Week of March 20,2017

ALGEBRA 1

Choosing a Factoring Method

Step 1 Check for a greatest common factor. 
Step 2 Check for a pattern that fits the difference of two squares or a perfect-square trinomial. 
Step 3 To factor x 2 + bx + c, look for two numbers whose sum is b and whose product is c. To factor ax 2 + bx + c, check factors of a and factors of c in the binomial factors. The sum of the products of the outer and inner terms should be b.
 Step 4 Check for common factors.

Use the following table to help you choose a factoring method.

First factor out a GCF if possible.

Then, Explain how to choose a factoring method for x2 − x − 30. Then state the method. • There is no GCF. • x2 − x − 30 is a trinomial.

 • The terms a and b are not perfect squares, therefore this is not a perfect square trinomial. • a = 1 Method: Factor by checking factors of c that sum to b. Explain how to choose a factoring method for 2x2 − 50. Then state the method.

 • Factor out the GCF: 2(x2 − 25) • x2 − 25 is a binomial. • a and b are perfect squares. This is a difference of squares.

 If binomial, check for difference of squares. yes no Use (a + b)(a − b). If no GCF, it cannot be factored. If trinomial, check for perfect square trinomial. yes no Factor using (a + b) 2 or (a − b) 2 . If a = 1, check factors of c that sum to b. If a ≠ 1, check inner plus outer factors of a and c that sum to b. If 4 or more terms, Try to factor by grouping.

ALGEBRA 2

Assessments: Unit project on Thursday

Monday
Today, we look at characteristics of polynomials ( intercepts, roots, extrema)
Help:http://www.coolmath.com/precalculus-review-calculus-intro/precalculus-algebra/12-relative-extrema-minimums-maximums-01
HW; None

Tuesday
It is a day of review for the characteristics of Polynomials and finding roots of polynomials, a full day all about polynomials.
Help:https://www.sophia.org/tutorials/characteristics-of-polynomials
HW; None

Wednesday
It is a day of review of all things polynomials and a look at end behavior, with discovery mixed in.
Help:https://www.khanacademy.org/math/algebra2/polynomial-functions/polynomial-end-behavior/v/polynomial-end-behavior
HW; Prepare for a unit test if you chose the unit test for the assessment

Thursday
Your choice, you can work on a project for the unit or take a unit test on the unit. Both items can count as a test grade.
Help:http://www.sosmath.com/algebra/factor/fac02/fac02.html
HW; None

Friday
Yes, it is Friday and we look at rational expressions, with a focus on factoring polynomials.
Help:https://www.khanacademy.org/math/algebra/polynomial-factorization
HW; None

Have a great weekend!
One week to Spring Break!

Week of March 13

Algebra 1

Formula For Factoring Trinomials(when a =1)

It's always easier to understand a new concept by looking at a specific example so you might want to do that first. This formula works when 'a' is 1. In other words, we will use this approach whenever the coefficient in from of x2 is 1. (If you need help factoring trinomials when a≠1$a\ne 1$, then go here.

1. identify a,b, and c in the trinomial ax2 + bx+c
2. write down all factor pairs of c
3. identify which factor pair from the previous step sums up to b
4. Substitute factor pairs into two binomials

Example of Factoring a Trinomial

Example 1

Factor x2 + 5x + 4

Step 1

Identify a,b, and c in the trinomial 
ax2 + bx+c

a= 1
b= 5
c= 4

Step 2

Write down all factor pairs of 4 
(Note: since 5 is positive we only need to think about pairs that are either both positive or both negative. Remember a negative times a negative is a positive. As the chart on the right shows you -2*-2 is positive 4...so we do have to consider these two negative factors. This is probably easier to understand if you watch our video lesson factoring trinomials)

Step 3

identify which factor pair from the previous step sums up to c

Step 4

Substitute that factor pair into two binomials

(x +4)(x+1)

Step 5

If you'd like, you can check your work by multiplying the two binomials and verify that you get the original trinomial

Algebra 2

Assessments: Unit Project given later in the week

Monday
It is the day to write equation of the polynomial from the roots and to solve polynomials with imaginary roots and real roots combined.
Help:https://www.varsitytutors.com/hotmath/hotmath_help/topics/polynomials-with-complex-roots
HW: none

Tuesday
Happy Pi Day! We will solve circle problems all day today! The work will be polynomial roots and writing equations using the roots.
Help: http://www.piday.org/

HW; Prepare for a mini quiz, a circle problem!

Wednesday
We confirm circle problems are good and there will be a mini quiz today. You will do a great job!
Help:http://www.purplemath.com/modules/fromzero2.htm
HW; None

Thursday
It is the day to look at characteristics of polynomials
Help:https://www.youtube.com/watch?v=epR8bf5PDmc
HW: None

Friday
Happy St. Patrick's Day! We continue with characteristics today
Help:http://www.sosmath.com/algebra/factor/fac03/fac03.html
HW: None

Have a great weekend! Two weeks to Spring Break!

Week of March 6, 2017

Algebra1

Factoring A GCF From an Expression Lessons

To best understand this lesson, you should make sure you know how to find the GCF of two or more terms. To learn how, see the lesson called Finding a GCF.

3x³ + 27x² + 9x

To factor out the GCF in an expression like the one above, first find the GCF of all of the expression's terms.

3	(1, 3)
27	(1, 3, 9, 27)
9	(1, 3, 9)

GCF = 3x

Next, write the GCF on the left of a set of parentheses:

3x(               )

Next, divide each term from the original expression (3x³+27x²+9x ) by the GCF (3x), then write it in the parenthesis.

3x³ / 3x = x²
27x² / 3x = 9x
9x / 3x = 3

3x(x² + 9x + 3)

The next expression we will be factoring is shown below.

36x² - 64y⁴

To begin factoring the GCF out of the expression, find the GCF of the two terms.

36	(1, 2, 3, 4, 6, 9, 12, 18, 36)
64	(1, 2, 4, 8, 16, 32, 64)

GCF = 4

As you can see, the two terms to do not have any variables in common, therefore the GCF is simply 4.
Now write 4, the GCF, on the left of a set of parentheses.

4(               )

Now divide each term 4, the GCF, and place the result inside the parentheses.

36x² / 4 = 9x²
-64y⁴ / 4 = -16y⁴

4(9x² - 16y⁴)

ALGEBRA 2

Assessments: Quiz on Wednesday and Summative assessment next week

Monday
Hope you had a great weekend! Today, we are working  on some polynomial cards  outside the classroom with varying questions on roots and factors.  Yes, you will think on this activity.
I have two colors posted, so ask me which color is the best fit for you today.  Also, you find roots of  4th degree polynomials and missing terms polynomials.
Help:http://math.stackexchange.com/questions/12787/finding-roots-of-the-fourth-degree-polynomial-2x4-3x3-11x2-9x-15
HW; None

Tuesday
Today, we look  at the Rational Root Theorem and  find all roots of a given polynomial.  I will have practice time for you today to master this concept and you will have choices on the problems.
HW; Study for quiz
Help:https://www.khanacademy.org/math/algebra2/polynomial-functions/finding-zeros-of-polynomials/v/finding-roots-or-zeros-of-polynomial-1


Wednesday
Moving along the week, you shall shine on the formative assessment (a quiz) in class today. I will have a quiz preview first to ensure you are ready this quiz. Also, we analyze the long division process to see a connection with the polynomial, the radical and the factors.
Help:https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/long-division-of-polynomials/v/dividing-polynomials-1
HW; None

Thursday
Almost Friday! It is the day to write the equation of a polynomial given the roots and to solve a couple of circle problems ( I created these problems, so I love them)
Help:https://www.youtube.com/watch?v=gyksK76Dg1c
HW; None

Friday
Yes, it is Friday! It is time for more circle problems today and some other polynomial cards filled with thinking questions.
Help:http://www.purplemath.com/modules/fromzero2.htm
HW; None

Week of February 27, 2017

Algebra 1

Function Notation and Evaluating Function

You've been playing with "y=" sorts of equations for some time now. And you've seen that the "nice" equations (straight lines, say, rather than ellipses) are the ones that you can solve for "y=" and then plug into your graphing calculator. These "y=" equations are functions. But the question you are facing at the moment is "Why do I need this function notation, especially when I've got perfectly nice 'y=', and how does this notation work?"

Think back to when you were in elementary school. Your teacher gave you worksheets containing statements like "[  ] + 2 = 4" and told you to fill in the box. Once you got older, your teacher started giving you worksheets containing statements like "x + 2 = 4" and told you to "solve for x".

Why did your teachers switch from boxes to variables? Well, think about it: How many shapes would you have to use for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h? The formula is as follows:

$A = \dfrac{h}{2}(a + b)$

If you try to express the above, or something more complicated, using variously-shaped boxes, you'd quickly run out of shapes. Besides, you know from experience that "A" stands for "area", "h" stands for "height", and "a" and "b" stand for the lengths of the parallel top and bottom sides. Heaven only knows what a square box or a triangular box might stand for!

Why did your teachers switch from boxes to variables? Well, think about it: How many shapes would you have to use for formulas like the one for the area A of a trapezoid with upper base a, lower base b, and height h? The formula is as follows:

$A = \dfrac{h}{2}(a + b)$

If you try to express the above, or something more complicated, using variously-shaped boxes, you'd quickly run out of shapes. Besides, you know from experience that "A" stands for "area", "h" stands for "height", and "a" and "b" stand for the lengths of the parallel top and bottom sides. Heaven only knows what a square box or a triangular box might stand for!

Algebra 2

Assessments: Quiz on Friday, unit test next Thursday, the projected test date.


Monday
Welcome back! Today, we start the class with some applications of polynomials, to ease back into the routine. Then, we move to long division with polynomials, you will do great once you see the pattern. Many of you know I will review the long division from elementary school for the steps.
Help:http://www.purplemath.com/modules/polydiv2.htm
HW; None


Tuesday
We review long division again and look at synthetic division. It is  tough at the beginning, yet you will master the process very quickly. We also review factoring today for the rest of the unit needs.
Help:http://www.purplemath.com/modules/synthdiv.htm
HW; None


Wednesday
It is finding roots of polynomials today with synthetic division and factoring. We will review our factoring skills too. I know all of you will love the lesson today.
Help:https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/synthetic-division-of-polynomials/v/synthetic-division-example-2
HW; None


Thursday
It is finding roots day three, more advanced with the type of factoring
Help:https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/synthetic-division-of-polynomials/v/synthetic-division
HW; Study for quiz tomorrow

Friday
Yes, we made it to Friday! It is quiz day, a day for you to shine. Also, we take notes on 4th degree polynomials.
Help:http://www.mesacc.edu/~scotz47781/mat120/notes/divide_poly/synthetic/synthetic_division.html
HW; None

Have a great weekend!

Week of February 13, 2017

Algebra1

Systems of Linear Equations:
  Solving by Addition / Elimination

The addition method of solving systems of equations is also called the method of elimination. This method is similar to the method you probably learned for solving simple equations.
If you had the equation "x + 6 = 11", you would write "–6" under either side of the equation, and then you'd "add down" to get "x = 5" as the solution.

x + 6 = 11
    –6    –6
x       =   5

You'll do something similar with the addition method.

• Solve the following system using addition.
2x + y = 9
3x – y = 16
Note that, if I add down, the y's will cancel out. So I'll draw an "equals" bar under the system, and add down:

2x + y = 9
3x – y = 165x      = 25
Now I can divide through to solve for x = 5, and then back-solve, using either of the original equations, to find the value of y. The first equation has smaller numbers, so I'll back-solve in that one:

2(5) + y = 9
  10 + y = 9
          y = –1
Then the solution is (x, y) = (5, –1).

It doesn't matter which equation you use for the backsolving; you'll get the same answer either way. If I'd used the second equation, I'd have gotten:

3(5) – y = 16
  15 – y = 16
        –y = 1
          y = –1

...which is the same result as before.

• Solve the following system using addition.
x – 2y = –9x + 3y = 16
Note that the x-terms would cancel out if only they'd had opposite signs. I can create this cancellation by multiplying either one of the equations by –1, and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the –1 through the entire equation. (That means both sides of the "equals" sign!)
I'll multiply the second equation.

The "–1R₂" notation over the arrow indicates that I multiplied row 2 by –1. Now I can solve the equation "–5y = –25" to get y = 5. Back-solving in the first equation, I get:

x – 2(5) = –9
x – 10 = –9
x = 1
Then the solution is (x, y) = (1, 5).

A very common temptation is to write the solution in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y-value first and then the x-value second, and of course in points the x-value comes first. So just be careful to write the coordinates for your solutions correctly. 


Algebra2

You can also evaluate compositions symbolically. It is simpler to evaluate a composition at a point because you can simplify as you go, since you'll always just be plugging in numbers and simplifying. Evaluating a symbolic compositon, where you're first plugging x into some function and then plugging that function into some other function, can be much messier. But the process works just as the at-a-number composition does, and using parentheses to be carefully explicit at each step will be even more helpful.

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find ( f o g)(x).
In this case, I am not trying to find a certain numerical value. Instead, I am trying to find the formula that results from plugging the formula for g(x) into the formula for f(x). I will write the formulas at each step, using parentheses to indicate where the inputs should go:

( f o g)(x) = f (g(x))     = f (–x² + 5)     = 2(             ) + 3     ... setting up to insert the input formula     = 2(–x² + 5) + 3     = –2x² + 10 + 3     = –2x² + 13

If you plug in "1" for the x in the above, you will get ( f o g)(1) = –2(1)² + 13 = –2 + 13 = 11, which is the same answer we got before. Previously, we'd plugged a number into g(x), found a new value, plugged that value into f(x), and simplified the result. This time, we plugged a formula into f(x), simplified the formula, plugged the same number in as before, and simplified the result. The final numerical answers were the same. If you've done the symbolic composition (the composition with the formulas) correctly, you'll get the same values either way, regardless of the value you pick for x. This can be a handy way of checking your work.
Here's another symbolic example:   Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Given f(x) = 2x + 3 and g(x) = –x² + 5, find (g o f )(x).
(g o f )(x) = g( f(x))     = g(2x + 3)     = –(           )² + 5    ... setting up to insert the input     = –(2x + 3)² + 5     = –(4x² + 12x + 9) + 5     = –4x² – 12x – 9 + 5     = –4x² – 12x – 4

---

There is something you should note from these two symbolic examples. Look at the results I got:

( f o g)(x) = –2x² + 13 (g o f )(x) = –4x² – 12x – 4

That is, ( f o g)(x) is not the same as (g o f )(x). This is true in general; you should assume that the compositions ( f o g)(x) and (g o f )(x) are going to be different. In particular, composition is not the same thing as multiplication. The open dot "o" is not the same as a multiplication dot "•", nor does it mean the same thing. While the following is true:

f(x) • g(x) = g(x) • f(x)            [always true for multiplication]

...you cannot say that:

( f o g)(x) = (g o f )(x)           [generally false for composition]

That is, you cannot reverse the order in composition and expect to end up with the correct result. Composition is not flexible like multiplication, and is an entirely different process. Do not try to multiply functions when you are supposed to be plugging them into each other.

Usually composition is used to combine two functions. But sometimes you are asked to go backwards. That is, they will give you a function, and they'll ask you to come up with the two original functions that they composed. For example: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

• Given h(x) = (x + 1)² + 2(x + 1) – 3, determine two functions f (x) and g(x) which, when composed, generate h(x).
This is asking you to notice patterns and to figure out what is "inside" something else. In this case, this looks similar to the quadratic x² + 2x – 3, except that, instead of squaring x, they're squaring x + 1. In other words, this is a quadratic into which they've plugged x + 1. So let's make g(x) = x + 1, and then plug this function into f (x) = x² + 2x – 3:

( f o g)(x) = f (g(x))     = f (x + 1)     = (         )² + 2(         ) – 3     = (x + 1)² + 2(x + 1) – 3
Then h(x) may be stated as the composition of  f (x) = x² + 2x – 3 and g(x) = x + 1.

• Given h(x) = sqrt(4x + 1), determine two functions f (x) and g(x) which, when composed, generate h(x).
Since the square root is "on" (or "around") the "4x + 1", then the 4x + 1 is put inside the square root. I need to take x, do "4x + 1" to it, and then take the square root of the result:

g(x) = 4x + 1,  f(x) = sqrt(x), and h(x) = ( f o g)(x).

© Elizabeth Stapel 2003-2011 All Rights Reserved