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)

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)

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?

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

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)

Algebra Honors (Period 6 & 7)

Factoring Pattern for ax² + bx + c Section 5-9

When a > 1
We used a different method than what is taught in the book. I showed you X box

2x² + 7x -9
Multiply the 2 and the 9
put eighteen in the box
Your controllers are
2x² and -9
THen using a T chart find the factors of 19 such that the difference is 7x
we found that +9x and -2x worked

so
2x² +9x -2x -9
Then separate them in groups of 2
such that


(2x² +9x) + (-2x -9)

Then realize you can factor a - from the second pair

(2x² +9x) - (2x + 9)
Then wht is the GCF in each of the hugs( )
x(2x +9) -1(2x +9)
look they both have 2x + 9
:)
(2x +9)(x-1)
But what if you said -2x + 9x instead to make the +7x in the middle
Look what happens
(2x² -2x) + (9x -9)
now, factor te GCF of each
2x(x -1) + 9(x -1)
now they both have x -1
(x-1)(2x +9)
SAME RESULTS!!

14x² -17x +5
remember the second sign tells us that the numbers are the same and the first sign tells us that they are BOTH negative

create your X BOX with the product of 14 and 5 in it
70

Place your controllers on either side

14x² and + 5

Now do your T Chart for 70
You will need two numbers whose product is 70 and whose sum is 17
that's 7 and 10

14x² -7x -10x + 5

Now group in pairs

(14x² -7x) + (-10x + 5)
which becomes

(14x² -7x) - (10x - 5)

FACTOR each
7x(2x -1) - 5(2x-1)
(2x-1)(7x-5)

10 + 11x - 6x ²

sometimes its better to arrange by decreasing degree so this becomes

- 6x ² +11x + 10

now factor out the -1 from each terms


- (6x ² - 11x - 10)

Se up your X BOX with the product of your two controllers :)
60 We discover that +4x and -15x are the two factors

-1(6x ² +4x - 15x - 10)

-1[(6x ² +4x) + (- 15x - 10)]
-1[6x ² +4x) - (15x +10)
-1[2x(3x +2) -5(3x+2)]
-(3x+2)(2x-5)


If you had worked it out as
10 + 11x -6x² you would have ended up factoring
(5 -2x)(2 + 3x)
and we all know that
5 -2x = -(2x-5) Right ?


Next, we looked at the book and the example of
5a² -ab - 22b²
We discussed the books instructions to test the possibilities and decided that the X BOX method was much better.... I need to check out hotmath.com... did you????

5a² -ab - 22b² Using X BOX method we have 110 in the box and the controllers are
5a² and - 22b²
What two factors will multiply to 110 but have the difference -1?
Why 10 and 11

5a² +10ab -11ab - 22b²

separate and we get
(5a² +10ab) + (-11ab - 22b²)
( 5a² +10ab) - (11ab + 22b²)

5a(a + 2b) -11b(a + 2b)
(a + 2b)(5a - 11b)



Factoring by Grouping 5-10


5(a -3) - 2a (3 -a)

a-3 and 3-a are OPPOSITES
so we could write 3-a as -(-3 +a) or -(a -3)
sp we have
5(a-3) -2a [-(a-3)]
which is really
5(a-3) + 2a(a-3)
wait... look... OMG they both have a-3
so
(a-3)(5 + 2a)

What about
2ab-6ac + 3b -9c

What can you combine...
some saw the following:

(2ab -6ac) + 3b -9c)
then
2a(b-3c) + 3( b-3c)
(b -3c)(2a + 3)

BUT others look at 2ab-6ac + 3b -9c and saw
2ab +3b -6ac -9c
which lead them to
(2ab + 3b) + (-6ac -9c)
b(2a +3) -3c(2a +3)
(2a +3)(b-3c)
wait that's the same!!
Hooray

What about 4p² -4q² +4qr -r²
First look carefully and you will see

4p² -4q² +4qr -r²
That's a trinomial square OMG

so isn't that
4p² - ( 2q -r)²

BUT WAIT look at

4p² - ( 2q -r)² That's the
Difference of Two Squares
Which becomes
(2p + 2q -r)(2p -2q +r)

Algebra Honors (Period 6 & 7)

Multiplying Binomials Mental 5-4

Look at
(3x - 4)(2x+5)

remember FOIL
First terms (3x)(2x)
Outer terms (3x)(5)
Inner Terms (-4)(2x)
Last Terms (-4)(5)

6x² + 15x -8x -20
6x² + 7x - 20
Or use the box method as we have done in class
This is a quadratic polynomial
The quadratic term is a term of degree two

Remember a linear term has a term of degree 1 such as y = 3x + 5

6x² + 7x - 20

The 6x² is the quadratic term
the +7x is the linear term
and the - 20 is the constant term

(x +1)(x +3) = x² + 4x + 3

(y + 2)( y + 5) = y² + 7y + 10
( t -2)( t -3) = t² -5t + 6

( u -4)(u -1) = u² - 5u + 4

What about
( u-4)(u +1) = u ² -3u -4

See the difference between the two?

(7 - k)(4 -k)

28 - 11k + k²


r + 3)(5 - 5)
r² - 25 - 15



(3x - 5y)(4x + y)
12x² - 17xy - 5y²

a + 2b)(a-b)

careful....
a² + ab - 2b²

n(n-3)(2n+1)
first distribute the n
(n² -3n)(2n +1)
2n³ - 5n² - 3n

Solve for
(x-4)(x +9) = (x +5)(x -3)
x² + 5x - 36 = x² + 2x -15
5x - 36 = 2x - 15
3x = 21

x = 7
or in solution set notation {7}

Algebra Honors (Period 6 & 7)

Dividing Monomials 5-2

There are 3 basic rules used to simplify fractions made up of monomials.
Property of Quotients
if a, b, c, d are real numbers with b≠0 and d ≠0

ac/bd = a/b ⋅c/d
Our example was 15/21 = (3⋅5)/(3⋅7) = 5/7
The rule for simplifying fractions follows ( when a = b)
(bc)/(bd) = c/d
This rule lets you divide both the numerator and the denominator by the same NON ZERO number.

35/42 = 5/6
-4xy/10x = -2y/5 which can also be written (-2/5)x as well as with out the (((HUGS)))

c⁷/c⁴ = c⁴c³/c4 = c³
another way we proved this was to write out all the c's
c⋅c⋅c⋅c⋅c⋅c⋅c⋅/c⋅c⋅c⋅c = and we realized we were left with
c⋅c⋅c = c³
In addition, we noticed that
c⁷/c⁴ = = c^7-4 = c³
THen we considered
c⁴/c⁷ =
c⋅c⋅c⋅c/c⋅c⋅c⋅c⋅c⋅c⋅c = 1/c⋅c⋅c = 1/c³ = c^-3

Since we all agreed that any number divided by itself was = 1
(our example was b⁵/b⁵ ), we proved the following
1 = b⁵/b⁵ = b^5-5 = b⁰

We finally arrived at the Rule of Exponents for Division

if m > n
a^m/aⁿ = a ^m-n

If n > m
a^m/aⁿ = 1/a ^n-m
and if m = n
a^m/aⁿ = 1

A quotient of monomials is simplified when
1)each base appears only once in the fraction,
2) there are NO POWERS of POWERS and
3)when the numerator and denominator are relatively prime, that is, they have no common factor other than 1.

35x³yz⁶/ 56x⁵yz
5z⁵/8x²

Finding the missing factor when you are given the following
48x³y²z⁴ = (3xy²z)⋅ (______)
we find that

48x³y²z⁴ = (3xy²z)⋅ (16x²z³)

Algebra Honors (Period 6 & 7)

Factoring Integers 5-1

When we write 56= 8⋅7 or 56 = 4⋅14 we have factored 56
to factor a number over a given set, you write it as a product of integers in that set ( the factor set).
When integers are factored over the set of integers, the factors are called integral factors.

We used the T- charts ( students learned in 6th grade) to first find the positive integer factors
56 = 1, 2, 4, 7, 8, 14, 28, 56

A prime number is an integer greater than 1 that has no positive integral factors other than itself and 1.
The first ten prime numbers are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

To find prime factorization of a positive integer, you express it as a product of primes. We used inverted division (again taught in 6th grade)

504
Try to find the primes in order as divisors.
Divide each prime as many times as possible before going on to the next prime
we found 504 - 2⋅2⋅2⋅3⋅3⋅7
which we write as 2³⋅3²⋅7
Exponents are generally used for prime factors
The prime factorization is unique--> and the order should be from the smallest prime to the largest.
A factor of two or more integers is called a common factor of the integers.
The greatest common factor (GCF) of two or more integers is the greatest integer that is a factor of all the given integers.

Find the GCF(882, 945)
First find the prime factorization of each integer Then form product of the smaller powers of each common prime factor.
The GCF is only the primes (and the powers) that they SHARE!!
882 = 2⋅3²⋅7²
945 = 3³⋅5⋅7
The common factors are 3 and 7
The smaller powers of 3 and 7 are 3² and 7
You combine these as a PRODUCT and get

the GCF(882, 945) = 3²⋅7 = 63

We also talked about listing ALL pairs of factors--> thus including negative integers
For example:
List all the pairs of factors of 20
(1)(20) but also (-1)(-20)
(2)(10) and (-2)(-10)
(4)(5) and (-4)(-5)

Listing all the factors of -20, we discovered
(1)(-20) but also (-1)(20)
(2)(-10) and (-2)(10)
(4)(-5) and (-4)(5)