SQRT 64 = 6 + 8 = 14
SQRT 100 = 10
Is it true that SQRT 36 + SQRT 64 = SQRT 100? No
You cannot add square roots in that manner.
However look at the following:
Evaluate
SQRT 225 = 15
(SQRT 9)(SQRT 25)= (3)(5) = 15
so
SQRT 225 = (SQRT 9)(SQRT 25)

Also notice that the SQRT 1600 = 40
But notice that SQRT 1600 = SQRT (16)(100) = 4(10) = 40

Try this:
Take an odd perfect square, such as 9. Square the largest whole number that is less than half of it. ( For 9 this would be 4). If you add this square to the original number what kind of number do you get? Try it with other odd perfect squares...

In this case, 9 + 16 = 25... hmmm... what's 25???

Math 6 Honors ( Periods 1, 2, & 3)

Tests for Divisibility 5-2

It is important to learn the following divisibility rules:
A number is divisibility by:

2 ... if the ones digit of the number is even
3 ... if the sum of the digits is divisible by three ( add the digits together)
4 ... if the number formed by the last two digits is divisible by by four ( Just LOOK at the last two numbers-- DON"T ADD them!!)
5 ... if the ones digits of the number is a 5 or a 0
6 ... if the number is divisible by both 2 and 3... (or if it is even and divisible by 3)
8 ... if the number formed by the last three digits is divisible by 8. (Like FOUR, just look at the last three digits-- divide them by 8)
9 ... if the sum of the digits is divisible by 9
10 ... if the ones digits of the number is a 0.

You will not need to know the divisibility rules for 7 or 11 but they are interesting...

You can test for divisibility by 7
Let's start with a number 959
Step 1: drop the one's digit so we have 95
Step 2: Subtract twice the ones' digit ( that you dropped) in this case we dropped a 9
so we double that and subtract 18 from 95
or 95-18 = 77. If the results, in the case, 77, is divisible by 7 --- so is the original number 959.
Step 3: If the number you get is still to big.. continue the process until you can determine if your number is divisible by 7.


To test for divisibility by 11
add the alternative digits beginning with the first
so let's try the following
4,378,396
Step 1: Add the alternate digits beginning with the 1st 4 + 7+ 3 + 6 = 20
Step 2: Add alternate digits beginning with the 2nd 3 + 8 + 9 = 20

Step 3: If the difference of the sums is divisible by 11 so is the original number.
In this case, 20-20 = 0 and 0/11= 0 so
4,378,396 is divisible by 11.


A good test for divisibility by 25 would be if the last two digits represent a multiple of 25.

A perfect number is one that is the SUM of all its factors except itself. The smallest perfect number is 6, since 6 = 1 + 2+ 3
The next perfect number is 28 since
28 = 1 + 2 + 4 + 7 + 14
What is the next perfect number?

Math 6 Honors ( Periods 1, 2, & 3)

Finding Factors and Multiples 5-1


You know that 60 can be written as the product of 5 and 12. 5 and 12 are called whole number factors of 60. A number is said to be divisible by its whole numbered factors.

To find out if a smaller whole number is a factor of a larger whole number, you divide the larger number by the smaller.--- if the remainder is 0, the smaller number IS a factor of the larger number.

We set up T charts to find al the factors of numbers.
For example. Find all the factors of 24
24
1--24
2--12
3--8
4--6

When you go down the left side and back up the right you have
1, 2, 3, 4, 6, 8, 12, 24
all the factors of 24 in order!!!

A multiple of a whole number is the product of that whole number and ANY whole number. You can find the multiples of given whole numbers by multiplying that number by 0, 1, 2, 3, 4, ...and so on
The first five multiples of 7 are
0, 7, 14, 21, 28
because 0(7) = 0 ; 1(7) = 7 ; 2(7) = 14; 3(7) = 21; 4(7) = 28
... and put in set notation it would be
{0, 7, 14 ,21, 28}



If you were to ask for the first four NON-ZERO Multiples of 6
the answer would be 6, 12, 18, 24.. and in set notation
{ 6, 12, 18, 24}
Whereas the first four multiples of 6 ( you would need to include 0)
{0, 6, 12, 18}



Generally, any number is a multiple of each of its factors. That is, 21 is a multiple of 7 and it is a multiple of 3!!

Any multiple of 2 is called an EVEN number
A whole number that is NOT an even number is called an ODD number
Since 0 is a multiple of 2 .. that is 0 = 0(2) 0 is an EVEN number

What number is a factor of every number? ONE
Is every number a factor of itself? YES
What is ( are) the only multilpe(s) of 0? 0
How many numbers have 0 as a factor? only one number What number(s)? ZERO



The word factor is derived from the Latin word for "maker" the same root for factory and manufacture. When multiplied together factors 'make' a number.
factor X factor = product.

Algebra Honors (Period 6 & 7)

Solving Equations by Factoring Section 5-12

a⋅0 = 0
and if a = 0 or b = 0
then we know that ab= 0
This is an if, then statement
conversely
if ab = 0 then either a= 0 or b = 0
THis Zero Products Property helps us solve equations.

(x +2)(x -5) = 0
either x + 2 must equal zero or x - 5 must equal zero
so set each expression equal to zero and solve
x + 2 = 0
x= -2
and x-5 = 0
x = 5

{-2. 5}

5m(m-3)(m-4) = 0
now you have three expressions so set each of them to zero
5m = 0 so m = 0
m-3 = 0 so m=3
m-4 = 0 so m=4
{0,3,4}

What happens with
3x²+ x = 2
It isn't the 2 products property but the ZERO products property so set the expression equal to ZERO
3x²+ x -2 = 0
Now factor
(x+1)(3x -2) = 0
set each of these equal to zero
x + 1 = 0 x = -1
3x -2 = 0 so x = 2/3

10x³ - 15x² = 0
factor
5x²(2x -3) = 0
again set each equal to zero
5x² = 0 so x = 0
and
2x -3 = 0 so x = 3/2

polynomial equation named by the term of highes degree

ax + b = 0 linear equation

ax² + bx + c = 0 quadratic equations

ax³ + bx² + cx + d = 0 cubic equation

2x² + 5x = 12
becomes
2x² +5x - 12 = 0
(x + 4)(2x-3) = 0
so x = -4 and x = 3/2
{-4, 3/2}

18y³ + 8y + 24y² = 0
Rearrange first
18y³+ 24y ² + 8y = 0
Then factor the GCF
2y(9y² +12y +4) = 0
WAIT--> its a trinomial SQ
2y(3y +2)² = 0
2y = 0 so y = 0
and 3y + 2 = 0 so y = -2/3

-2/3 is a double or multiple root but you only list it once in solution set.
That is,
{-2/3, 0}

y = x² + x - 12
solve for the roots means you set this quadratic equal to ZERO
so
x² + x - 12 = 0
(x+4)(x -3) = 0

x = -4 and x = 3

Math 6 Honors ( Periods 1, 2, & 3)

Dividing Decimals 3-9 cont'd

For word Problems use the 5 step plan found on Page 18 of our textbook

296.06 ÷ (18.7 + 3.9)
Following Aunt Sally ( or PEMDAS... remember our singing...
we do the operation inside the hugs!! ( )using a sidebar
18.7 + 3.9 make sure to stack them lining up the decimals and you will get 22.6

296.06 ÷ 22.6

When dividing by a decimal remember the rule from yesterday, multiply the divisor ( 22.6) by a power of ten which makes it a natural number

then use that same power of ten and multiply the dividend,
WHen you divide you have
2960.6 ÷ 226
Please do that problem and your quotient should be 13.1

(47.1 - 16.9) ÷ (21.9 -6.8)
Again you need to do the operations inside the ( ) first. Using a side bar and lining up the decimals
47.1 - 16.9 = 30.2
and 21.9 - 6.8 = 15.1
Just take a look at those two numbers and you will notice a relationship!!
30.2 ÷ 15.1
BUT... practice your division skills and confirm what you can tell...
30.2 ÷ 15.1 = 2



At an average rate of 55 km/hour how long will it take to drive 225 km to the nearest tenth of an hour?

d = rt
What must we find and what are the clues? Well, how long... is usually time and the fact that we need to round to the nearest tenth of an hour indicates we are finding TIME as well.
So what is the distance? 225 km and what is the rate? 55 km/h
so plug into the formula
225= 55t
Now, how do we solve this one step problem?

divide both sides by 55
225/55 = 55t/55

do the division as a side bar

225/55 ≈ 4.09 so
t ≈ 4.1
and the answer is 4.1 hour

Algebra Honors (Period 6 & 7)

Using Several Methods of Factoring Section 5-11

1) Always factor the GCF first
2) look for he difference of 2 SQ's
3) Look for a Perfect SQ trinomial
4) If trinomial is NOT SQ look for a pair of factors
5) If 4 or more terms-- look for a way to group the terms into pairs or into a group of 3 terms that is a perfect SQ trinomial
6) Make sure each binomial or trinomial factor is PRIME
7) check your work
-4n⁴ + 40 n³ -100n²
GCF
-4n²(n² -10n + 25)
-4n²(n-5)²

What about
5a³b² + 3a⁴b - 2a²b³
Again factor the GCF
a²b(5ab + 3a²-2b²)
reorder this and use either XBOX or factor pairs to solve

a²b(3a²+5ab-2²)


a²b(3a²+6ab-ab-2b²)


a²b[(3a²+6ab)+(-ab-2b²)]

a²b[3a(a+2b)-b(a + 2b)]
a²b(a+2b)(3a-b)


a²bc -4bc + a² -4b
b(a²c-4c+a²-4)
b(a²c+a²-4c-4)
b[(a²c+a²) -(4c+4)]
b[a²(c + 1) -4(c +1)]
b(c+1)(a²-4)
but we aren't finished...
b(c+1)(a+2)(a-2)



6c²+18cd+12d²
6(c²+3cd+2d²)
6(c+2d)(c+d)

3xy²-27x³

3x(y²-9x²)
3x(y+3x)(y-3x)


-n⁴-3n²-2n³

-n²(n²+3+2n)
-n²(n²+2n+3)

Its factored completely!!


16x²+16y -y²-64
16x²-y²+16y-64
16x²-y²+16y-64
16²-(y²-16y+64)
16x²-(y-8)²
becomes the difference of two squares
(4x+y-8)(4x-y+8)

x¹⁶ -1
(x⁸+1)(x⁸-1) =
(x⁸+1)(x⁴+1)(x⁴-1)=
(x⁸+1)(x⁴+1)(x²+1)(x² -1) =
(x⁸+1)(x⁴+1)(x²+1)(x +1)(x-1)


2(a +2)² + 5(a +2) - 3
Think of this as letting a+ 2 = x

2x² +5x -3
Factoring that is easy
(2x-1)(x +3)
so substitute in a + 2 for each x
[2(a+2) -1]{a+2 +3]


2a + 4 -1)(a +5)
(2a +3)(a +5)

Math 6 Honors ( Periods 1, 2, & 3)

Dividing Decimals 3-9

According to our textbook-
In using the division process to divide a decimal by a counting number, place the decimal point in the quotient directly over the decimal point in the dividend.

Check out our textbook for some examples!!

When a division does not terminate-- or does not come out evenly-- we usually round to a specified number of decimal places. This is done by adding zeros to the end of the dividend, which as you know, does NOT change the value of the decimal. We then divide ONE place beyond the specified number of places.

Divide 2.745 by 8 to the nearest thousandths.
See the set up in our textbook on page 89. Notice that they have added a zero and the end of the dividend ( 2.745 becomes 2.7450) because you want to round to the thousandths and we need to go ONE place additional.
DIVIDE carefully!!

the quotient is 0.3431 which rounds to 0.343


To divide one decimal by another

Multiply the dividend and the divisor by a power of ten that makes the DIVISOR a counting number


Divide the new dividend by the new divisor

Check by multiplying the quotient and the divisor.