Quadratic Factoring and More

Through these classes, there has been a collection of vocab words:
Factoring Quadratics: finds the roots  of a quadratic equation. Factoring quadratic equations in standard form, , can often be accomplished by finding two numbers that add to give b, and multiply to give ac
Factoring by Grouping: means that you will group terms with common factors before factoring as well as common factors inside of a parenthesis. 

Over the past few classes, D block covered and reviewed a couple key topics. First we looked another way or secret to factor an expression. By adding and subtracting a specific factor (which is only adding zero to the expression) you can absorb one of the added factors and continue to solve and factor.

Example #1:

X^4+X^2+1 hint: add and subtract X^2

X^4+X^2+1+X^2-X^2

Now you have to absorb one of the factors that you just added to the expression.

X^4+2X^2+1-X^2

(X^2+1)^2-X^2

Now this is just difference of two squares

A=X^2+1

B=X

=(X^2+1+X)(X^2+1-X)

Example #2:

X^4+4 hint: add and subtract 4X^2

X^4+4+4X^2-4X^2

A=X^2

B=4

(X^2+4)^2-4X^2

(X^2+4-2X)(X^2+4+2X)

Another topic was introduced to the class called Factoring Quadratics.

Here is a video on how to factor a quadratic a basic but helpful way

Click Here to watch video 

There are a couple of ways that you can speed the process up and the way that was taught in class is in a table form. But first here are three example that I think give a good range as to how to factor quadratics with using the table method to help.


Example #1: This is a simple and great star as were get into factoring quadratics
X^2+8X+12
In this case a= (X^2), b=(8X), and c=(12) but we are only going to use the actual consistence of each term listed and bring the variables back later.
These terms finally become a=1, b=8, and c=12
Now is how you use those terms:
Make a table like the one below, substitute the letters (a,b and c) for the terms that equal them, and multiply as shown.
a*c=12
b=8

As you can see, for the first column, I looked for all the factors of 12 and saw if any could be added up to 8. 
So now here comes the math part :)
(X+6)(X+2)
And if you are unsure about your solution, you can FOIL out the problem

Example #2: This is a great one that gets a little more challenging from switching up the negitives and positives, but also just makes sure that you are doing the steps correctly
X^2-12X+20

Now, if you look at the chart that I made, you can look for the once that match up. And in this case the factors will be 2*10. But, as you can see there can be different answers when adding or subtracting these numbers. If you look at the original equation, you can see that (+20) and that means that both of the signs will be the same so the numbers will be -2*-10
(X-2)(X-10)

Example #3: As you can see, there is a coefficient before the first term, this does complicate things just a bit but also can give you reassurance on what you need to work on and what you are good at with these steps.
12X^2+X-1
 This is exactly the same procedure as the previous two Examples. So let's first check out the signs. (-1) this means that the signs will be different. 

So using these numbers we can figure out the factoring (3x-1)(4X+1)

You can also do something called factoring by grouping.

I will continue to use this specific example.

So from the step of using the chart. We will be using -3,4 as because that will give us 1. 

=12X^2-3X+4X-1

now we are going to use factoring by grouping

3X(4X-1)+1(4X-1)

=(4X-1)(3X+1)

This, comes out to the same answer but if you feel more comfortable using this method so you can see what is going on and why.

Some of these, you will need to be guessing and checking the factors that you think would work. But using this method, you will always get your answer.
Here is a version of the examples but may take a bit more time to complete or find the correct factors.

 This might be a little hard to figure out what is exactly going on in this picture due to the lack of quality. This photo is merely the same steps that we were using in example #3 but with possibly a little more complicated numbers.

If you would like to have some practice problems here are some great problems to review. Hint: There will be problems most likely similar to these following on a test. 

A huge hint that I got from these classes  and I think it plays a major role in why some of these might not seem to work is that if numbers are factors in a specific parenthesis, you can take out numbers so that each parenthesis is the same and you can factor by grouping. 

 

Just follow the step for any of the examples and remember to make a table. I think that the table helps a lot.

For the Review in the Book Look Through Page 176-178
For some more information on factoring quadratics and why Click HERE
For some more information for factoring quadratic equations Click HERE
And for my personal favorite site that has great information and helps a lot for reviewing, check this out and Click HERE!!!!

I hope that this summarized the first two classes of this week

Thanks guys, for reading and I hope you liked it :) 

Remember to comment!!!

Have a good weekend (hopefully with snow)

GCF, LCM, and Factoring by Gail

--> Hello Class!

In the last two days of class we have mostly covered GCF, LCM and Factoring.

GCF-This means the greatest common factor of the two or more numbers that you are comparing.

LCM-This is the least common multiple of the two or more numbers that you are comparing

We started the class off with finding the GCF and LCM of the following three numbers:

225p^3qr^2 135p^2q^2r^2 80p^3r^3

We first start with doing a factor tree for these numbers to find common factors, of the three numbers.

We will first start with the coefficients

Next we will create a venn diagram, showing what the common factors are between all three numbers. The center of the diagram, where all three numbers overlap is where you will find the GCF, by multiplying the middle numbers together. By multiplying all of the numbers on the venn diagram (counted once), equals the LCM. In this venn diagram, we will add in the variables.

Looking at this graph, and what I stated before, we can come to the conclusion that:

GCF- 45p^2r^2

LCM-2,700q^2r^3p^3

Next, we began to talk about factoring, there are three main types of factoring that we learned:

Here are the formulas that you plug the equations numbers into.

Trinomial Squares: a^2+2ab+b^2=(a+b)^2

a^2-2ab+b^2=(a-b)^2

Difference of Squares: a^2-b^2=(a-b)(a+b)

Sum and Difference of Cubes: a^3+b^3=(a+b)(a^2-ab+b^2)

a^3-b^3=(a-b)(a^2+ab+b^2)

Here are some examples of the three types of factoring, these are the examples that we did in class.

16a^2b+24ab^2=

8ab(2a+3b)

Steps:

Look for the GCF, then pull out the GCF putting it in front of the numbers that are left over after pulling out the GCF.

Trinomial Square:

32x^2-48xy+18y^2=

2(16x^2-24xy+9y^2) <---This matches the trinomial squared sequence (plug in numbers to formula)=

2((4x)^2-24xy+(3y)^2)=

2(4x-3y)^2

Trinomial Square:

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

(x^3-y^2)^2

Plug in the numbers from the equation into the variables in the formula ex: a=x^3//b=y^2)

Difference of Squares:

81x^2-16y^2=

(9x-4y)(9x+4y)

We also talked about another type of factoring, which is factoring by grouping. You usually try this method when there are more than three terms.

Factoring by Grouping:

Is a way to group terms of an equation to factor out the common factor.

Example:

ab+3a-2b-6

The two groups are ab+3a and -2b-6

here in these two groups you pull out the common factor-which is (b+3)

a(b+3)-2(b+3)

Since the two common factors are the same, you can pull it out and multiply it by the left over numbers.

(b+3)(a-2)

In an equation with four terms you can also group three terms together, and leave one out.

Example-

x^2-4y^2-4x-4=

x^2-4x+4-4y^2<--In pink is a trinomial squared

(x-2)^2-4y<--Difference of two Squares

(x-2-2y)(x-2+2y)

Hope That Helps And Re-Caps Our Past Two Classes!!

:)

Examples and further explanation for these types of problems can be found on pages: 168-175

Exponent Review and Some More

In class on Thursday we reviewed the topic of exponents. There a several rules that are applied when using exponents and will be stated below:

RULES

1. The definition of something being raised to a power is just being that it is multiplied by itself as many times as the exponent requires. Ex. 2^3 = 2  * 2 * 2 = 8
2. When you have a power with a power you must multiply the exponents.                                    Ex. (x^4)^12 = x^48 How is was done: (x^4)^12 take the power that x is raised to and multiply it with the power the parenthesized equation is: 4 x 12 = 48 48 is the number that the parenthesized equation is raised to.
3. When multiplying powers with the same base, you add the exponents.                                                                                               Ex. x^4 * x^7 = x^11 How this was done: x^4 * x^7 take the two (or more) exponents and add them together: 4 + 7 = 11 Once you get the exponent, you raise it to the base, which is x (in this equation, you get x^11
4. Distributing the exponent through out each variable. This relates to rule one. You distribute the exponent and add it to each other exponent to get the answer. Ex. (5x^3st^2)^2 = 25x^6s^2t^4 This may look confusing, but no fear! Lets pick it apart: take each variable or number and distribute the exponent. 5^2 = 25 (x^3)^2 Remember to multiply the exponents and you should get: x^6  For S since there was no corresponding exponent the answer is: s^2 And lastly (t^2)^2 multiply the exponents and the answer is: t^4 If you put all this distributing together, the answer to this equation is: 25x^6s^2t^4 

MORE EXAMPLES

(z^x-1 / z^2) ^3 =   z^3x-3 / z ^ 6

-Just remember to distribute throughout fraction!

(-8z^4)^2x = -8^2 z^8x = 64z^8x

-Don't forget rule # 3 when doing this problem!

Some Exponent Websites

http://www.mathsisfun.com/algebra/exponent-laws.html

http://www.algebra-class.com/image-files/lawsofex5.gif

Where to look in the book!

Page 161 Chapter 4 Section 1

Mr. Exponent here to save the day...

http://www.youtube.com/watch?v=sr0AcO7HQHs

=^.^=

Wednesday we reviewed how to solve and absolute value equation and inequalities. I will show one example to refresh your minds.

Ex. 1/2|d| + 5 (greater than or equal to) 2|d| - 13 

How to: 

1. Subtract 1/2|d| from both sides
2. You should be left with: 5 (greater than or equal to) 1.5|d| - 13 
3. Add 13 to both sides
4. Now you should have: 18 (greater than or equal to) 1.5|d|
5. Divide 1.5|d| from |d| 
6. Now: 18/1.5 (greater than or equal to) |d| OR 12 (greater than or equal to) |d|
7. Since d is absolute value, 12 can either be negative or positive thus getting two answers: d (less than or equal to) 12 or -12

This problem is found on pg. 73: 30

There are many other problems on page 73 in our textbook that you can practice your ability to solve absolute value equations and inequalities.

Solving Absolute Value Equations and Equalities

Hello Everyone :)

This week in class we discussed how to solve absolute value equations and equalities.

   Example:

                              |2x-6|= 4 

                 2x-6= 4 or       2x-6= -4
                       .                      .
                       .                      .
                    x = 5              x = 1

We have learned 3 different "kinds" of equations that are each solved and graphed differently

1) Equal To ("OR")

2) Greater Than/Greater Than Equal To ("OR")

3) Less Than/Less Than Equal To ("AND")

( p.71)

1) | 4s + 5 | = 2

     4s + 5 = 2                                                           or                                                    4s + 5 = -2
     4s = -3                                                                                                                       4s = -7
                                       s = - 3/4                                               s= -4/7
In this problem the answer is either of the two answers but nothing else

2) | 2y + 3 | > 11

      2y + 3 > 11                                                         or                                                   2y + 3 < -11                                   
      2y > 8                                                                                                                       2y < -14
                                       y > 4                                                   y < -7
In this problem the answer is everything either above or below the two solutions to the equation excluding the solutions.


3) | k - 7/2 | < 5/2

  k - 7/2 < 5/2                                                           and                                                k - 7/2 > - 5/2
                                      k < - 1                                                  k > -6
In this problem the answer is in between the two solutions to the equation excluding the answers

Reminder: In all of these, the original equation is written out into two different equations with the absolute value, the signs are switched in one of them, and they are solved separately.

For problems like these: 2 < | x | < 4

Separate the equation into two different equations

| x | > 2

and

| x | < 4

Then solve them separately resulting most of the time in 4 answers instead of 2

x > 2                                                                                                     x < 4
                                                       and                                  
x < -2                                                                                                    x > -4



Tips and Tricks:

|4f +18| = 0  - This is a trick problem because there will be only one answer because 0 is neither negative nor positive

- Always write the absolute value first in the order of the equation




This can all be found in the book from pages 71-73

Hope this was helpful :)

Elise

Functions and Absolute Values

In class we talked about how absolute value and functions can be related and/or can be found in the same equation and we learned how to graph absolute values and how to graph inequalities with absolute values. 

Functions and Absolute Value:

-Functions are a specific type of relation because they have only one input value for each output value 

-Absolute Value is the distance a number is away from 0 on the number line

• When using absolute value the number is always in "| |" 
• Inside the brackets (| |) the number CAN be NEGATIVE but because it is the amount away from 0 the absolute value of a number is ALWAYS POSITIVE.

Ex. |5|=5 and |-5|=5

Problems with Absolute Value:

1. y=|x|

      Because x is in the absolute value sign it means that it can be either positive or negative, even if y is positive. That is so because the absolute value of x will always came out positive! But looking at the problem we can conclude that y cannot be negative because the absolute value of x will negative, even if it's negative in the brackets. 

ANSWER: x can be any number negative or positive, and 0, while y can only be 0 and all positive numbers.  And this equation is a function because there is only one input for each output, even though x=-2 and y=2 and then x=2 and y=2

2. A more complex one that is problem number 8. on page 151:

{ (x,y): |x|+|y| <= 1 }

      So like in the first problem x and y can be both negative and positive numbers, and they have to be less than or equal to 1. 

An easy number to always try first is 0 so if we plug in 0 for x and y they equation would look like: |0|+|0|<=1 and that statement is TRUE

Then we can plug in 1 for both x and y: |1|+|1|<=1 and that is TRUE also!

ANSWER: domain would be {x|-1,0,1}

Graphing with Absolute Value and Inequalities:

Problem #27 on page 151

27. y>=|x|

There are 3 things we need to figure out/do here:

1. What are the points that work? We need to find the boundaries.

2. Graph the inequality

3. Decide if this is a function.

Step 1:

When trying to find the boundaries it is always easiest to set the two side of the inequalities equal to each other. So the new equations would be:

y=x or y=-x   because x was in the absolute value brackets it can be negative or positive.

Step 2:

Now we need to see which points on the graph/which section solves the equation. We do this by picking a point from each section.

So if we plug in (3, 2): 2>=3 and that statement is FALSE!

Now if we plug in (0,-3): -3>=0 and that statement is also FALSE! 

If we plug in (-4,-2): -2>=|-4| = -2>=4 and that statement is again FALSE!

Finally if we plug in (1,4): 4>=1 and that statement is TRUE!!

The Final Graph:

The shaded are are all the values that work!!

Step 3:

This is NOT a function because for some inputs there is more that one output.

This information can be found in the textbook on pages 148-151.

And last but not least the homework for the weekend was on page151 #16 and 29!!

To Be or Not to Be.... a Function

In class today we explored the differences between relations and functions and learned about function notation, domain, and range.  Here are the top ten eight take aways from class:

8. A relation is a connection between input values and output values (can be an equation, a table, a graph, a verbal description)

7. A function is a specific type of relation.  In a function, each input value maps to just one output value.  The examples below are from our class worksheet.

6. You can tell if a graph is a function by using the vertical line test.  Does a vertical line intersect the graph in more than one location?  NO: it is a function,  YES: it is not a function

5.  f(x) is function notation.  It reads "f of x" or "f in terms of x."  f(x) is the output of your function, just like y.

4. Every relation and every function has a set of input values and a set of output values.  The set of input values is called the domain.  The sent of output values is called the range.

3.  Tip!  To remember which is which, don't forget that it's alpahbetical:

2. The sets of numbers in the domain and range can be written a few different ways:

• If your domain or range is finite, you can list it inside squiggly brackets, like this:  {x | x = -2, 0, 4 }  (Which relation, on the sheet above and from class, matches with this domain?)
• If your domain or range in infinite, you can represent it with inequalities inside squiggly brackets, like this:

 

(Which function, on the sheet above and from class, matches with this range?)

•  If your domain or range includes all real numbers, you can denote this with:

 

1.  FUNctions!

For a more thorough re-teaching of this topic, you can check out the video below:

This material can also be found in your book, Section 3-10, Relations.

It's 2013, and we're blogging!!

I'm very excited about the ideas, questions, concepts, debates, theories, etc. you will all post on this site this semester!  For inspiration, take a moment to view some of the best math scribe blog posts in The Scribe Post Hall of Fame!