You have no items in your shopping cart.

Sale ends in

$299.00

$140.00

This is a video that talks about what the course is about and why you should purchase it.

Every student is different and comes to this class with varying abilities and knowledge. It does not matter if students take the pretest and check their answers in the resouce section and skip unit 1 or do unit 1 as a refresher.

It is important that the student has the knowledge and skills in unit 1 to be successful in the course.

The student will apply the rules of adding, subtracting, multiplying and dividing signed numbers to operations involving integers, fractions, and decimal numbers.The rules:

Addition - If the signs are the same then you find the sum and use the sign

If the signs are difference then you find the difference and use the sign of the larger.

Subtraction - We do not subtract we add the opposite sign and follow the rules for addition.

Multiplication and Division - If the number of negative operators is odd then the answer is negative

If it is even then the answer is positive.

To add or subtract fractions you find a common denominator and use the sign rule.

To multiply fraction you multiply the numerators, multiply the denominators, then reduce the answer remembering to use the sign rules.

To divide fraction you do not divide you multiply by the inverse and follow the multiplication rules.

The nature of an exponent tells you if it is a power or a radical(root). Positive integer exponents are called powers. Radicals or roots is when you have a fractional exponent with 1/#. A power tells you how many times you multiply a number by itself. A radical or root is asking you how many times you mulitply some number times itself to get the number. **Negative exponents flip the base. Remember to follow the sign rules for multiplication and division.**

**ex 5**^{2} = 5 x 5 = 25 3^{4} = 3 x 3 x 3 x 3 = 81

-6^{2} = - (6 x 6) = -36 (-7)^{2} = -7 x -7 = 49

4^{-2} = 1/4 x 1/4 = 1/16

ex √49 = 7 or -7 for now we will only worry about the positive answer

ex ^{3}√64 = 4 or ^{3}√-64 = -4 because three negatives make a negative

ex √-16 = ∅ has no answer at present but later on we will address how to handle this

ex^{ 5}√-32 = -2 because 5 negatives make a negative

Students will use PEMDAS to simplify mathematical expressions.

P - do what is inside the parenthesis first

E - next do all exponents (radicals are consider exponents)

MD - multiply **or** divide from left to right

AS - add **or** subtract from left to right

A mathematical expression is a numerical sentence that can contain numbers, mathematical operators, and variables but does not contain an equality or inequality sign.

To simplify expressions at this portion of the course is only to combine like terms using the rules for signed numbers.

Note: 3x and 5x are alike, 4x and 2y are not alike, and 5x and 3x² are not alike

To evaluate and expression, you stick a numerical value in for the variable in the expression.

Note: 3x + 5 when x = 4 is 3(4) + 5 is 17

To be able to write a mathematical expression we need to know common vocabulary for mathematical operators.

Addition - sum, plus, add, total, all together, more than (8 more than a number)

Subtraction - difference, minus, take away, subtract, less than (4 less than a number)

Multiplication - product, times, twice, doubled, tripled

Division - quotient, divided by, into, ratio, over

Equals - is, is the same as, means

Others are squared, cubed, square root, cube root, greater than, greater than or equal to, less than, less than or equal to, is at least, at the most, etc

EG the sum of three times a number and seven is 3x + 7

EG four more than the quotient of a number and five is x/5 + 4

There are 5 steps to solving equations with one varaible. Thes problems are the ones that just require students to do the opposite until we get the varaible by itself called **DTO. **

**5 steps**

**Remove parenthesis RP - **remove all parenethesis by using the distributive property

**Remove all fractions RF - **remove all fractions by mulitplying every term by the LCD(lowest common denominator)

**Combine like terms CLT - **students will combine like terms on both sides of the equation

**Isolate the variable ITV - **if there are variables on both sides of the equation you need to get them moved to one side of the equation

**Do the opposite DTO - **do the opposite operation until you get the variable by itself.

Remember there are 5 steps to solving an equation with one variable.

Remove Parenthesis by distributing

Remove Fractions by multiplying every term by the common denominator

Combine Like Terms on each side of the equation to make it simpler

Isolate The Variable if there are variable on both sides move one of them

Do The Opposite operation until you get the variable by itself

Students will use the steps on solving equations with one variable to solve inequalities with one variable with the following exception. **If you multiply or divide by a negative number in the DTO step you must change the direction of the inequality.** *In addition to solving inequalities your are often asked to graph them on a number line and write the answer in interval notation. This is the focus of the next lesson.*

Inequalities

greater than > x>6 reads x greater than 6

less than < x< 4 reads x less than 4

greater than or equal to ≥ m ≥ 5

reads m is greater than or equal to 5

less than or equal to ≤ n ≤ 3

reads n is less than or equal to 3

Ex 2x - 5 < 9 2x < 14 2x < 7

Ex 8 - 3x > 17 -3x > 9 x < -3 **the direction of the inequality switched because we divided by a negative number**

To graph or write an inequality in interval notation you must understand the number line.

<--|----|----|----|----|----|----|----|----|----|--->

-∞ -20 -15 -10 -5 0 5 10 15 20 ∞

**Graphing**

> parenthesis at point going right (

< parenthesis at point going left )

≥ bracket at point going right [

≤ bracket at point going left ]

**Interval notation**

x > 7 goes right (7 , ∞)

x < 12 goes left (-∞, 12)

x ≥ -2 goes right [ -2 , ∞)

x ≤ 3 goes left ( -∞ , 3]

In real life the problems we are given to do are usually in some word format or diagram or instruction. We are expected to take that information and come up with a solution. One way to do that is to use the information, diagram, or instruction to write and equation or inequality and solve it. Students will use skills developed in unit 1 lesson 5, and the previous lessons in this unit to write and solve equations or inequalities.

In order to help us understand this concept it is useful to think of a numberline.

ex x > 4 or x < 1 this means all numbers larger than 4 and all numbers less than 1. Many numbers fit one or the other but none fit both of them. **Or statements are also called unions. **On a graph or statements will face out.

ex x < 8 and x > 3 this means all numbers that are less than 8 and also greater than 3. Only numbers between 3 and 8 meet both conditions at the same time. This statement can also look like this 3<x<8. **And statements are also called intersections. **On a graph and statements will face in.

Many times you are required to graph compound inequalities on a number line as part of the solution.

Solving an and statement written like -4<x+5<12. You still get x by itself but whatever you do to the middle term you must do to both end terms.

To determine if a relation is a function:

If it is a table or order pair then if x does not repeat then it is a function.

If it a graph then a function exist if no vertical line will cross it more than once.

Domain describes all the possible values of x, horizontal values.

Range describes all the possible values of y, vertical values.

Domain and range can be described in set builder notation or interval notation.

Set builder notation example {x | x > 8 } reads x such that x is greater than 8

A function desribes the relationship between two variables (usually x and Y).

One is called the independent variable (domain) and the other is the dependent variable (range).

Ex If you are an hourly employ then hours is the independent variable and pay is the dependent variable. Hours determines pay.

Functional notation replaces y with f(x).

If f(x) = 2x + 5 then f(2) means when x is 2 what is y y = 2(2) + 5 = 9

If f(x) = 3x - 1 then f(x) = 10 means what is x when y is 10 10 = 3x +1 x = 3

Students that are given a relationship between two variables can write the function or equation that describes the relationship and solve.

Students given a table will recognize the relationship and write an equation.

y | 4 | 7 | 10 | 13 |

x | 1 | 2 | 3 | 4 |

In the above table y is increasing by 3 while x is increasing by 1. This means that y is 3 times x.

Y = 3x but 3 time 1 is 3 not 4 so we need to add one.

y = 3x + 1 works for all cases.

In direct variation, variables move the same way. As one increases the other increases. An example is the more you exercise the more you sweat. Standard equation is y = kx where k is a constant of variation.

In inverse variation, variables move the opposite way. As one increases the other decreases. An example is as the diameter of a pipe decreases then the pressure in the line increases. Standard equation is y = k/x where k is a constant of variation.

Joint variation is a combination of the two. An example is as demand goes up, price goes up, but supply goes down. p = kd/s

Companies use variation constants to determine price, cost, etc.

In the documents section you will find notes that can be made into a foldable that in detail explain the parent functions.

**Name Parent Equation Translation**** Equation**

Linear y = x y = a(x-h) + k

Absolute Value y = |x| y = a|x-h| + k

Quadratic y = x² y = a(x-h)² + k

Square root y = Γx y = aΓ(x-h) + k

Rational y = 1/x y = a/(x-h) + k

Cubic y = x³ y = a(x-h)³ + k

Exponential y = ab ^{x} y = a(b) ^{(x-h)} + k

Logarithmic y = logx y = a log(x-h) + k

**Details can be found in the document section of this lesson.**

Most functions have the variables of a, h, and k.

** value of a** - if a > 1 then the function is vertically stretched or gets steeper, if a < 1 then the function is vertically compressed or gets flatter, if a is negative then the function is reflected across one of the axis

**value of h -**** **moves the function left or right, since most equations have negative h it is the opposite of what you might think, positive value moves it left and negative moves it right

**value of k -**** **moves the function up or down, positive value moves it up and negative value moves it down

**There is free application called desmos graphing calculator that is very useful for this course and will give you great insight in the translation of parent functions.**

**Exponential functions and logarithmic functions have a variable called b (for base) which we will discuss in more detail when we get to those functions.**

**Important: I mentioned that I would add a video for Desmos Graphing Calculator but there are numerous ones onYou Tube. There is one that talks about translations of parent functions specifically.**

**The first document in this lesson is a set of notes that covers all linear equations. I would suggest that you print it to help you in this unit.**

A linear equation is the equation of a line. It has either x or y or both. Two descriptors, slope and y-intercept, tells us a lot about a linear equation.

The standard equation of a line is ax +by = c . The slope intercept form is y = mx + b (where m is the slope and b is the y-intercept).

To change from standard to slope intercept you just get y by itself.

Ex 4x - 3y = 12 or -3y = -4x + 12 or y = 4/3 x - 4

To determine if a table is linear there must be a constant rate of change between x and y. (change in y/change in x also called slope)

y | 4 | 7 | 10 | 13 | 16 |

x | 1 | 2 | 3 | 4 | 5 |

y is changing by 3 while x is changing by 1. rate of change or slope = 3/1

**For a table to be linear there must be a constant rate of change.**

There are 3 ways to graph a linear equation.

First you can make a table or set of ordered pairs and plot the points on a coordinate plane. Pick values for x and find the y values by plugging the x value into the equation.

Second you can find the x-intercept by putting a value of zero in for y. Then you can find the y-intercept by putting a value of zero in for x. Then draw a line through the intercepts.

Finallly and most useable is to get the equation into y = mx +b and use the slope and y-intercept to graph the line.

Using the following equation 2x - y = 4 we will walk through all methods.

First If x = 2 then y = 0, if x = 3 then y = 2, if x = 1 then y = -2 Plot the points (2,0) (3,2) (1,-2) draw a line

Second If x = 0 then y = -4 and if y = 0 then x = 2 Plot the intercepts (0,-4) and (2,0) draw line

Finally 2x - y = 4 -y = -2x + 4 y = 2x - 4 Put a point at -4 on the y-axis. **Slope is rise over run of change in y over change in x. **From the point -4 on the y-axis we move up 2 and right 1 and put another point. Move up 2 and right 1 from the point you just made and make another point. Connect the points.

Slope is m in the equation y = mx +b. It is the rise/run on a graph. It is change in y over change in x given ordered pairs or a table. It is also known as rate of change.

On a graph you look at the rise/run. How much does a line go up between points over how much it is moving right or left. **Positive slopes go up and right. Negative slopes go up and left. **If you look on a graph and line is going up 5 and right 2 then the slope is 5/2. If you look on a graph and the line is going up one and left three then the **sope** is -1/3.

Given two points you can find the slope by change in y over change in x or **(Y2 - Y1)/(X2 -** X1)

Given the points (3,7) and (5, 10) the slope is 3/2 or (10-7)/(5-3) = 3/2

Verical lines have no slope because rise/0 is not possible.

Horizontal lines have a slope of zero because 0/run is zero

**Parallel lines have the same slope. Perpendicular lines have negative reciprocal slopes.**

Given a table you can find the slope and the y-intercept.

y | 7 | 9 | 11 | 13 | 15 |

x | 2 | 3 | 4 | 5 | 6 |

The change in y is and the change in x is 1. So the slope it 2. If we use first pair and the slope in y = mx + b 7 = 2(2) + b b is 3 y = 2x + 3

Given a graph you can use the information on the graph(slope, y-intercept, horizontal, vertical) to write the equation.

Given the slope and intercept you just plug that into y = mx + b

Given the slope and a point you plug into y = mx + b to find b

Given two points you can use the points to find the slope. Then you can use the slope and one of those points and plug into y = mx + b

Given a point and information about the slope we can plug that into y = mx +b

Example What is the equation of a line parallel to Y = 3x + 7 going through the point (5 , 6)

Since it must parallel then it has to have a slope of 3. 6 = 3(5) + b b = -9 answer y = 3x - 9

Greater than or less that are dotted lines.

Greater than or equal to or less than or equal to are solid lines.

If you get y by itself it will tell you which way to shade. Greater than above the line and less than below the line.

**If a point is a solution is will be in the shaded portion of the graph of plugged into the equation will make it true.**

Example 6x + 3y > 9 3y > -6x + 9 y > -2x + 3

You would graph the line just as normal but it will be a dotted line not a solid line.

Since it is y > then you would shade above the line.

Is (6 , 2) part of the solution? Yes it is in the shaed region or is 2> -2(6) + 3 and that is yes

The following is an example of the same problem solved by elimination and substitution.

**Elimination**

2x + y = 10 multiply this equation 2 to eliminate y 4x + 2y = 20

3x - 2y = -6 3x - 2y = -6

7x = 14 or x = 2 plug into one equation 2(2) + y = 10 y = 6 solution (2,6)

notice it makes both equations true when you plug it in

**Substitution**

2x + y = 10 get y by itself y = -2x + 10 plug into second equation 3x -2(-2x + 10) = -6

3x - 2y = -6 3x + 4x - 20 = -6

7x - 20 = -6

7x = 14 x = 2 plug it into one equation 2(2) + y = 10 y = 6

solution (2,6) notice it makes both equations true when you plug it in

2x + y = 1

3x - 2y = -6

**Matrices**

| 2 1 | | 10 |** d = | 2 1 | = 2(-2) - 3(1) = -4 - 3 = -7** to find dx you replace the x column with the answers column

| 3 -2 | |-6 | ** | 3 -2 | dx = | 10 1 | = 10(-2) - (-6)(1) = -20 + 6 = -14 ** to find dy you replace the y column with the answers

**|-6 -2 | dy = | 2 10 | = 2(-6) - 3(10) = -12 - 30 = - 42**

** | 3 -6|**

x = dx/d = -14/-7 = 2 y = dy/d = -42/-7 = 6 answer is (2,6)

**Graphically**

get y by itself and graph the lines and see where they meet

2x + y = 1 y = -2x + 1

3x - 2y = -6 -2y = -3x - 6 y = 3x/2 + 3

they will meet at (2,6)

Word problems usually have some form of double relationship. Find the two relationships and write the equations and solve them.

Example

One number is 4 more than another number. The sum of the two numbers is 84. What are the two numbers?

x + 4 = y x + y = 84

Example

The snack bar sold 151 drinks at lunch. If they sold 7 more mediums than larges, then how many of each size did they sell?

M + L = 151 M = L + 7

Example

George is twice the age of his brother. If the total of their ages is 117, then what is the age of his brother?

G = 2b G + b = 117

Below is an example of the same problem worked with elimination and substitution. x + 2y - z = 2 , 2x + y + z = 7, 2x - y + 2z = 6

Elimination Substitution

I will chose to eliminate z I will get x by itself in the first equation and substitute it into equations 2 and 3

E1 x + 2y - z = 2 2( E1) 2x + 4y - 2z = 4 E1 x + 2y - z = 2 x = 2 - 2y + z

E2 2x + y + z = 7 E3 2x - y + 2x = 6 E2 2(2 - 2y + z) + y + z = 7 4 - 4y + 2z + y + z = 7 NE2 -3y + 3z = 3

NE1 3x + 3y = 9 NE2 4x + 3y = 10 E3 2(2 - 2y + z) - y + z = 6 4 - 4y + 4z - y + z = 6 NE3 -5y + 4z = 2

-1(NE1) -3x -3y = -9 5(NE2) -15y + 15z = 15

NE2 4x + 3y = 10 -3(NE3) 15y - 12z = -6

x = 1 3z = 9 z = 3

NE! 3(1) + 3y = 9 3y = 6 y = 2 NE2 -15y + 15(2) = 15 y = 2

E1 1 + 2(2) - z = 2 5 - z = 2 -z = -3 z = 3 E1 x + 2(2) - 3 = 2 x = 1

Systems of equations is based on multiple relationships between two or three variables. If you can write an equation that models those relationships then you can just solve them using substitution, elimination, and martices.

Mathethematic example: The sum of two numbers is 87. One number is 11 more than 4 times the other number E1 x + y = 11 E2 x = 4y + 11

Financial example: usual form is x + y = # and ax + by = c$ The theater sold 232 tickets today for $1463. Adult tickets are $8, student tickets are $5, and senior tickets are $6.50. They sold 30 more student tickets than adult tickets.

E1 a + s + r = 232 E2 8a + 5s + 6.5r = 1463 E3 s = a + 30

Chemistry example: A pharmacist needs to make 400 ml of 15% acidic solution. He only has 12% and 20% solution. How much of each solution should he use? E1 x + y = 400 E2 .12x + .20y = .15(400)

Physics example: D=RT A boat is making a roundtrip excursion on a river. The current of the river is 15 mph. It takes the boat 4 hours downstream and 6 hours upstream. E1 D = (S + 15)(4) E2 D = (S - 15)(6)

You would simply graph both inequalities and where they overlap would be the solution set. Be careful if you have dotted lines then point on those lines are not part of the solution.

To verify if a point is a solution then you look on the graph or see if it makes both inequalities true.

Standard equation is y = |x| or y = a|x - h| + k. The variable a makes it wide or narrow and if negative makes it face down. The variable h moves it right or left and the variable k moves it up or down.

Ex1 Y = 2|x + 3 | - 2 2 makes it more narrow, moves it left 3 and down 2

Ex2 y = -.4|x - 4| - makes it face down, .4 makes it wider, moves right 4

Absolute value equations have a vertex where they turn either up or down.

Ex1 has a vertex at (-3, -2) , D:{x|x is all real numbers}, R: {y|y≥ -2}

Ex2 has a vertex at (4,0), D: {x|x is a real numer}, R: {y|y≤ 4}

**Functions that are abolsute value inequalities are graphed the same except for the solid or dashed line and you shade above or below.**

**Y> |x| dotted and shaded above**

**y< |x| dotted and shaded below**

**Y≥ |x| solid and shaded above**

**y≤ |x| solid and shaded below**

**Rule | x | = a then solve x = a or x = - a**

**Process is to get the absolute value by itself and the follow the rule.**

**Ex | 2x + 3 | -4 = 8**

** | 2x + 3 | = 12 so 2x + 3 = 12 or 2x + 3 = - 12**

** 2x = 9 2x = - 15**

** x = 9/2 x = -15/2**

**Rules |x| > a same for ≥ x > a or x < -a**

** |x| < a same for ≤ - a < x < a**

**Important remember to get the absolute value by itself before you try to use the rules.**

**Ex 1 | x - 3 | + 5 > 8**

** | x - 3 | > 3 x - 3 > 3 or x - 3 < -3**

** x > 6 or X < 0 (-∞, 0) u (6, ∞) <---------------}--------------(--------->**

** 0 6**

**Ex 2 2|x + 4 | ≤ 14**

** |x + 4 | ≤ 7 -7 ≤ x + 4 ≤ 7**

** -11 ≤ x ≤ 3 [-11, 3] < --------[--------------------------------]---------->**

** -11 0 3**

**Note: | x | < negative number has no solution**

** | x | > negative number see the notes in document section**

To add polynomials you combine like terms. Ex1 2x + 6 - 7x + 8 is -5x + 14 Ex2 4x + 8y - 7x + 3y is -3x + 11y

Ex3 (5x^{2} - 4x + 9) +( 3x^{2} + 8x - 1) is 8x^{2} + 4x + 8

To subtract polynomials you change the signs of the one that is subtracting and then combine like terms

(13x^{2} + 9x - 15) - (7x^{2} - 4x + 8) is 13x^{2} + 9x - 15 - 7x^{2} + 4x -8 is 6x^{2} + 13x - 23

When you multiply polynomials you multiply the coefficients(the numbers in front of variiables) and you add the exponents

EX1 4x(7x) ia 28x^{2} Ex2 3x(5x + 9) is 15x^{2} + 27x EX3 3x^{2}(2x^{2} - 5x + 7) 6x^{4} - 15x^{3} + 21x^{2}

If you are multiplying binomials you multiply every term in the first binomial by every term in the second binomial also called **FOIL**

EX1 (x + 5)(x - 4) is x^{2} -4x + 5x - 20 is x^{2} + x - 20 EX2 (2x - 9)(3x + 2) is 6x^{2} + 4x - 27x - 18 is 6x^{2} - 23x - 18

**Special cases ( a ± b) ^{2} is a^{2} ± 2ab + b^{2} Ex (x - 8)^{2} ix x^{2} -16x + 64 or Ex (x + 3)^{2} is x^{2} + 6x + 9**

**(a + b)(a - b) is a ^{2} - b^{2} Ex (x + 4)(x - 4) is x^{2} - 16 or EX (2x + 7)(2x - 7) is 4x^{2} - 49**

When you divide you divide the coefficients and subtract the exponents or you can look at it as reducing a fraction

EX x^{4}/x^{2} = x^{2} or 12x^{5}/4x = 3x^{4} or (7x^{5} - 4x^{3})/4x = 7x^{5}/4x - 4x^{3}/4x = 7x^{4}/4 - x^{2}

One way to solve polynomial equations is to factor out the Greatest Common Factor.

The GCF is what each term has in common with each other

EX 4x + 12 we can factor out 4 and we get 4(x + 3) Ex 12xy - 3x we can factor out 3x and we get 3x(4y - 1)

EX 15x^{2}y^{3} + 9xy^{2} we can factor out 3xy^{2} and we get 3xy^{2}(5xy + 3)

The coefficients are numbers and students are use to finding the GCF given numbers

If you have the same variables in each expression then you factor out the lowest exponent

EX 16x^{4}y^{7} - 8x^{3}y^{2} + 10x^{2}y^{3} we can factor out 2x^{2}y^{2} and we get 2x^{2}y^{2}(8x^{2}y^{5} - 4x + 5y)

**You will use GCF to solve equations**

**Ex 5x - 20 = 0 so 5(x-4) = 0 so x -4 = 0 or x - 4**

**EX 4x ^{2} - 12x = 0 so 4x(x-3) = 0 so x = 0 or x = 3**

**Difference of squares a ^{2} - b^{2} = (a + b)(a - b)**

Ex x^{2} - 49 - (x + 7)(x - 7) Ex x^{2} - 81 = (x + 9)(x - 9) Ex x^{2} - y^{2} = (x + Y)(x - y) Ex 4x^{2} - 25y^{2} = (2x + 5y)(2x - 5y)

**Difference of cubes a ^{3} - b^{3} = (a - b)(a^{2} + ab +b**

Ex x^{3} - 27 = x^{3} - 3^{3} so a is and b is 3 (x - 3)(x^{2} + 3x + 9)

Ex m^{3} - 64 = m^{3} - 4^{3} so a is m and b is 4 (m - 4)(m^{2} + 4m + 16)

**Sum of cubes a3 + b3 = (a + b)(a2 - ab +b2) always write as a sum of cubes**

Ex x^{3 }+ 125 = x^{3} + 5^{3} so a is x and b is r (x + 5)(x^{2} - 5x + 25)

Ex 8 + m^{6} = 2^{3} + (m^{2})^{3} so a is 2 and b is m^{2} (2 +m^{2})(4 + 2m^{2} + m^{4})

**These are the typical binomials that factor plus the one that has a GCF.**

Using the premise that to factor you reverse the multiplication emerges from studying the product of two binonials.

(x + 4)(x+5) = x^{2} + 4x + 5x + 20 = x^{2} + 9x + 20 The sign in front of 20 is plus so we are looking for a sum of factors. What two numbers multiplied together give you 20 and add up to 9. The numbers are 4 and 5. Since it is a sum they must be the same sign and since the sign in from of 9 is positive the both must be positive.

(x + 7)(x - 2) = x^{2} -2x + 7x -14 = x^{2} + 5x -14 The sign in front of 14 is negative so we are looking for a difference of factors. What two numbers multiplied together give you 14 and have a difference of 5. The numbers are 7 and 2. Since it is a difference the signs must be different. The sign in front of 5, which is positive, tells you the sign of the largest number which is 7.

Ex Factor x^{2} - 11x + 24 The + in front of 24 says we are looking for a sum. What numbers multiplied together give you a product of 24 and a sum of 11. They are 8 and 3. Since it is a sum the signs are the same and the sign in front of 11 is negative so they both must be negative. (x - 3)(x -8)

Ex Factor x^{2} - 7x - 30 The - in front of 30 says we are looking for a difference. What numbers multiplied together give you a product of 30 and a difference of 7. They are 3 and 10. Since it is a difference then the signs must be different and the sign in front of 7 tells you the sign of the largest number which is 10. (x + 3)(x - 10)

Please look at the example for when there is a number in front of the x^{2} because now you must use a combination of the factors of the first and last number to find a sum or difference that equals the miiddle number.

If you have four terms it is easy to spot the special difference of squares because the 1st, 3rd and 4th terms must be perfect squares. In addition the 2nd term must 2 times the sqaure root of the first and third terms. Finally there must be a negative sign in front of the 4th term.

**a ^{2} ± 2ab + b^{2} - c^{2} = (a ± b)^{2} - c^{2} = (a ± b + c)(a ± b - c)**

Ex x^{2} - 12x + 36 - y^{2} = (x - 6)^{2} - y^{2 }= (x - 6 + y)(x - 6 - y)

Ex x^{2} + 10x + 25 - y^{2} = (x + 5)^{2} - y^{2} = (x + 5 + y)(x + 5 - y)

If you four terms that are not the special difference of squares format then you will use grouping. You look for a GCF in the first two terms and then you look for GCF in the last two terms.

Ex 4x + 8 + ax + 2a = 4(x + 2) + a(x + 2) = (x + 2)(4 + a)

Ex 7m -21 + mn - 3n = 7(m - 3) + n(m - 3) = (m - 3)(7 + n)

Quadratics have two forms: **Standard form y = ax ^{2} ± bx ± c where the vertex is (-b/2a, f(-b/2a)) axis of symmetry is x = -b/2a**

** Vertex form y = a(x - h) ^{2} + k where (h, k) is the vertex and axis of symmetry x = h**

**In addition a makes it narrow or wide and also tells you way it opens, h moves it left or right, and k moves it up or down.**

Finding the vertex and graphing in standard form

Ex 1 y = x^{2} - 6x + 8 x = -b/2a = -(-6)/2(1) = 3 y = (3)^{2} -6(3) + 8 = -1 Vertex (3, -1) since vertex is at 3 we will go three places each way

x | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

y | 8 | 3 | 0 | -1 | 0 | 3 | 8 |

Notice the function is symmetric to the axis of symmetry x = 3 and it opens up **notice you have roots at (2, 0) and (4, 0)**

Finding the vertex and graphing in vertex form

Ex 2 y = -(x + 1)^{2} + 4 vertex is (h, k) or (-1, 4) since the vertex is at -1 we will go three places each way

Notice the functon is symmetric to the axis of symmetry x = -1 and it opens down **notice you have roots at (-3, 0) and (1, 0) in table below**

x | -4 | -3 | -2 | -1 | 0 | 1 | 2 |

y | -5 | 0 | 3 | 4 | 3 | 0 | -5 |

The process is very simple. Get the quadratic to one side of the equation and set it equal to zero. Factor and then set each factor equal to zero to find the roots.

Ex1 Y = x^{2} + 7x +10 x^{2} + 7x + 10 = 0 (x + 5)(x + 2) = 0 x+5 = 0 or x = -5 and x+2=0 or x = -2

Ex2 Y = 4x^{2} + 8x 4x^{2} + 8x = 0 4x(x + 2) - 0 4x=0 or x = 0 and x+2=0 0r x = -2

Ex3 f(x) = x^{2} -5x - 14 x^{2} - 5x - 14 = 0 (x - 7)(x + 2) = 0 x-7=0 or x = 7 and x+2=0 or x = -2

Ex4 f(x) = x^{2} - 16 x^{2} - 16 = 0 (x + 4)(x - 4) = 0 x+4=0 or x = -4 and x-4= 0 or x = 4

Ex5 y = x^{2} - 12x + 27 x^{2} - 12x + 27 = 0 (x - 3)(x - 9) = 0 x-3=0 or x = 3 and x-9=0 or x=9

Ex6 f(x) = 2x2 -x - 6 2x2 - x - 6 = 0 (2x + 3)(x - 2) = 0 2x+3=0 or x = -3/2 and x-2=0 or x = 2

**This method is based on the fact that (a ± b) ^{2} = a^{2} ± 2ab + b**

Get terms with variables by Ex1 y = x^{2} + 6x - 7 Ex2 y = x^{2} - 8x + 3 Ex3 y = x^{2} +4x + 11

themselves. Divide the x^{2} + 6x - 7 = 0 x^{2} - 8x + 3 = 0 x^{2} + 4x + 11 = 0

coefficent in front x by 2. x^{2} + 6x = 7 x^{2} - 8x = -3 x^{2} + 4x = -11

Square that number and x^{2} + 6x + 9 = 7 + 9 x^{2} -8x + 16 = -3 + 16 x^{2} + 4x + 4 = -11 + 4

add to both sides. (x + 3)^{2} = 16 (x - 4)^{2} = 13 (x + 2)^{2} = -7

Complete the square. x + 3 = ±√16 x - 4 = ±√13 x + 2 = ±√-7

Take the square root x = -3 ± 4 x = 4 ±√13 x = -2 ±√-7

of both sides. Solve for x. x = -3 + 4 or -3 - 4 x = 4 + √13 or 4 - √13 x = -2 + √-7 or -2 - √-7

x = 1 or x = -7

**You cannot take the square root of a negative number so you put i in front of the radical x= -2 + i√7 or -2 - i√7**

**Vertex form Ex1 y = (x + 3) ^{2} - 16**

**You can use the quadratic equation when you have the format ax2 ± bx ± c y =( - b ± √b ^{2} - 4ac )/ 2a**

Ex1 y = x^{2} + 6x -3 y = ( - 6 ± √6^{2} - 4(1)(-3))/2(1) = ( - 6 ± √36 + 12)/2 = ( -6 ± √48)2 = ( -6 ± √16·3)/2

= ( -6 ± 4√3)/2 = -3 ± 2√3

Ex2 y = x^{2} -5x + 4 y = ( -(-5) ± √(-5)^{2} - 4(1)(4))/2(1) = (5 ±√25-16)/2 = (5 ± √9)/2 = (5 ± 3)2 = 8/2 or 2/2 = 4 or 1

Ex3 y = x^{2} + 3x +6 = (-3 ± √3^{2} -4(1)(6))/2(1) = (-3 ± √9-24)/2 = (-3 ± √-15)/2 = (-3 ± i √15)/2

Ex4 y = 3x^{2} -x - 4 = (-(-1) ± √(-1)^{2} -4(3)(-4))/2(3) = (1 ± √1 + 48)/6 = (1 ± √49)/6 = (1 ± 7)/6 = 8/6 or -6/6

= 4/3 or -1

**If you have x ^{2} and a number then get the x^{2} by itself and take the square root of both sides**

Ex5 y = x^{2} - 12 x^{2} = 12 x = ± √12 x = ± √4·3 x = ± 2√3

Ex6 y = x^{2} + 7 x^{2} = -7 x = ± √-7 x = ± i √ 7

Answers in the resource section

**Course Overview**

- 8 units
- 8 unit exams with answer keys
- 1 pretest skills test for the first unit
- 7 unit reviews with video solutions
- 39 lessons and videos with exercises with answer keys
- over 400 problems to solve
- 39 lesson video solutions explaining the steps to solve each problem
- 10 additional detailed notes to the document sections of the lessons
- 1 final exam with answer key
- over 14 hours of instructional and solution videos
- Solutions to all exams in the resource section

**Course Goals**

Upon completion of this course students will be prepared to tackle the more difficult concepts found in Algebra II and better prepared to future math courses.

**Target Audience**

This video-course is designed for students that need mathenatical concepts explained simply and then reinforced through practice and application.

**Course Requirements**

Students taking this course should have completed Algebra I or an equilvatent course and be familiar with graphing and the coordinate plane.

**Course Topics**

**Unit 1 - prerequisite knowledge and skills**

Operations with signed numbers

Powers, exponents and radicals

Order oof operations

Simplifying and evaluating mathematical expressions

Reading and writing mathematical expressions

**Unit 2 - Solving equations and inequalities**

Solving one and two steps equations

Solving multiple step equations

Solving inequalities

Inequality graphs and solution notations

Writing and solving word/stated problems

Compound inequalities

**Unit 3 - Introduciton to functions**

Introduction to functions

Writing and recognizing functions

Variation

Intoduction to parent functions

Translation of functions

**Unit 4 - Linear functions and equations**

Defining a linear equation/function

Graphing linear equations

Slopes and their meaning

Writing linear equations

Graphing linear inequalities

**Unit 5 - Linear systems of equations**

Solving 2 x 2 systems of equations

Writing and solving 2 x 2 systems of equations

Solving 3 x 3 systems of equations

Real applications of systems of equations

Graphing systems of linear inequalities

**Unit 6 - Absolute value equations and functions**

Absolute value functions

Solving absoute value equations

Solving absolute value inequalities

**Unit 7 - Working with polynomials and factoring**

Operations with polynomials

Factoring GCF

Factoring binomials

Factoring trinomials

Factoring polynomials with 4 terms

**Unit 8 - Quadratic equations and functions**

Characteristics of quadratics/parabolas

Solving quadratics by factoring

Solving by completing the square

Solving using the quadratic equation

- Teacher: Scott
- Areas of expertise: Mathematics, Business, instruction, teaching, training, and management
- Education: AS mathematics, BS mathematics with teacher certification in the state of Texas, MBA emphasis MIS, Mid-management Certification
- Interests: puzzles, music, history, ancestry
- Skills: experienced manager in oil and gas exploration, writing programs and running jobs on a computer system, teaching and training, writing curriculum and teaching materials, good awareness of target audience
- Associations: State certified teacher (Texas) Founding member of Beta Theta Chapter of Alpha Kappa Lambda Youth sports coach
- Issues I care about: concern about the general decline in morality and attacks on Christianity, decline in social skills, reduction in the amount of quality time families spend together, the lack of work ethic that is becoming accepted by society and business, education of students that will be an asset in their future development and job opportunities

The courses I design are focused on giving students skills that can be carried forward. Education is not magical. It is based on learning the concepts and principals, practicing the use and application of those concepts and principals, and then assessing to see if you understand and can apply them. Athletic teams and extra curricular activities only succeed if the participants practice and get good at what they are required to do. Leaning is a partnership that involves the students, the teacher, and the family working together to achieve a common goal.

This pretest is provided because some students already have the prerequisite skills in unit 1 and do not need to do it. Results on this test will let them know if they can skip unit 1.

This the answers to the Unit 1 test. There is also a document that says Unit 1 test explanation.

This exam will be very challenging for the students. If they do not do well have them go back and look at the notes and retake it. Because knowing how to factor is critical to success.