Math 7 ( Period 4)

Solving Multi Step Equations 4.2

Before you use inverse operations to solve an equation, you should check to see whether ONE or BOTH sides of the equation can be simplified by combining like terms!

2x + 3x - 4 = 11

5x -  4 = 11   ( Combine like termsà 2x + 3x = 5x)

we used the  addition property of equality ( +  prop=)

5x + 0 = 15    
Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

5x = 15

Now we need to divide both sides by 5 and the Division Property of Equality ( ÷ prop =)  allows us to do that

5x/5 = 15/5  Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 3  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 3 
and we BOX our answer.

Solving

-13 = 3n + 3 + n

Involves a similar process:

we need to combine like terms first

-13 = 4n + 3 ( combining 3n + 1n= 4n)

  we used the Subtraction Property of Equality ( -prop=)

-16 = 4n +0   
Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

-16 = 4n

Now we need to divide BOTH sides by 4 and the Division Property of Equality (÷prop=) allows us to do that.

-16/4 = 4n/4   

Multiplicative Inverse or the Inverse Property of Mult (Inv×)

-4 =1n   we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

-4 = n  

and we BOX our answer. 

Algebra Honors ( Periods 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 called an  “ if, then”  statement

 conversely

if ab = 0 then either a= 0 or b = 0

That is the converse.

Not all converse of true statements are true. For example

"If a polygon is a square, then it is a rectangle," is a true statement but its converse

"If a polygon is a rectangle, then it is a square"—is NOT true.

The words “ if and only if” are used to combine a statement and its converse when BOTH are true.

The Zero Products Property  is one such statement

This Zero Products Property helps us solve equations.

For all real numbers a and b,

ab= 0 if and only if a = 0 and b = 0

A product of factors is zero if and only if one or more of the factors is zero.

For example: 

(x +2)(x -5) = 0

either x + 2 must equal zero or x - 5 must equal zer0

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 zer0

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" --> we need to use the ZERO Products Property,  so set the expression equal to zer0

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

{-1, 2/3}

10x³ - 15x² = 0

factor

5x²(2x -3) = 0

again set each equal to zer0

5x² = 0 so x = 0

and

2x -3 = 0 so x = 3/2

{0, 3/2}

Polynomial equations are named by the term of highest degree

(where a≠ 0)

ax + b = 0  is a linear equation

ax² + bx + c = 0 is a quadratic equations

ax³ + bx² + cx + d = 0  is a cubic equation

Many polynomial equations can be solved by factoring and then using the zero-products property. The first step is often to transform the equation into STANDARD FORMà in which one side is zer0. The other side should be a simplified polynomial arranged in order of decreasing degree of the variable.

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 PERFECT 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}

Find an equation in standard form with integral coefficients that has the following solutions set: {2/3, -4}

This is just working backwards…

Since we know that the solution set must be {2/3, -4}, we know that

( x – 2/3)(x + 4) = 0  because we get the solution set from those two sets of hugs by using the Zero Products Property

( x – 2/3)(x + 4) = 0   is NOT in standard form so we need to use the BOX METHOD   However… since we have a fraction in one of the two sets of parenthesis.. we may want to clear that BEFORE we start using FOIL, or the BOX method. If we multiply BOTH SIDES by 3, we haven’t changed the value so  3( x – 2/3)(x + 4) = 0   Now distribute the 3( x - 2/3) becomes (3x -2)  so now we have (3x – 2)(x + 4) = 0   Now FOIL, or BOX, or FIREWORKS…. and we get 3x²+ 10x – 8 = 0

(x-1)(x+3) = 12

We can’t immediately solve this—there isn’t a “12 products property.” That is, we need to set this to “zer0.”

(x-1)(x+3)  - 12 = 0

Now FOIL, or BOX, or FIREWORKS…. à  x² -2x – 3 -12 = 0

which is x² -2x – 15 = 0

FACTOR

(x -3)(x + 5) = 0

so

x = 3 and x = -5

{-5, 3}

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

Math 7 (Period 4)

Solving Two Step Equations 4.1

Solving equations may involve using more than one inverse operation to isolate the variable (putting the variable alone on one side of the equation)

Since we are undoing the equation, we literally do PEMDAS in reverse!

3x + 8 = 23

   

 we used the  subtraction property of equality ( -  prop=)

3x + 0 = 15    

Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity Property of Addition (ID+) allows us to write

3x = 15

Now we need to divide both sides by 3 and the Division Property of Equality ( ÷ prop =)  allows us to do that

3x/3 = 15/3   Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 5  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 5 

and we BOX our answer.

Solving

x/4 – 12 = 1

Involves a similar process:

We need to add 12 to both sides

  we used the Addition Property of Equality ( +prop=)

x/4 + 0 = 13  

Here we have the Additive Inverse or Inverse Property of Addition  (Inv +)

We don’t want to write with “+ 0”

The Identity property of Addition (ID+) allows us to write

x/4 = 13

Now we need to multiply BOTH sides by 4/1 and the Multiplication Property of Equality (×prop=) allows us to do that.

Notice where the 4/1 is located and how it has  HUGS ( ) surrounding it! That’s required in my class! It shows that you are understanding the process of multiplying by the reciprocal of ¼.

  

Multiplicative Inverse or the Inverse Property of Mult (Inv×)

1x = 52  we don’t want to write 1x ->Using the Identity Property of Multiplication (ID×)

x = 52 

and we BOX our answer. 

Math 6A (Periods 1 & 2)

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!!

Math 6A ( Periods 1 & 2)

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

Fundamental Theorem of Arithmetic

Every composite number can be written as a product of prime factors in exactly one way

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.

We watch Brain Pop
http://www.brainpop.com/math/numbersandoperations/primenumbers/
If you need to log in
user name: linderocanyon
password: lcms

You should be able to view the above-- any time in the next week-- just as long as you view it BEFORE 5PM.

Algebra Honors ( Periods 6 &7)

Using Several Methods of Factoring 5-11

A polynomial is factored completely when it is expressed as the product of a monomial and one or more prime polynomials.

Guidelines for Factoring Completely

1.     Factor out the GCF first

2.    Look for a difference of two squares

3.    Look for a perfect square trinomial

4.    If a trinomial is not a square, look for a pair of binomial factors.

5.    If a polynomial has four or more terms, look for a way to group the terms in pars or in a group of three terms that is a PERFECT SQUARE TRINOMIAL

6.    Make sure that each binomial or trinomial factor is prime.

7.    Check your work by multiplying the factors

Math 6A ( Periods 1 & 2)

Square Numbers & 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???

Algebra Honors ( Periods 6 & 7)

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)

Math 6A (Periods 1 & 2)

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 (Periods 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 first showed you what I call the "Matrix" method

2x² + 7x - 9

Consider the last sign... in this case the negative.
What does that tell us?
"That the signs in the two sets of (  )(  ) are different."
What does the first sign tell us?  in this case we have a positive.
That the positive " wins."

Now consider all the factors of 2
That's easy  just 2 and 1
Set them in a column
2
1

Now consider the factors of 9
Hmm... that's  1 and 9 as well as 3 and 3
Now you need to set up a matrix
You can try out all the different combinations
2 3
1 3
or
2  1
1  9

or
2 9
1 1

What you do at this point is multiply diagonally
that is with the first matrix
2 3
1 3
You would multiply the upper left number (2) with the lower right number (3) = 6
You would then take the upper right number (3) and multiply it by the lower left (1) = 3
Ask yourself, is there anyway to get a difference of 7 (the middle term in your problem above)?
NO-- so that matrix is not correct.



2  1
1  9
Try the same with this
You would multiply the upper left number (2) with the lower right number (9) = 18
You would then take the upper right number (1) and multiply it by the lower left (1) = 1
Ask yourself, is there anyway to get a difference of 7 (the middle term in your problem above)?
NO-- so that matrix is not correct.

But with the last matrix
2 9
1 1
Try it
You would multiply the upper left number (2) with the lower right number (1) = 2
You would then take the upper right number (9) and multiply it by the lower left (1) = 9
Ask yourself, is there anyway to get a difference of 7 (the middle term in your problem above)?
Yes-- so that matrix is correct but which product needs to be + so we end up with +7? The Lower left product. Travel up the arrow and place the + in front of the number on the upper right! Then place the opposite sign on the number below it. AS this shows:
2 +9
1  -1

Now just read across..  and return the variable
(2x +9)(x - 1)
You should ALWAYS FOIL, Double DP, BOX and make sure you have factored correctly!

Tomorrow I will show 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)

Math 6A ( Periods 1 & 2)

 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 (Periods 6 & 7)

 Factoring Pattern for x² + bx+ c  where c is  negative 5-8 

Goal- to factor quadratic trinomials whose quadratic coefficient is 1 and whose constant is negative

The method used in this lesson is  very similar to that used to factor x² +bx + c , c is  positive, except instead of the sum of the two factors you  find their difference.
Remember  with x² +bx + c , c is positive, you find two numbers whose product is c and whose sum is b.

This time find two numbers whose product is c (which is negative)—so ONE of the TWO factors must be negative. You will have either  (x + )(x - )  or  (x - )(x + ) Since c is negative, one of the two factors MUST be negative.

The first sign in x² + bx + c , c is negative determines “Who wins!”  Let’s rewrite x² +bx + c , c is negative as either   x² +bx - c ,  or x² -bx - c   to see how this works.

We started with x² –x – 20
Set up your hugs…. (x -  ) (x +  )… with the winning sign going in the first set of hugs

Then using the X method find two numbers whose product is 20 and whose difference is 1 ( and in this case actually -1) We found 5 and 4 works and the 5 must be negative to get  -1

so (x -5)(x + 4)

If the quadratic was x²+x-20 you would still have the same factors 5 and 4 but this time the difference is +1 so you would have ( x +5)(x -4)

How about x² + 29a – 30

The factoring pattern is ( x + )(x -  )
Use the X method and find two numbers whose product is 30 and whose difference is 29
We find it has to be 30 and 1  (x+30)(x -1)

To check just FOIL, FireWorks, use the BOX method or just double-distribute to get back to where you started!!

x²-4kx +12k²

This time we have another variable on the last two terms so the factoring pattern starts out as
(x- _k)(x+ _k)
But again we just need to find two numbers whose product is 12 and the difference is 4
6 and 2 work so its ( x -6k)(x + 2k)

Find all the integral values for k for which the given polynomial can be factored.

c²-kc-20

For this exercise, set up a T chart with all the factors that multiply to 20

we found 1 and 20, 2 and 10, and 4 and 5.  Now taking their differences, we find that ± 19 ± 8 ± 1 all work.

Find two negative values for k  for which the given polynomial can be factored. (There are many possibilities—the class found several)

y² + 4y  + k

We found -5, -12, -77,  45, -21, … and ....

Algebra Honors ( Periods 6 & 7)

Factoring Pattern for x² + bx+ c  where c is  positive 5-7 

In this lesson we will be factoring trinomials that can be factored as a product of ( x +r)(x + s)
where r and s are both positive OR both negative integers.

x² + ( r + s)x + rs  

(x +3)(x+5) = x² + 8x + 15

(x – 6)(x -4) = x² -10x + 24

where -10 is the sum of -6 and -4

and

24 is the product of -6 and -4

Our book suggests that you list all the pairs of integral factors whose products equal the constant term Then, find the pair of integral factors whose SUM equals the coefficient of the linear term (remember your new vocab)

For factoring x² + bx + c, where c is positive  you only need to consider factors WITH THE SAME SIGNS as the linear term!!

In class, I showed the diamond method of calculating products and sums… and even recommended an app to practice!!
Here is the link for the iphone,  ipad app…

y² + 14y + 40

Since the linear term ( +14y) is positive you know to set up two sets of HUGS with + in the middle

(  +  )(  +  )

then you can add the single y’s since y² = y·y

(y + )(y + )

Now either list all the integral factors or use the diamond method and you discover that 10 and 4 are the two factors that multiply to 40 AND also sum to 14

(y + 10 )(y + 4)

To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

y2 – 11y + 18

This time notice that the linear term is -11y so you will be looking for a pair of numbers whose product will be positive  ( so both need to be negative)

Set  up your  HUGS  similarly—EXCEPT both signs need to be NEGATIVE

(y -   )( y -  )

Since -11 is negative, think of the negative factors of 18 using the book’s method or the diamond method

hmmm… -9 and -2 work

(y - 9  )( y -  2)

Again To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

A polynomial that cannot be expressed as a product of polynomials of lower degree is said to be irreducible. An irreducible polynomial with integral coefficients whose greatest monomial factor is 1 is a PRIME POLYNOMIAL.

Factor x² -10x + 14

Setting up your HUGS   you start to think what two factors multiply to 14 and SUM to 10… hmmm. NOTHING…

Therefore x² -10x + 14  cannot be factored and it is a prime polynomial

Find all the integral values of k for which the trinomial can be factored

x² + kx + 28

28 can be factored as a product—using  a T chart

list all the factors

(1)(28) (2)(14) (4)(7)

Taking the corresponding sums you get 29, 16 and 11

BUT… remember you can also have the negatives here

so the values of k can be ± 29,   ± 16,   ± 11

{-29, -16, -11, 11, 16, 29}

Algebra Honors ( Period 6 & 7)

Squares of Binomials 5-6

(a +b)² =(a+b)(a+b)

You could use foil, fireworks, the box method… to find the results of multiplying the two binomials

a² + 2ab + b²

What happens when you square the binomial difference a-b?

(a-b)² = (a-b)(a-b)

a² - 2ab + b²

Notice you have the square of the first term,  twice the product of the two terms and finally the square of the last term.

The textbook states that it is helpful to memorize these patterns for writing squares of binomials as trinomials.

(a +b)²     = a² + 2ab + b²

(a - b)²     = a² -2ab + b²

My comment-  MEMORIZE … you need to be able to see these patterns

(x + 3)² = x² + 6x + 9

( 7u -3)² = 49u² -42u+ 9 

(4s – 5t)² = 16s² – 40st + 25t²

(3p² – 2q²)²

=9p⁴-12p²q² + 4q⁴

When we need to factor… we need to realize these patterns in reverse…that is,

a² + 2ab + b^{2  =}    (a +b)²     

a² -2ab + b²   =  (a - b)²     

a² + 2ab + b²     and  a² -2ab + b²       are called perfect square trinomials

because each expression has a three terms and is the square of a binomial.

To test whether a trinomial is a perfect square… ask these three questions:

1)  Is the first term a square?

2) Is the last term a square?

3) Is the middle term twice the product of √( 1^st term)  and √(last term).

For example, is

4x² – 20x + 25 a perfect square trinomial?

 1)  Is the first term a square? YES 4x² = (2x)²

2) Is the last term a square? YES  25= (5)²

3) Is the middle term twice the product of √( 1^st term)  and √(last term).

YES  2[√( 4x²) ∙ √25)]. =  2(2x∙5) = 20x

Yu may need to rearrange the terms of a trinomial BEFORE you test whether it is a perfect square.

For example, x² + 100 -20x must be rewritten as

x² -20x + 100  so that we can answer YES to all three questions.

Take a look at

63n³ – 84n² + 28n

It doesn’t look like a perfect square BUT if you factor out the GCF or the Greatest Monomial factor…

we end up with à 7n(9n² -12n +4)

Now… that becomes  7n(3n-2)²

What about,

8u3 -24u2v + 18uv2

2u(4u² -12vu +9v²)

2u(2u-3v)²

What do we need to do to solve the following:
(x + 2)² – (x -3) ² = 35

First multiply each of the binomial squares

x² + 4x + 4 – (x² -6x+ 9) = 35

Make sure to properly employ the inverse property of a sum

x² + 4x + 4 – x² + 6x -  9 = 35

Combine like terms

10x – 5 = 35

10x = 40

x = 4

or using set notation {4}