lesson 11 excessive expressions - ogburn

STUDENT MANUAL ALGEBRA II / LESSON 11

Lesson 11 Excessive Expressions

One of the most important things that you can do in order to be successful in Algebra II is to review and familiarize with common terms used in Algebra I; this way, you’ll be better able to expand on those previously learned concepts. Today we’re going to review some basic Algebra I terms and learn some new ones. Let’s review some basic terms and expand on them by learning some new terms that are related:

Monomial:

Whether you realize it or not, you’ve been working with monomials during your recent Algebra II studies. A monomial is defined as an algebraic expression that consists on of term, such as one number/coefficient with one variable, or such as the product of the two. It’s basically one set of a number of variable put together.

Here are some examples of common monomials:

3

x2

5yz (Here the coefficient is 5 and the variables are y and z)

2p3 (Here the coefficient is 2 and the variable is p)

Also, something such as x ÷ 2 can be a monomial as well because it can also be written as one term expressed as: ½ x

Here are some examples of expressions that are NOT monomials:

√36

23(x – 5x)

3 ÷ y

4y + 4

Any monomials must have something known as a degree. The degree is the sum of the exponents of the variables in an expression.

t4 u3


For this problem, the degree is equal to 7, the sum of the two exponents of t and u (4 +3).

If there is no exponent listed for a variable, then it is assumed that degree of that variable is 1.

For example:

t4 u3 v has a degree of 8 (4 + 3+ 1).

Also, when referring to degrees, it’s important to note that coefficients or any whole numbers are known as constants. Constants are numbers or values that have a constant or consistent value; we know what their values are. Any constants are assumed to have a degree with a value of 0.

For example:

4t4 u3 still has a degree of 7 because the number 4 is a constant; we know that its value is 4. Since 4 is a constant, we know that we count its degree as zero.

So basically, any positive whole number can also be referred to as a constant monomial (1, 2, 3, 4, etc.).

Binomials:

Next, let’s look at binomials. Binomials are basically two monomials put together, or an expression with two separate terms.

82 – x

4 + 8xy

x2 – 3y

t + 7t

4xyz + x8

A quadratic binomial is a binomial expression that has a term that is raised to the second power:

x2 – 3y

Trinomials:

Next, let’s look at trinomials. Trinomials are basically three monomials put together, or an expression with three separate terms.

x2 + x – 9

t2 + 8 – 5uv

4xyz + 6 + x8


A perfect square trinomial refers to an expression that is the result of a perfect square of a binomial:

For example:

x2 + 3x + 9

Can be also be represented as a binomial squared:

(x + 3) 2

= (x + 3) (x + 3)

= x2 + 3x + 9

Polynomials:

Next, let’s look at polynomials. Polynomials are basically expressions with two or more terms—both binomials and trinomials are polynomials. A monomial is not a polynomial because it is represented as one term. Oftentimes, polynomials may even have more than three terms.

Here are some examples of polynomials:

x8 + x + y - 66

4x4 + 3x3 - 2x2 + x

4x4 + 3x3 - 2x2 + x

x8 + xyz3 + 9

z20 + xyz3

Generally, polynomials are placed in order of the highest degree; this is also referred to as the polynomials leading term of the polynomial.

For example in: 4x4 + 3x3 - 2x2 + x

4x4 is the leading term.

In the same respect, each polynomial has a leading coefficient, which is the coefficient for the leading term.

For example in: 4x4 + 3x3 - 2x2 + x

4x4 is the leading term, so the leading coefficient, which is also a constant, is 4.

If there is no coefficient given, such as in: x2 + 3x + 9, then the leading term is x2 and the assumed coefficient is 1, since the product of 1 and any number will not change the value of that number.


Next, the degree of a polynomial is determined by the degree of the leading term in that polynomial.

For 3x4 + 2x3 - 6

The degree of the polynomial is 4 because the highest degree in the polynomial is 4.

The degree of the polynomial y – 3 is 1 because 3 has a degree of zero since it is a constant, and the degree of y must be 1 because there is no other exponent listed.

Finally, the constant term of a polynomial is the constant term or number that has a degree of 0.

For 3x4 + 2x3 – 6

The constant term would be -6


Lesson 11 Excessive Expressions

Name:__________________________________________Date:___________________

Now, on your own, please answer the following questions based on the information given today.

For problems 1-10, determine the degree of each polynomial listed.

1. y4 x3 v + 14 __________________________________________________________________

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2. 2x + 6 __________________________________________________________________

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3. 5y10 – 6x7y3 – 4y – 5 __________________________________________________________________

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4. x3 +4y - 9 __________________________________________________________________

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5. 60x2 y + 8 __________________________________________________________________

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6. 5y3 – 4y – 1 __________________________________________________________________

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7. 4x4 y3 + v y + 7 __________________________________________________________________

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8. 8n5 – 4n2 – x __________________________________________________________________

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9. 134x10 y3 + x2 v y + 7 __________________________________________________________________

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10. 1,234u15 v12 w10x8y6 z4 + 4 __________________________________________________________________

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For problems 11-15 determine the leading term, leading coefficient, and constant term of each polynomial

11. 5y3 – 4y – 1

Leading term: _______________________________________________________ Leading coefficient: __________________________________________________

Constant term: ______________________________________________________

12. 4x4 y3 + v y + 7


Constant term: ______________________________________________________

13. 8n5 – 4n2 – 2



Constant term: ______________________________________________________

14. 134x10 y3 + x2 v y + 123


Constant term: ______________________________________________________

15. 1,234u15 v12 w10x8y6 z4 + 4


Constant term: ______________________________________________________

For problems 16-20 determine if the binomial is a perfect square binomial. If so, show the original monomial squared. You may have to remember or refer back to the rules for multiplying exponents.

16. x8y + x4y + 2 __________________________________________________________________

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17. y2 + 4y + 16

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18. 9y2 + 6y + 1 __________________________________________________________________

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19. x2+ 16 __________________________________________________________________

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20. x4 + 12x2+ 36 __________________________________________________________________

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Lesson 12 Pretty Simple Polynomials

Polynomials:

Remember from the previous lesson that Polynomials are basically expressions with two or more terms—both binomials and trinomials are polynomials. Oftentimes, polynomials can be simplified by combining like terms

For example, if we have:

3x4y3 + 2x4y3 + 3y2 + 4y2- 89 +6

We can easily simplify this polynomial be combining the like terms or the monomials that have the same variables with the same exponents. In like terms, the coefficients are generally the only portion of the monomial that can be different.

If we look in the example polynomial above, we can easily see that there are 6 different monomials in the expression. We can simplify this expression by combining the like terms.

Let’s look at the first two terms in the expression; they are like terms:

3x4y3 + 2x4y3

Since both 3 and 2 have x4y3 equally attached to them, then we can combine these terms. If the x was not to the forth power in one term or the y was not to the third, then these two terms would not be considered like terms, and we could not combine them. Each variable must be raised to the same power in order to combine like terms.

In order to combine these like terms: 3x4y3 + 2x4y3

We simply just add the 3 + 2, and the place the variable portion of the expression beside the answer to the coefficients that were added. So, 3x4y3 + 2x4y3 = 5x4y3

Now, we can combine the other like terms in the original polynomial: 5x4y3 + 3y2 + 4y2- 89 +6

3y2 + 4y2 = 7y2

Now, the polynomial becomes 5x4y3 + 7y2 + 89 +6

And finally we can combine the last two numbers of the expression, which are also like terms: 89 + 6 = 95

So the original polynomial: 3x4y3 + 2x4y3 + 3y2 + 4y2- 89 +6

Has been simplified by combining like terms to: 5x4y3 + 7y2 + 95


3x4y3 + 2x4y3 + 3y2 + 4y2- 89 +6 = 5x4y3 + 7y2 + 95

Once you’ve combined all the like terms in a polynomial, and you go to place them in order, it’s always important to remember that a polynomial should start with the leading degree of each expression that goes in descending order based on each terms degree. This is known as its simplest form.

For example:

x + 14xyz3 + 7x3 + 5xy7 + 5x2

Should be ordered by the highest degree of each individual term or monomial to the lowest. The value of the coefficient is not relevant.

Let’s first determine the degree of each individual term in the polynomial.

x has a degree of 1

14xyz3 has a degree of 5

7x3 has a degree of 3

5xy7 has a degree of 8


Now, based on the degree of each term, we can put the polynomial in its simplest form by the highest degree to the lowest.

5xy7 + 14xyz3+ 7x3 + 5x2+ x

These methods can be used when adding multiple polynomials together.

For example,

(3xyz3 + 7xy7 + 7x3 + 4x2) + (1 + x + 5xy7 + 8xyz3)

For this example, first we must combine like terms, only adding the coefficients:

8xyz3 + 3xyz3 = 11xyz3

7xy7+ 5xy7 = 12xy7

If there are no like terms for some of the terms in each polynomial, then it just carries over and remains the same. There are no like terms to be combined for these terms in the addition problem: 7x3, 5x2, 1, x, so these terms just carry over.

Now, we have the following terms:

11xyz3 +12xy7 + 7x3+ 4x2+ 1 + x


Now, we must determine the degree of each term, and then place them in descending order based on their value.

11xyz3 has a degree of 5

12xy7 has a degree of 8



1 has a degree of 0

x has a degree of 1

No we can put the sum of the two polynomials in simplest form:

(3xyz3 + 7xy7 + 7x3 + 4x2) + (1 + x + 5xy7 + 8xyz3) = 12xy7 + 11xyz3 + 7x3 + 4x 2+ x + 1


Lesson 12 Pretty Simple Polynomials

Name:____________________________________________Date:__________________

Now, on your own, please add the following polynomials by combining like terms. Do not worry about putting them into simplest for just yet. Be sure to show all your work.

1. (10b8 + 3ab7 + c + 9abc + 4b2) + (1 + c + 4ab7 + 9b2)

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2. (13yx7 + 11yz9 + x + 9 + 8xy2) + (yz9 + 13xy7 + x + 10 + xy2)

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3. (5 + n + 4mo2 + 8n8) + (19 + 16mo2 + 4n10 + 9mno2) + (45 + n) __________________________________________________________________

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4. (15yz3 + 13xy7 + x + 9 + 4x2) + (1 + x + 4xy7 + 8yz3) __________________________________________________________________

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5. (9ab7 + bc + 14ab3 + 9a + 4ab2) + (1 + a + 16ab3+ 4ab7 +bc + 9ab2) __________________________________________________________________

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6. (12 + r + 6s4t2 + 8st2) + (19 + 16s2 + 4st7 + 6s4t2 + st2) + (8s4t2 + 15 + r) __________________________________________________________________

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7. (3xy2+ 6yx5 + 11yz9 + y + 87) + (13yx5 + 13xy2 + y + 10 + yx5) __________________________________________________________________

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8. (n2 + 4 + 4o3 + n) + (4 + n + 4m2n2 + 8n2) + (18 + 9m2n2 + 4o3 + 9n2) __________________________________________________________________

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9. (12 +6s4t2 +s) + (10 + s + 6s4t2 + 8s2) + (21+ 6s4t2 + 7s4t2) __________________________________________________________________

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10. (13yx5) + (13xy2 + 19y + 3z9) + (x + yz9+ 8xy5) + (yz9 + 43xy2 + 11y + 6z9+ xy2) __________________________________________________________________

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Now, put each of the following polynomials, which you’ve already added together, into simplest form by ordering the degrees of each term correctly. Hint: you should use your answers from the above problems.

11. (10b8 + 3ab7 + c + 9abc + 4b2) + (1 + c + 4ab7 + 9b2) __________________________________________________________________

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12. (13yx7 + 11yz9 + x + 9 + 8xy2) + (yz9 + 13xy7 + x + 10 + xy2) __________________________________________________________________

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13. (5 + n + 4mo2 + 8n8) + (19 + 16mo2 + 4n10 + 9mno2) + (45 + n) __________________________________________________________________

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14. (15yz3 + 13xy7 + x + 9 + 4x2) + (1 + x + 4xy7 + 8yz3) __________________________________________________________________

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15. (9ab7 + bc + 14ab3 + 9a + 4ab2) + (1 + a + 16ab3+ 4ab7 +bc + 9ab2) __________________________________________________________________

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16. (12 + r + 6s4t2 + 8st2) + (19 + 16s2 + 4st7 + 6s4t2 + st2) + (8s4t2 + 15 + r) __________________________________________________________________

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17. (3xy2+ 6yx5 + 11yz9 + y + 87) + (13yx5 + 13xy2 + y + 10 + yx5) __________________________________________________________________

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18. (n2 + 4 + 4o3 + n) + (4 + n + 4m2n2 + 8n2) + (18 + 9m2n2 + 4o3 + 9n2) __________________________________________________________________

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19. (12 +6s4t2 +s) + (10 + s + 6s4t2 + 8s2) + (21+ 6s4t2 + 7s4t2) __________________________________________________________________

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20. (13yx5) + (13xy2 + 19y + 3z9) + (x + yz9+ 8xy5) + (yz9 + 43xy2 + 11y + 6z9+ xy2) __________________________________________________________________

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Lesson 13 Polynomials in Action with Subtraction

In the previous lesson, we reviewed how to combine like terms while adding polynomials and you placed them into simplest form based on each term’s degree. Today we’re going to look at how to subtract polynomials. Subtracting polynomials can often be a bit trickier than adding them because you have to pay close attention to the positive and negative values of each term. There is a very simple solution that can be used when subtracting polynomials.


(6x4y3 + + 3y2 + 8) – (2x4y3 + 4y2 – 6)

You can simply find the opposite value of each term after the subtraction sign to subtract. Basically, the negative sign needs to be distributed into the second polynomial.

So for the second polynomial in this equation, you have:

- (2x4y3 + 4y2 – 6)

The negative sign must be applied to each term in the polynomial: - 2x4y3 - 4y2 + 6

As you can see, each term turned into the opposite of it’s original value. Be sure to do this whenever subtracting any polynomials. Be sure to pay close attention to this.

So now the problem becomes:

6x4y3 + 3y2 + 8 – 2x4y3 - 4y2 + 6

Finally, combine like terms:

4x4y3 - y2 + 14


Lesson 13 Polynomials in Action with Subtraction

Name:___________________________________________Date:___________________

Now, on your own, please subtract the following polynomials. Be sure to place the polynomials in their simplest form based on the leading degree. Show all your work.

1. (10b2 + 3abc + c) - (abc + 4b2) __________________________________________________________________

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2. (7yx7 + 11y - 8xy2) - (yz9 + 17xy7 - xy2) __________________________________________________________________

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3. (34 + n - 4mo2 - 8n8) - (16 + 16mo2 + 12mno2) - (16 + n) __________________________________________________________________

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4. (- 23xy7 + x + 9 + 8z3) - (5 + 6x) - (7xy7 + 8z3) __________________________________________________________________

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5. (4ab7 - 22ab3 + 3a - 4ab2) - (19 + 3a + 16ab3+ 4ab7 - 9ab2) __________________________________________________________________

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6. (172 - r + 9s4t2 + 8st2) - (21st7 + 6s4t2 + 8st2) - (8s4t2 + 10 - 5r) __________________________________________________________________

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7. (7xy2+ 9yx5 + 7y + 13) - (13yx5 - 93xy2 + 8y - 17 + yx5) __________________________________________________________________

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8. (m2 + 14 - 4o3 + m) - (14 + m + 14m2n2 + 8m2) - (14 + 14m2n2 - 4o3 + 4m2) __________________________________________________________________

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9. (21+ 14s4t2 - 7s4t2) - (t2 +5s4t2 +s) - (10 t2 - 6s4t2 + 7s2) __________________________________________________________________

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10. (x - z9+ 2xy2) - (8z9 + 43xy2 - 16y + 10z9+ 3xy2) - (3yx5) - (13xy2 - 19y + 3z9) __________________________________________________________________

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Now, for problems 10 – 15, first square each of the binomials and place them into perfect square trinomial form, and then subtract the trinomials. Show all your work.

11. (x + 3) 2 - (x - 4) 2

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12. (2y - 5) 2 - (y + 6) 2

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13. (3m - 4) 2 - (m + 4)2

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14. (7b2 - 7) 2 - (2b2 - 11) 2

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15. (5ab2 + 3) 2 - (6ab2 - 7) 2__________________________________________________________________

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Now, for problems 16 – 20, create 5 different polynomial subtraction problems, and show the solution. For each problem, subtract a binomial by a trinomial. Be sure to include like terms in your problems.

16.

Problem:

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Solution:

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Problem:

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Solution:

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Problem:

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Solution:

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Problem:

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Solution:

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Problem:

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Lesson 14 Products of Polynomials

In the previous lessons, we’ve reviewed how to add and subtract polynomials, and in this lesson, we’re going to look at how to multiply polynomials. When multiplying polynomials, we must always use the distributive process.


(3y2 – 6) (2y3 + 5y2 – 7)

First, let’s multiply each of the terms in the second polynomial by 3y2

3y2 • 2y3 = (3 • 2= 6) Multiply all coefficients together.

(y2 • y3 = y(2+3)) Remember, when you are multiplying exponents, you just add them together.

So 3y2 • 2y3 = 6y5

No, let’s multiply 3y2 by the remaining terms in the second polynomial:

3y2 • 5y2 = 15y4

3y2 • -7 = -21y2 Don’t forget to pay close attention to negative and positive signs.

So, so far we have:

(3y2) (2y3 + 5y2 – 7) = 6y5 +15y4 -21y2

Now, that we’ve distributed the first term, we must now distribute the second term.

(-6) (2y3 + 5y2 – 7)

-6 • 2y3 = - 12y3

-6 • 5y2 = - 30y2

-6 • -7 = 42

So we have: - 12y3 - 30y2 + 42


Now, we must combine nay like terms in the two separate trinomials that came from the distribution of the binomial in the original problem, and then order them correctly based on their degree:

6y5 +15y4 -21y2- 12y3 - 30y2 + 42 =

6y5 +15y4- 12y3- 51y2+ 42

So, our original problem:

(3y2 – 6) (2y3 + 5y2 – 7) = 6y5 +15y4- 12y3- 51y2+ 42


Lesson 14 Products of Polynomials

Name:______________________________________________Date:________________

Now, on your own, please multiply the following polynomials. Be sure to place the polynomials in their simplest form based on the leading degree. If there is more than one term with the same degree, then place them in the order with which they first appear in the solution. Show all your work.

1. (5xy + 6) (7xy2 + 8) ________________________________________________________________________

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2. (4y + y2) (7xy2 + 6) ________________________________________________________________________

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3. (ab + 2b2) (5b2 + 3ab) ________________________________________________________________________

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4. (9xyz + 3xy2) (xy2 + 6) ________________________________________________________________________

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5. (4b + 2b2) (5b2 + 3b) ________________________________________________________________________

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6. (7yx7 + 11)(yz9 - 17xy4 + xy2) ________________________________________________________________________

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7. (4mo2 - 8n2) (16mo2 + 12mno2 – 5m) ________________________________________________________________________

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8. (- 3xy7 + y) (3 + 2z3 – 17) ________________________________________________________________________

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9. (3ab5 - 12ab3) (7a - 4ab2 – 18) ________________________________________________________________________

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10. (3a + 15ab3) (5ab7 - 3ab2- 5) ________________________________________________________________________

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11. (5r + 6s4t2 - 4st2)(2st5 + 8s4t2 - 4st2) ________________________________________________________________________

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12. (4xy2+ 4y + 14) (14yx4 + 4xy2 + 8y) ________________________________________________________________________

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13. (5m2n2 + 5o3 + m)(5m2n2 + 5m2 + 5m) ________________________________________________________________________

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14. (14s4t2 - 7s4t2 + 12) (t2 +6s4t2 +s) ________________________________________________________________________

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15. (3xy2 + 2xy2+8z2) (4xy2 - 16y + 10z2) ________________________________________________________________________

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Now, for problems 16 – 20, create 5 different polynomial multiplication problems, and show the solution. For each problem, multiply a binomial by a trinomial. Show all your work.

16.

Problem:

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Solution:

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Answer:

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Problem:

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Solution:

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Lesson 15 The Greatest Factor

In the previous lessons, we’ve reviewed how to multiply polynomials, and in this lesson, we’re going to look at how to find the greatest common factor in two polynomials in order to divide them.

When trying to determine the greatest common factor, or the factor that is in all the terms of a polynomial, you must make sure that each separate term includes all of the factors in your greatest common factor.

For example, if you have the following polynomial:

3y4 + 12y3 + 9y

We can easily determine the greatest common factor between the

If we look at the example, first let’s think about the leading coefficient: 3

Does 3 go into all the other coefficients in the polynomial (12 and 9)? Yes! But 12 is not a factor if 3, and neither is 9, so the greatest common coefficient that goes into all three terms in 3.

Next, let’s look at the variable: y

We have the following uses of y in the polynomial:

y4, y3 , y

The greatest common factor in these is y. No other variable goes into all three options.

So, the greatest common factor of 3y4 + 12y3 + 9y is 3y.

Now, let’s look at when a polynomial has terms that don’t use the same variable or factored coefficients.

For example, if we had: x3y4 + 7x2y3 + 4x

Let’s look at the leading term: x3y4

What factors do x3y4 and 7x2y3 have in common? They both have x2y3 but the first term does not have a factor of 7, so 7 can’t be a common factor.

Now, let’s take a look at our greatest common fact so far x2y3 and the final term 4x.

The greatest common factor between the two is x because there is no 4 in the first term, and no y in the second therefore, the greatest common factor if our original polynomial:


x3y4 + 7x2y3 + 4x is simply x. It is the only factor that appears in all three terms in the polynomial.

Now that we have good foundation for how terms relate to one another in monomials and polynomials, you we easily divide a polynomial by a monomial.

For example, if you have:

6y3 + – 21y2 + 15y ÷ 3y

Then you would divide each individual term in the polynomial and the monomial; we divide each term separately. It doesn’t matter what the greatest common factor between each of them is because we treat them as separate terms.

First:

6y3 ÷ 3y = 2y2

You basically subtract the exponents of the variables, as learned in previous lessons, and divide the coefficients.

Next, you have:

21y2 ÷ 3y = 7y

Finally

15y ÷ 3y= 5

So

6y3 + – 21y2 + 15y ÷ 3y = 2y2 + 7y + 5

Now try some practice problems that relate to the concept of greatest common factor and dividing polynomials by monomials.


Lesson 15 The Greatest Factor

Name:___________________________________________Date:__________________

Now, on your own, find the greatest common factor in the following polynomials. Show all your work.

1. 5xy +10xy2 + 5x ________________________________________________________________________

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2. 4y + xy2 + 6y ________________________________________________________________________

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3. ab + 2a2b2 + 3ab ________________________________________________________________________

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4. 9xyz + 3xy4z+ 3xy2 + 6xz ________________________________________________________________________

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5. 4ab + 5ba2 + 3b ________________________________________________________________________

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6. 2yx7 + 10yz9 + xy4 + 5xy2

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7. 4mn7o2 + 8n2o+ 16mn3o2 + 12mn3o2 – 20mn2 o ________________________________________________________________________

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8. - 3xy7z + xyz3 + 2xyz3 + z ________________________________________________________________________

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9. 2ab5 - 12ab3 + 4ab2 + 18 b2 ________________________________________________________________________

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10. 3ab5 + 15ab3 + 5ab2 - 3ab2- 5b ________________________________________________________________________

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Now, divide the following polynomials by the monomials listed. Don’t worry about listing your answers in descending order. Show all your work.

11. 6s4t2 +10st - 4st2 ÷ 2st ________________________________________________________________________

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12. 4xy2+ 4xy4 + 12xy2 ÷ 4xy2

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13. 5m2n2 + 5mno3 + 5mno2 ÷ 5mn ________________________________________________________________________

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14. 14s4t2 - 7s4t2 + 12t2 ÷ t ________________________________________________________________________

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15. 12xy5 + 24xy4 + 8xy3 ÷ 4xy2

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16. 20yx7 + 10xyz6 + xy4 + 30xy2 ÷ 2xy ________________________________________________________________________

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17. 4mn7o2 - 16mn3o2 + 12mn3o + 8mn2o ÷ 4mn2 o ________________________________________________________________________

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18. - 3xy7z + xyz3 + 2xyz ÷ yz ________________________________________________________________________

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19. 84ab5 - 12ab3 + 4ab2 + 18b2 ÷2b2

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20. 30ab5 + 15ab3 + 5ab2 - 105ab2- 75ab ÷ 5ab ________________________________________________________________________

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Lesson 16 It’s been a Long Time…

In the previous lessons, we’ve reviewed how to divide polynomials by monomials, but now we’re going to do something just a bit more complicated. We’re going to divide a polynomial by another polynomial. To do this, it gets a little bit trickier than when we were just dividing a monomial by each separate term in the polynomial. One of the best ways to do this type of division is to use the tested and true method of long division. Long division is great for when you’re dividing large numbers or large groups of numbers and variables. It may have been a while since you’ve done some long division so here is a quick review just using basic numbers:

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5 | 2510

First, we determine how many times 5 goes into 25: 5

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5 | 2510 So we place the 5 up top, and then subtract 5 x 5 from the dividend of 2510

- 25

0 10 There is no remainder, so we simply bring the 10 down

5 goes into 10 two times, so we now place a 2 up top:

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5 | 2510 So we place the 2 up top, and then subtract 5 x 2 from 10

- 25

0 10

- 10

0

Now there is nothing left to divide 5 into, and lucky us there is no remainder!

Now, let’s learn how to use the process of long division with polynomials:


Let’s say we want to simplify for the following division problem:

y + 5 ÷ 2y2 + 7y - 15

Let’s put this into long division form, with the second polynomial as the dividend and the first as the divisor.

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y + 5 | 2y2 + 7y - 15 So we place the 2 up top, and then subtract 5 x 2 from 10

First, divide the divisor by the first term in the dividend:

How many times does y go into 2y2

Remember the division method in the previous lesson, you simply subtract for division of monomials:

So 2y2 ÷ y = 2y

Now, let’s put 2y on the top of the division sign:

2y_________

y + 5 | 2y2 + 7y - 15

Now, we need to multiply 2y(y + 5), and then carry our answer down to be subtracted from the dividend.

2y(y + 5) = 2y2 + 10y

So now the division problem looks like this:

2y_________


y + 5 | 2y2 + 7y - 15

-(2y2 + 10y) Be sure to ALWAYS put parenthesis around the polynomial that

you’re subtracting so that you don’t forget to distribute the

negative sign! In this problem, the + 10y will become -10y after

the negative sign is distributed.

Follow through with the long division method, distribute the negative sign, and subtract to get:

2y_________

y + 5 | 2y2 + 7y - 15

-2y2 - 10y

- 3y

Now, bring down the last term in the dividend: -15:

2y_________

y + 5 | 2y2 + 7y - 15

-2y2 - 10y

- 3y -15

Now, we must divide -3y by y, and we get -3

We add the – 3 to the top of the division sign

2y -3________

y + 5 | 2y2 + 7y - 15


-2y2 - 10y

- 3y -15

Then, we multiply -3 by the divisor:

-3(y + 5)

After we distribute, we get: -3y – 15

So we bring that down to be subtracted from the remaining portion of the dividend.

2y - 3________

y + 5 | 2y2 + 7y - 15

-2y2 - 10y

- 3y – 15

-3y – 15

0

There are no remainders, so our solution becomes:

y + 5 ÷ 2y2 + 7y – 15 = 2y – 3

Remember, it’s also a good idea to use multiplication to double-check your answers!


Lesson 16 It’s been a Long Time…

Name:_____________________________________________Date:________________

Now, on your own, use the long division method to divide the following polynomials. Remember, the first term in the dividend, and should go under the division sign.

Be sure to go through each step carefully, and refer back to the example above as you complete your work. Show all your work. In the next work set, problems #11-20, you will be ask to use multiplication to re-check your answers. You may want to complete that work as you simultaneously divide the original polynomials; that way you can check your answers for correctness right away.

1. (x2 + 2x -15) ÷ (x -3) ________________________________________________________________________

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2. (a2 - a -30) ÷ (a - 6) ________________________________________________________________________

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3. (b2 - 6 + 9) ÷ (b -3) ________________________________________________________________________

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4. (2y2 – 9y + 10) ÷ (y -2) ________________________________________________________________________

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5. (2m2 – 3m - 9) ÷ (m + 3) ________________________________________________________________________

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6. (4a2 + 22a - 12) ÷ (a + 6) ________________________________________________________________________

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7. (x3 + 4x3 + x -6) ÷ (x + 2) ________________________________________________________________________

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8. (6b3 – 24b -6b2 + 24) ÷ (b2 - 4) ________________________________________________________________________

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9. (5a2b +4ba -15a - 12) ÷ (ab - 3) ________________________________________________________________________

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10. (p5 + 4p4 - 2p3 - 16p2 -8p) ÷ (p3 – 4p) ________________________________________________________________________

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Now, for problems 11-20, check your work from above and multiply the quotient that you determined by the divisor to see if you come up with the dividend as your answer. Show all your work for the multiplication of the two polynomials, and then rewrite the original division problem as a multiplication problem with the solution.

The original multiplication problems have been listed for you. If you find that you have completed one of the above problems in error after you use multiplication to check it, then go back and show your work as you simplify it correctly.

11. (x2 + 2x -15) ÷ (x -3) = ________________________________________________________________________

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12. (a2 - a -30) ÷ (a - 6) = ________________________________________________________________________

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13. (b2 - 6 + 9) ÷ (b -3) = ________________________________________________________________________

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14. (2y2 – 9y + 10) ÷ (y -2) = ________________________________________________________________________

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15. (2m2 – 3m - 9) ÷ (m + 3) = (2m - 3) ________________________________________________________________________

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16. (4a2 + 22a - 12) ÷ (a + 6) = (4a - 2) ________________________________________________________________________

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17. (x3 + 4x3 + x -6) ÷ (x + 2) = (x2 + 2x -3) ________________________________________________________________________

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18. (6b3 – 24b -6b2 + 24) ÷ (b2 - 4) = (5b + b -6) ________________________________________________________________________

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19. (5a2b +4ba -15a - 12) ÷ (ab - 3) = (5a + 4) ________________________________________________________________________

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20. (p5 + 4p4 - 2p3 - 16p2 -8p) ÷ (p3 – 4p) = (p2 + 4p +2) ________________________________________________________________________

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Lesson 18 Cumulative Review for Lessons 11-17

Name:__________________________________________Date:____________________

Please complete the following cumulative review. Ask you instructor if you have any questions. You may want to review some of the instructional sections of previous lessons to help you remember certain processes as you complete the work below. These will be the same concepts covered in the upcoming assessment.

Determine the total degree of each polynomial listed.

1. 5y10 – 6x7y3 – 4y – 5 __________________________________________________________________

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2. x3 +4y - 9 __________________________________________________________________

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3. 60x2 y + 8 __________________________________________________________________

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Determine the leading term, leading coefficient, and constant term of each polynomial.

4. 5y3 – 4y – 1

Leading term: _______________________________________________________ Leading coefficient: __________________________________________________ Constant term: ______________________________________________________

5. 4x4 y3 + v y + 7

Leading term: _______________________________________________________ Leading coefficient: __________________________________________________ Constant term: ______________________________________________________

Determine if the binomial is a perfect square binomial. If so, show the original monomial squared.

6. x8y + x4y + 2 __________________________________________________________________

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7. y2 + 4y + 16

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Add the following polynomials by combining like terms, and place them into simplest form.

8. (10b8 + 3ab7 + c + 9abc + 4b2) + (1 + c + 4ab7 + 9b2)

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9. (13yx7 + 11yz9 + x + 9 + 8xy2) + (yz9 + 13xy7 + x + 10 + xy2)

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10. (5 + n + 4mo2 + 8n8) + (19 + 16mo2 + 4n10 + 9mno2) + (45 + n)

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Subtract the following polynomials. Be sure to place the polynomials in their simplest form based on the leading degree. Show all your work.

11. (7yx7 + 11y - 8xy2) - (yz9 + 17xy7 - xy2)

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12. (34 + n - 4mo2 - 8n8) - (16 + 16mo2 + 12mno2) - (16 + n)

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13. (- 23xy7 + x + 9 + 8z3) - (5 + 6x) - (7xy7 + 8z3)

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First square each of the binomials and place them into perfect square trinomial form, and then subtract the trinomials. Show all your work.

14. (2y - 5) 2 - (y + 6) 2

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15. (3m - 4) 2 - (m + 4)2

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Multiply the following polynomials. Be sure to place the polynomials in their simplest form based on the leading degree. Show all your work.

16. (4y + y2) (7xy2 + 6)

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17. (ab + 2b2) (5b2 + 3ab)

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18. (9xyz + 3xy2) (xy2 + 6)

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19. (4b + 2b2) (5b2 + 3b)

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20. (7yx7 + 11)(yz9 - 17xy4 + xy2)

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21. (4xy2+ 4y + 14) (14yx4 + 4xy2 + 8y)

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Find the greatest common factor in the following polynomials. Show all your work.

22. ab + 2a2b2 + 3ab

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23. 9xyz + 3xy4z+ 3xy2 + 6xz

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24. 4ab + 5ba2 + 3b

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25. 2yx7 + 10yz9 + xy4 + 5xy2

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Divide the following polynomials by the monomials listed. Show all your work.

26. 4xy2+ 4xy4 + 12xy2 ÷ 4xy2

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27. 5m2n2 + 5mno3 + 5mno2 ÷ 5mn

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28. 14s4t2 - 7s4t2 + 12t2 ÷ t

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Use the long division method to divide the following polynomials.

Show all your work. Use multiplication to re-check your answers. Show all your work for the multiplication of the two polynomials, and then rewrite the original division problem as a multiplication problem with the solution.

29. (2y2 – 9y + 10) ÷ (y -2)

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30. (2m2 – 3m - 9) ÷ (m + 3)

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lesson 11 excessive expressions - ogburn

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