Factoring and Quadratics

This Tuesday we worked on problems with an x**2 coefficient of more than one. After going over the homework, which were mostly problems with an x**2 coefficient of one. Lisa gave us this problem to try out:

20x**2 + 39x + 18

While she told us that we could try to trial and error the factors, there was a much more efficient way of solving the problem.

By setting up a table like the one above and plugging in AxC and B you can find the two numbers that add up to B.

       A + B + C

After finding the two terms that work, Lisa told us to then substitute them in for B, like so:

20x**2 + 39x +18

becomes

20x**2 + 15x + 24x +18

We figured out from here that this problem is now perfectly set up for factoring by grouping.

5x(4x + 3) 6(4x + 3)

(4x + 3)(5x +6)

*poof like magic*

On Wednesday we spent a few minutes going over some similar problems.

Such as:

Find the Common GCF of

z**3 - 4z = z(z**2 - 4) = z(z - 2)(z + 2)

z**3 - 2z**2 = z**2(z - 2)

z**3 - z**2 - 2z = z(z**2 - z - 2) = z(z - 2)(z + 1)

All have z(z - 2)

After going over the homework Lisa introduced us to quadratics by drawing a picture and asking us a question.

She asked us "What does the triangle have to equal for this to be true? What about the square?".

After thinking it over we decided that one or both of them had to equal zero.

Lisa then took it up a notch and showed us a problem like this:

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

She then asked us what x would have to equal for this to be true

We decided that since x is + 3 it must equal -3 to have the equation equal zero

and since it is also + 5, x must also equal -5 for this to be true.

Now that we had our answer we had two ways to write it either;

x = -3,-5

or

{-3,-5}

After completing more simple quadratic problems, we began on word problem like this:

Find two consecutive odd integers whose product is 99

(although many of us guessed this one we still had to write it out)

We first came up with a simple equation

x(x + 2) = 99

We made this equation because its a product so its multiplying, and its two consecutive odd integers so x and x +2

Now that we had our equation we went into solving it

x(x + 2) = 99

x**2 + 2x = 99

     -99         -99

x**2 - 2x - 99 = 0

(x + 11)(x - 9)

{-11,-9}

However we are not done yet. Since -11 and 9 are not x and (x + 2) we know that we must take them separately and have 9 and 11 as one set and -11 and -9 as the other.

Hope this was helpful!

Group and Quadratic Factoring

Peter Michalakes

Group Factoring (and any other variation of which) is essentially taking a problem and expanding it, eventually bringing the equation to this format: (      ) (      ) in which each set of parentheses is multiplied by each other. Addition, subtraction and division cannot be present outside the parentheses when a problem is completely factored.

In this post, I will explain how to factor by grouping and how to factor quadratic equations.

Factoring Equations

When factoring, it is good to reference equations which illustrate what combination of variables equal what combination of parentheses. 

Difference and Trinomial of Squares:

Sum and Difference of Cubes:

These equations are extremely helpful when factoring. For example, when you are asked to factor a + 2ab + b, it’s already done for you in the Trinomial of Squares equation, as a + 2ab + b = (a+b)^2.

Group Factoring

Factoring is a puzzle. And like any puzzle, there is no strict guideline to solving it; instead, the solver must rely on less-common mathematical methods to factor an equation. Again, referencing the previous equations are extremely helpful.

An Example:

When asked to factor x^2 + 8x +12, the first step would be to determine what factors each term has.

Listed in the picture above are the factors of 12:

6, 2;

3,4;

1, 12.

With this information, the next step is to determine the value of two of the integers in the (    ) (    ) guideline.

Bear in mind that the factors must add up to 8 from “8x,” so therefore, the factors 6 and 2 take their place in the parentheses as none of the others add up to 8. Therefore, we are left with (    +  6) (    + 2).

Now we must take the variable into account. Because x is squared, we automatically need two x’s, and since 6x + 2x = 8x, (foiled) then x is the missing term. 

And so x^2 + 8x +12 factors out completely to (x+6)(x+2).

Recap

Factoring by grouping can be tedious at times. But it can become less tedious if the Trinomial of Equations, Differences of Equations, Difference of Squares and Sum and Difference of Cubes equations are used.

The general steps for factoring by grouping is:

-identify common factors.

-identify correlating equation (if possible).

-factor.

Factoring Quadratics

It can generally be said that factoring quadratics takes less time than that of equations, even more so if (I know, I need to stop bringing up the guideline equations) the Trinomial of Squares equation is used as a reference.

An Example:

Quadrics can be solved similarly to those equations involving grouping, with a similar mindset of breaking down the puzzle of an equation on the paper in front of you.

A method we had discussed in class involved making a t-chart representing the Greatest-Common-Factors between the two groups of a quadratic. The results of which are described in the picture above.

Another Example:

In class on Friday we factored several quadratic equations, many of which are listed above.

For example, in x^2 - 12x + 11, the Trinomial of Squares reference takes its form. The method of which is described earlier in this post.

And so.

Thursday and Friday’s math classes focused solely on Group Factoring and Quadratic Factoring, both of which, while requiring more thinking than some of the other units we have studied, are very helpful in expanding our mathematical education.

Prime Factorization.

DAY 1: The almost snow day. 

Despite the snow day calculator's prediction for a 99% chance of a snow day, we had no snow come tuesday morning and instead we had school. I suppose we couldn't complain because we had already had a long weekend due to Martin Luther King day, but it didn't stop a few complaints when we arrived in math class bright and early. However, the almost snow day was forgotten as we resumed  work from the previous class on friday, finishing up the problems that we had started about 

MULTIPLYING POLYNOMIALS. 

We had already reviewed multiplying polynomials;

And the work we did on page 165 in the book built on that; 

After we checked our answers we went on to review PRIME FACTORIZATION by doing a factor tree for 972: 

We also quickly reviewed least common multiples and greatest common factors: 

LCM- least common multiple or the lowest number or term that is a multiple of a set of numbers. 

GCF- greatest common factor or the largest number or term that divides evenly into all numbers or terms. 

And the last thing we learned about in tuesday's class was VENN DIAGRAMS, and how they can help us find the GCF and LCM. 

Even though we had not necessarily expected to have class, overall we covered a lot tuesday morning! 

DAY 2: Day of puzzles. 

This time around we were all a lot more prepared for class, although still disappointed from the lack of snow. We started class slowly by reviewing our homework from page 170 of the book and ran into our first puzzle of the day on problem number 11, which looked something like this: 

Although the book said that the answer for the LCM was 216, we were not sure if that was actually correct because of the negative 108. 

As we moved on to other problems from the homework we found another, slightly more amusing, puzzle to think about. We had run into our first Venn Diagram with three circles instead of two and wondered if you could have more than three. (Eventually I think someone googled the answer but unfortunately we needed to continue going over the homework). 

We continued with problem 28 from the homework and learned that a perfect number is a number that is equal to the sum of its positive divisors excluding the number itself. This is where we came across our last puzzle for the day; whether or not a perfect number can be negative. After looking it up we discovered that it is possible yet nobody has been able to find proof of a negative perfect number. 

Lastly we started talk about FACTORING (the opposite of multiplying polynomials) and did a few examples checking to see if they followed certain patterns.  

The class ended before we could ponder any more puzzles, however, I hope these notes help! 

~Renata

Exponents!

Day 1

We kicked class off on Thursday with a quick single problem quiz on absolute value inequalities just to wrap up from Wednesday. 

The absolute value inequality problem that was our quiz

And then we began a BRAND NEW CHAPTER! We started with a quick review of exponents. And worked together on a few simple problems that were up on the board.

The eight exponential problems we solved together

Here is a photo of the problems along with my solutions which are a little more visual.

I hope this is helpful and doesn't confuse anybody!

Our homework Thursday night was very similar to what we did in class.

Our homework from Thursday night

To help keep things straight in my head I thought it would be helpful to write up a list of the rules for exponents. Mine was really incomplete and just useful for the homework we had but I found this website that has a really nice table of the laws.

Day 2

After going over homework and discussing solutions, we moved on to the main topic of class. 

MULTIPLYING POLYNOMIALS

There were a few problems on the board which we worked through together as a class, with some help from Lisa, so that we could get an idea of what we were learning.

Here I wrote down the examples along with the solutions that were on the board.

What I was able to take away from the examples was that

• Multiplying polynomials is like a bigger application of the distributive property

• Our review of exponents came in handy for simplifying our answers

                 and

• It closely related to the Binomial Theorem I just haven't quite figured out how yet

We then finished up class with a few group work problems and were free for the weekend!

I was thinking about inserting a Kahn Academy video but I thought I would just leave you with this gem! :)

Absolute Value in Open Sentences

Our homework for Tuesday started a new topic, which was absolute values.  All of the problems given for homework involved absolute values, either in equations or inequalities.  By themselves, absolute values are very easy.  The absolute value of a number is how far away a number is from zero.  A few examples:  
| –3 | = 3
| 28| = 28
|-8.4 | = 8.4

However, in the homework absolute value signs generally involved splitting a problem into two different problems, and solving the problem through these, as shown in the example below.  As for what happened during class on Tuesday, the majority of class involved going over the homework, but after this we tried more problems, and for these problems we paired up and worked on the board, allowing each other to compare are work to others, and find where we may have made mistakes.  Here is an example of a problem from the board from Tuesday in class:
 

The first step in this problem was splitting it into two different inequalities, and knowing that there would be an "and" in between, because of the less than sign.  Each of these two inequalities now splits into two more, leaving four inequalities which need to be solved.  The way to find this second set of two inequalities is to write the problem as is but without absolute value for the first one, and then switch the sign and multiply the side without the variable by negative one. After solving for the variable in these four inequalites, we know that the line we graph has to be less than or equal to five AND greater than or equal to negative four, as well as greater than or equal two one OR less than or equal to zero.  At the end of the problem, the line is graphed and appropriately shaded.

The homework for Tuesday night was similar, involving solving for variables by splitting absolute value problems into two equations or two inequalities.  Most of class Wednesday was spent putting the homework problems on the board, and then discussing them.  While most of the homework given was very similar to the above problem we did in class, the last problem was an extension of what we did in class.  The way the last problem was different was that there was an absolute value on each side of the inequality.  Here is the problem from when it was put on the board in class, after we discussed it.

 Although this problem looked different, solving it involved the same general steps.  First, the problem was split into two inequalities, with an and in between, and then each of these two inequalities became two more, the first pair having an and in between, the second pair with an or in between.  The way to solve this kind of problem, as we learned in class, was to find the two pairs by pretending the absolute value sign was only on one side, and then only on the other.  After using that trick to find the inequalities, all that was left was solving for the variable for each one, and after doing this we found that this line has to be greater than or equal to five AND less than or equal to fifteen, as well as less than or equal to fifteen OR greater than or equal to fifteen.  Although we got less than or equal to fifteen twice, it is still important to always find all the inequalities, because generally they are all different.  At the end of the problem you can see what this looks like graphed on a line.

One part of these problems that kept confusing me is whether to put "and" or "or" in between two inequalities.  While this may seem simple, it was actually one of the most tricky parts of the problem for me, and often where I made mistakes.  Usually, after finding the inequalities, I would draw a quick sketch of the line to try to figure it out, but Wednesday in class I had an aha moment when I realized it was as simple as this:

≤ - and
≥ - or

Here is a video I found on absolute value in inequalities.  We have not covered some things he talks about near the end of the video, but up until that point there is lots of useful information.  This video would be a great way to review this topic before a test or exams.  Also, problems exactly like the ones above can be found in the math book on page 73.  

- Ben Freedman

Functions continued

Thursday, January 10th-
In the two days we took to review homework for most of class and sort out any problems students had with the steps to solve and graph functions, a lot of new information turned up. In problem 17 (p.151) of the first night's homework- Determine whether or not f  and g are equal f: x --> x + 1; g: x --> 3 - x; D= {-1, 0, 1, 2, 3 } a new format of writing functions was introduced. In problem 27 (p.151)- Graph the relation { (x, y): y ≥ |x| } the graph of the equation (to the right) seemed to be one that tripped people up. To find out the correct area to shade in a problem like this, a point has to be selected (say, 1,0) and if it fits the equation, the area that includes the point that is within two intersecting lines should be shaded.

Friday, January 11th-
Absolute value was combined with inequalities in the past class. When that happens, a case system (creating two equations) for the possible negative and possible results, as well as shading a region. An example would be: (p.151) #30 { (x,y): x < |y + 2| }. To solve this, the < sign would have to be taken as an equals sign. Then it would have to be solved for both x and -x.

The lines y=x-2 and y=-x-2 intersect at 0,-2, and the inside of the v is shaded, because the point (3, -2), for instance, works in this problem. A dotted line is used because the sign doesn't include points found on the line itself, only inside the shaded region.

-Reilly

FUNctions!!!

This khan academy video was helpful for me to understand what makes a relation a function or not a function:

Math blog 1
FUNctions!
Tuesday, Jan 8:

Class started out with a couple verbal descriptions of functions:
“The time we arrive at Mt. Katahdin is a function of the time we leave Portland.”
“When we get to Acadia is a function of how fast we drive.”
“Our enjoyment of this hike is a function of the weather.”

Next, we had to match up a series of function descriptions to some graphs

We then looked at some examples of relations that are functions, and relations that are not functions. Based on those examples, we brainstormed what made a function, and this is what we came up with:
-A function exactly one value of y for each value of x
-If a vertical line goes through two or more points on the graph of a relation, than the relation is not a function.
-If it has a y is raised to a higher power than the power of 1, it will not be a function.

After we came up with these things, we went back to the examples of functions and non-function, looking at why each was which.
x=-9 is not a function because it is a vertical line, therefore there are infinite values of y for only one value of x, breaking the just one value of y for each value of x rule.

y=-9 is a function because it if you were to draw a vertical line through it at any place, it would not intersect more than once, because there is only one value of y though there are infinite values of x.

Sideways parabolas are not functions because if you put a vertical line through a function then it would intersect not once, but twice, meaning that there are two values of y for one value of x.


NOTATION:
Function notation: f(x)
-It reads “f of the x”
-NOT to be confused with algebraic notation; f is not a variable

After we talked a bit about function notation, we were given a function for which we were to find the f of x when x equaled different numbers:
Function: f(x)=3x2+2x

Find:
f(0)=0
f(1)=5
f(a)=3a2+2a
f(a+b)=3(a+b)2+2(a+b)
  ⇑
DON’T distribute here like you would with algebra

To wrap up class, we learned two definitions:
Domain: The set of all input values for a function
Range: The set of all output values for a function

And then found the domain and the range of a couple of the examples of functions:

x	f(x)
1	7
2	13
3	19
4	18

Domain: {x|1,2,3,4}
Range: {f(x)|7,13,18,19}
                  ⇑
This line means “such that”
This IS a function because it has just one value of y for each value of x

Wednesday, Jan 9:

Today we talked about absolute values and inequalities in relations.

Absolute value: The absolute value of a number is the number's distance from zero.

This is how absolute value looks:

|x|

Lets say that |x|=5

In that case, x could equal either 5 or -5, because those are the two numbers that are 5 away from zero. Absolute values are always positive because it is a distance, and a distance cannot be expressed in a negative number

|-3|=3

|2|=2

After we went over the meaning of absolute value, we started to look at absolute values in functions and relations:

The graph of y=|x|

This is a graph of the function y=|x|. The domain is all real numbers because it goes endlessly to both the left and the right. The ray going up to the right is for if x is positive, and the one going to left is for if x is negative, both are possible. We know it is a function because if you were to draw a vertical line through it, it would intersect only once. Its domain is all real numbers, and its range is y≥0.

Here is an example problem from class of absolute value relations:

{x,y:|y|=x and x≤3}

If you were to graph the integer points of this graph, it would look like this:

The domain would be {x|0,1,2,3} instead of just being all integers greater than or equal to zero because x is restricted by the inequality x≤3. It is not a function because if you were to draw a vertical line through it, it could go through 2 points.

Another khan academy video that might help explain absolute value inequalities. It has some expansions of ideas, especially towards the end:

Hope this helps!

-Isabel