Rational Expressions: Excluded Values & Simplifying

No post yesterday. It was just the quiz.

Today, we worked with rational expressions. Rational expressions are expressions that can be written as a fraction. However, even when the fraction is made with polynomials, we have to remember that the denominator cannot equal 0. So, we have to list any exclude values.

Example: Find the excluded values from the following rational expression.

1. Set the denominator equal to zero: 25u² - 36 = 0
2. Solve the equation to find the excluded values. Look back at Tuesday's notes to remember how to solve by factoring.

Here: (5u - 6) (5u + 6) = 0

So, 5u - 6 = 0 or 5u + 6 = 0

5u = 6 5u = -6

u = 1.2 or u = -1.2

Excluded Values: 1.2 and -1.2

We also discussed how to simplify rational expressions. This concept was taught last Friday, so today's lesson reviewed and reinforced that previous understanding.

Homework: Page 163, #2 - 14 even and #19; Page 164, #3 - 8 and #11 - 13

There will be a test next Friday covering ALL of unit 2: Radicals, Polynomials, Factoring, & Rationals.

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Applications & Review

We went over the homework on solving quadratic equations. These problems required students to first factor (same methods that we have been working with for a week now) and then solve easier equations. Then we looked at some application/word problems. We discussed how the students can do guess-and-check or a quadratic equation to solve the problems.

Homework: Review worksheet for tomorrow's quiz.

There is a review session tomorrow morning in room 4303. The students should bring their completed review worksheet.

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Solving by Factoring

Today's lesson revisits the concept of solving quadratic equations. Last time, we used square roots to solving quadratics (September 30th). This time, we are using the Zero Product Property. This property states that the only way to multiply 2 (or more) factors together for a product of 0 is if at least one of the factors is already 0. (Zero times any number results in zero!)

1. Set the equation equal to zero.
2. Completely factor the polynomial expression.
3. Set each factor equal to zero and solve the equations.

Example: Solve r² + 12r - 15 = 30

Step 1 - Set equal to zero, here subtract 30 from both sides: r² + 12r - 45 = 0

Step 2 - Factor the polynomial: (r + 15) (r - 3) = 0

Step 3 - Set each factor equal to zero and solve: r + 15 = 0 or r - 3 = 0. So, r = -15 or r = 3.

Here, we find out that there are 4 vocabulary terms that are very similar in use/definition: solutions, roots, zeros, and x-intercepts. The methods are almost the exact same for finding any one of those terms. However, when we are presented with a function in function notation, step 1 - Set equal to zero simply becomes rewrite f(x) as 0.

Example: Find the zeros of f(x) = 5x² + 15x - 50.

Step 1 - Set equal to zero, here rewrite f(x) as 0: 0 = 5x² + 15x - 50

Step 2 - Factor the polynomial, here a gcf and then a quadratic:

First, 0 = 5 (x² + 3x - 10)

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

Step 3 - Set each facgtor equal to zero and solve: 5 = 0 (not possible!) or x + 5 = 0 or x - 2 = 0. So, x = -5 or x = 2.

Homework: Page 79, #2 - 12 even and #22 - 30 even. Page 84, #10 - 24 even.

Remember: Quiz Thursday. Review session Thursday morning, 7:45am, in room 4303

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Monday's Notes ...

I was out sick. However, thanks to the help of my colleagues, the students didn't miss out on the lesson.

Monday's lesson went over the last two types of factoring that we will use this semester.

Differences of two squares: This type of quadratic polynomial has two perfect square terms that are being subtracted. The factors of this type look like the conjugates that we have seen several times in the last month and a half. To make the middle (linear) term of the quadratic drop out, the factors must have had the exact same terms, but opposite signs.

a² - b² = (a - b) (a + b)

Example: Factor 16x² - 25 = (4x - 5)(4x + 5)

Factor by grouping: This method of factoring is often used when there are 4 (or more) terms in the polynomial.

1. Create groups of the terms in such a way that each grouping has a gcf.
2. From each group, factor out the gcf.
3. The remaining factors in the parenthesis must be the same to continue. Now that the parenthesis have the exact same terms, that is a gcf of both parts of our polynomial expression. Factor the parenthesis out to the front and create another parenthesis with what is left behind.

Students saw a similar problem to Step 3 in notes from last Wednesday: Factoring by Using a Greatest Common Factor, example "l".

Example: Factor 12x³ - 8x² - 18x + 12

Step 1 - Group: (12x³ - 8x²) + (-18x + 12)

Step 2 - Find GCFs in the groupsing: 4x² (3x - 2) + -6 (3x - 2)

Step 3 - Factor out the GCF Parenthesis: (3x - 2) (4x² - 6)

Homework: Finish the Handout

Remember - there will be a quiz on Thursday covering all uses of Factoring! There will also be a binomial expansion problem (from last Monday). We will have a review session Thursday morning at 7:45 am in room 4303 to prepare for this quiz.

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Bringing it together ...

Today's guiding notes worksheet brings together the concepts from last week (dividing polynomials by cancelling common factors) and this week (doing the factoring on our own). In last week's problems, the division problems were already factored. Today's lesson requires us to factor first, and then cancel the common factors.

NOTE: Many students experience difficulty with today's homework because they are not reading the directions. The steps are clearly spelled out and nothing is new on its own.

1. Remember, it may be helpful to rewrite a division problem as a fraction. To do this, the dividend (what is being divided into) goes in the numerator. The divisor (what is doing the dividing) goes in the denominator.
2. Factor the polynomials. First, look for a gcf. If there is one, pull it to the front. Then, if the polynomial is a quadratic, factor it like yesterday's lesson.
3. Cancel any common factors that appear in both the numerator and the denominator.
4. Write the simplified answer.

Example: Simplify the following by first factoring out GCFs.

Example: Simplify the following by first factoring the quadratic polynomials.

Example: Simplify the following by first factoring out GCFs and then factor any remainging quadratic polynomials.Homework: Finish the handout.

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Factoring Quadratic Trinomials

We continued factoring polynomials today, but we kept them to the form ax² + bx + c. These are called quadratic polynomials because of the ax² term.

Example: Factor x² + 5x - 36

1. Consider factors of the constant term (c, or the last term). Here: -36 is -1*36; -2*18; -3*12; -4*9; -6*6; -9*4; -12*3; -18*2; -36*1.


2. Choose the pair of factors whose sum is the coefficient of the linear term (b, or the middle term). Here: -4 + 9 = 5


3. Create your factors (2 sets of parenthesis) by filling in the variable and the factors chosen in step 2. Here: (x - 4) (x + 9)

For our Hopewell Middle School students, we made the link from today's lesson to the Diamond Puzzles that they did several times last year. They were presented with a diamond with two numbers. They were instructed to find two numbers that multiply to make the upper number and add to make the lower number. The following diamond puzzle is similar to the above example.

The two numbers needed to complete the puzzle are -9 and 4, as shown below.

Homework: Page 83: #1 - 9; Page 84: #1 - 6; Page 91: #1, 7, and 13

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Wednesday's Notes ...

Yesterday, we reviewed how to find the greatest common factor (gcf). We added to this concept how to factor it out of the polynomial.

Example: Factor 9a³ + 12a² - 15a

1. Look at each number and identify the largest number that can go into each evenly. Here, 3 goes into 9, 12, and -15 evenly.
2. Look at each variable and identify how many each term has in common with the others. The students were told to find the smallest exponent because that term limits how many the others can have in common. Here, each term has one a-variable. While the first and second terms have more a-variables, the last term only has one so it limits the common factor.
3. Write the GCF and then start parenthesis after that: 3a ( ______ )
4. To find what polynomial goes inside the parenthesis, determine term by term either "what is left when we divide by the gcf" or "what multiplies with the gcf to make the term". Here, 3a * 3a² makes the first term of 9a³. 3a * 4a makes 12a². And, 3a * -5 makes -15a. So, the final answer after factoring out the gcf is 3a (3a² + 4a - 5).
5. We can check every factoring problem by multiplying the factors back together. Their product should be the original polynomial. Remember, add exponents when multiplying bases. Here: 3a * 3a² + 3a * 4a + 3a * -5 = 9a³ + 12a² - 15a.

Homework: Page 79: #13 - 21 and Page 80: #10 - 18

There will be a quiz on Thursday, October 29 on factoring and solving by factoring (for the next week).

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Grades ...

Today in class, after going over Monday's homework, all graded papers were passed back: Quiz #3, Quiz #4 and the EOCT Practice Test. We took a good deal of class time to go over these assessments. The EOCT Practice Test is a good indicator of how well each student understands and retains the concepts from the first 9 weeks of school. This test will not hurt a student's average. I entered these grades in parent connect in a manner that won't affect the grades, but it allows parents and students to see the grades. If a student has a quiz grade lower than this EOCT practice test grade, then it replaced one lower quiz grade.

Students received a grade print out with all of their grades. Most of these progress reports also have notes on them about the EOCT Practice Test and the quiz replacement. These papers need to be signed by the parents and returned to school.

Notice: We were going to have a test on Thursday. That test is cancelled. We will move on to the concept of factoring polynomials tomorrow. More information to come soon!

Homework: Square Root and Polynomial Review Worksheet

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Notes from Monday ...

Yesterday's lesson was on Binomial Theorem, which is a short cut for raising binomials to exponents. We simplify this process as much as we can. One way we simplify this is to use Pascal's Triangle. This "triangle" is an arrangement of numbers that continues forever. Each number is made by adding together the two numbers above it. Here is the triangle through the first 5 rows (the top number is considered to be row 0). The 6th row would be 1, 6, 15, 20, 15, 6, 1.


Example: Simplify (x + y)⁴

This problem would require us to multiply out (x + y)*(x + y)*(x + y)*(x + y). The students saw through a problem in class that this is a lengthy process with many places to have errors. The Binomial Theorem and Pascal's Triangle makes this much simpler.

1. Note the exponent and go to that row of Pascal's Triangle. Here, the exponent is 4 so we will use row 4: 1, 4, 6, 4, 1. These numbers become our coefficients (numbers in front of the variables). We write the numbers down followed by 2 blanks each: 1_ _ 4 _ _ 6 _ _ 4 _ _ 1_ _
2. To start filling in the blanks, use the first term inside the parenthesis (here, x). This term will start with the exponent of the problem (here, 4) and then count down as it fills in the blanks from left to right: 1 x⁴ _ 4 x³ _ 6 x² _ 4 x¹ _ 1 x⁰ _
3. To finish filling in the blanks, use the second term inside the parenthesis (here, y). This term starts with the exponent of 0 and counts up to the exponent of the problem as it fills in the blanks from left to right: 1 x⁴ y⁰ 4 x³ y¹ 6 x² y² 4 x¹ y³ 1 x⁰ y⁴ . Notice that each term has exponents that add together to 4 (the exponent in the problem).
4. Simplify, if possible. Remember, any non-zero value raised to the 0 power is 1, so we remove that from our answer. If there are any numbers raised to powers, this is the step to simplify those powers and multiply times the coefficients. Here: 1 x⁴ 4 x³ y¹ 6 x² y² 4 x¹ y³ 1 y⁴
5. Enter the signs throughout the polynomial answer. If the binomial has addition, then all signs in the answer are plus signs. If the binomial has subtraction, then all signs alternate plus and minus, with the first term being positive. Here: 1 x⁴ + 4 x³ y¹ + 6 x² y² + 4 x¹ y³ + 1 y⁴

Homework: Page 75, #2 - 16 even and #24

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Quiz #4

Today, the IAA classes took a quiz covering polynomials. Students had the option to come in this morning for a review session. Because there is no handout at the review sessions, students need to be sure to have their own paper and copy down the notes/examples that Ms. Como provides as review.

Students will not have math class tomorrow. Instead they will take the PSAT in their homerooms. I will be out Thursday and Friday of this week at a math teacher conference. Thursday, students will have a guide to working with polynomial division. Friday, students will take a practice EOCT. This practice test will be in a multiple choice format. It will be cummulative with problems from the last 9 weeks. This practice test cannot hurt a student's average. However, it may be used to raise a grade.

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Quiz #4 Review

We played a jeopardy game to review polynomials today. Students should study the polynomial vocabulary as well as the steps to add, subtract and multiply polynomials. All of these concepts were in the lessons last week. Tomorrow we will have a quiz.

There will also be a review session tomorrow morning at 7:45 in my classroom (4303).

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Multiplying Polynomials

After going over last night's homework assignment (to finish yesterday's packet), we compared 4 different ways to multiply polynomials: tiles, Punnet Square, FOIL, and distribution. Students are encouraged to use whichever method works the best for them. We discussed how to multiply monomials (one term times another).

Example: 4x²y * -6x³y²

1. Multiply the coeffiecients together: 4 * -6 = -24
2. Multiply the variables together by adding exponents: x² * x³ = x⁵ and y * y² = y³
3. Write the product: -24x⁵y³

We also practiced multiplying polynomials. See yesterday's blog for examples of the tiles and Punnet square methods. FOILing and distributing requires us to multiply each term in the first polynomial times each term in the second polynomial. Finish by collecting like terms. Remember, do not change the exponents when collecting like terms (adding/subtracting). [FOILing stands for First, Outer, Inner, and Last. Those are the pairs of terms based on location to multiply. It ONLY works when multiplying a binomial (2 terms) times another binomial (2 terms) such as (4x + 2) (5x - 8).]

Example: (4x + 3) (5x² - 2x + 10)

= 4x (5x²) + 4x (-2x) + 4x (10) + 3 (5x²) + 3 (-2x) + 3 (10)

= 20x³ - 8x² + 40x + 15x² - 6x + 30

= 20x³ + 7x² + 34x + 30

Homework: Page 67: #2 - 20 even AND Page 71: #2 - 18 even

Remember, there will be a quiz next Tuesday covering this week's lessons.

Enjoy your 3 day weekend from school!!!

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Multiplying with tiles

We revisited the algebra tiles today. This time, the times set up a rectangle of given dimensions and we found the area inside the rectangle by counting up the tiles inside the rectangle. Because area is also found by multiplying length times width, this model finds the product of the two polynomial dimensions.

Example: Find the product of (x - 3) * (x + 2).

1. Set up the dimensions to the side of the work space (as shown above and to the left of the diagram below).


2. Fill in the space with shapes matching the dimensions. Or, continue the lines down and across that are started in the dimensions.


3. Record the values of the shapes in the resulting rectangle as the product. Simplify zero pairs.

Product = x² - 3x + 2x - 6 = x² - x - 6

We also found the product of the two polynomials by using a chart that is similar to the Punnett squares that the students learned in Biology in middle school (and will see again this year).

Example: Find the product of (x + 2) * (x² + 3x + 1)

1. This process again requires the students to set up the polynomials on the top and side of the work space (this time as collected terms rather than individual tiles).


2. Each box in the grid is found by multiplying the term above and to the left.


3. Collect like terms for the answer (usually on diagonals).

We will continue working with this concept tomorrow. There is a quiz scheduled for Tuesday covering adding, subtracting, and multiplying polynomials.

Homework: Finish today's packet

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Adding & Subtracting Polynomials

Our lesson today covered polynomial vocabulary. Then we again found sums and differences with polynomials. This builds on yesterday's lesson with the tiles, but we just used symbols today. Remember, for subtraction problems we change it to addition and change all signs in the second polynomial. This step distributes the minus through the second polynomial. To add the polynomails, collect like terms. Do NOT change the value of the exponent!

Examples:
(4x² + 5x - 8) + (2x² - 3x - 4)
= 6x² + 2x - 12

(9x² - 2x - 5) - (7x² - 6x + 4)
= (9x² - 2x - 5) + (-7x² + 6x - 4)
= 2x² + 4x - 9

Homework: Page 61: #1 - 14 and Page 62: #14 - 19

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Catch up post! (Sorry!)

So, Thursday we used word problems to set up more quadratic equations. We solved these equations using the same steps as we learned on Wednesday. There was no homework.

Friday, we had a quiz covering square roots. There was no homework.

Today (Monday), we used colored tiles called Algebra Tiles to learn the processes of adding and subtracting polynomials.
For adding polynomials:

1. Group/Collect like terms - the terms that have the same variable(s) to the same exponent
2. Add the coefficients, keep the term like what they started as - same variable(s) to the same exponent
3. Write the resulting polynomial

For subtracting polynomials (KFC or "add the opposite"):

1. Keep the first polynomial the same
2. Flip the signs of the second polynomial (positive to negative, and vice versa)
3. Change the problem to addition
4. Follow the steps for addition

Homework: finish today's packet. Students could checkout a packet of Algebra Tiles from me. They may use the tiles on next week's quiz (but they do not have to use them).

Quiz: Next Tuesday, October 13th covering adding, subtracting, and multiplying polynomials.

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