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Foundations of Math 9 Updated June 2019
Adrian Herlaar, School District 61 www.mrherlaar.weebly.com
Section 6: Polynomials
This book belongs to: Block:
Section Due Date Date Handed In Level of Completion Corrections Made and Understood
π. π
π. π
π. π
Self-Assessment Rubric
Learning Targets and Self-Evaluation
L β T Description Mark
π β π Understanding terms, degree, coefficients, and constants
Grouping like terms
Pictorially demonstrating terms using algebra tiles
π β π Applying integer fundamentals in addition and subtraction of polynomials
Applying exponent laws in the multiplication and division of polynomials
Performing combined Operations of polynomials
Comments:
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
Category Sub-Category Description
Expert (Extending)
4 Work meets the objectives; is clear, error free, and demonstrates a mastery of the Learning Targets
βYou could teach this!β
3.5 Work meets the objectives; is clear, with some minor errors, and demonstrates a clear understanding of the Learning Targets
βAlmost Perfect, one little error.β
Apprentice (Proficient)
3 Work almost meets the objectives; contains errors, and demonstrates sound reasoning and thought
concerning the Learning Targets
βGood understanding with a few errors.β
Apprentice (Developing)
2 Work is in progress; contains errors, and demonstrates a partial understanding of the
Learning Targets
βYou are on the right track, but key concepts
are missing.β
Novice (Emerging)
1.5 Work does not meet the objectives; frequent errors, and minimal understanding of the Learning
Targets is demonstrated
βYou have achieved the bare minimum to meet the learning outcome.β
1 Work does not meet the objectives; there is no or minimal effort, and no understanding of the
Learning Targets
βLearning Outcomes not met at this time.β
Foundations of Math 9
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Competency Evaluation
A valuable aspect to the learning process involves self-reflection and efficacy. Research has shown that authentic
self-reflection helps improve performance and effort, and can have a direct impact on the growth mindset of the
individual. In order to grow and be a life-long learner we need to develop the capacity to monitor, evaluate, and
know what and where we need to focus on improvement. Read the following list of Core Competency Outcomes
and reflect on your behaviour, attitude, effort, and actions throughout this unit.
4 3 2 1
I listen during instruction and come ready to ask questions
Personal Responsibility
I am on time for class
I am fully prepared for the class, with all the required supplies
I am fully prepared for Tests
I follow instructions keep my Workbook organized and tidy
I am on task during work blocks
I complete assignments on time
I keep track of my Learning Targets
Self-Regulation
I take ownership over my goals, learning, and behaviour
I can solve problems myself and know when to ask for help
I can persevere in challenging tasks
I am actively engaged in lessons and discussions
I only use my phone for school tasks
Classroom
Responsibility and
Communication
I am focused on the discussion and lessons
I ask questions during the lesson and class
I give my best effort and encourage others to work well
I am polite and communicate questions and concerns with my peers and teacher in a timely manner
I clean up after myself and leave the classroom tidy when I leave
Collaborative Actions
I can work with others to achieve a common goal
I make contributions to my group
I am kind to others, can work collaboratively and build relationships with my peers
I can identify when others need support and provide it
Communication
Skills
I present informative clearly, in an organized way
I ask and respond to simple direct questions
I am an active listener, I support and encourage the speaker
I recognize that there are different points of view and can disagree respectfully
I do not interrupt or speak over others
Overall
Goal for next Unit β refer to the above criteria. Please select (underline/highlight) two areas you want to focus on
Rank yourself on the left of each column: 4 (Excellent), 3 (Good), 2 (Satisfactory), 1 (Needs Improvement)
I will rank your Competency Evaluation on the right half of each column
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Section 6.1 β Polynomials
Vocabulary
There will be a lot vocabulary necessary to accurately understand Polynomials
Term: Any variable, constant, or product of the two
Example: 3, 4π₯, π‘, 2π2, π₯π¦π§
Like Terms: Terms that have the same variable(s) to the same exponents
Example: π₯2 πππ 4π₯2, 7π‘ πππ 3π‘, 4 πππ 9
Degree of a Term: The exponent on the variable or sum of exponents on different variables
of one term
Example: 3π₯ ππ π·πππππ 1, 4π₯2 ππ π·πππππ 2, 5π₯π¦π§ ππ π·πππππ 3
Polynomial: Any term or terms separated by addition or subtraction where all
exponents on the variables are whole numbers
Example: 5π‘2 + 2π‘ β 7
Leading Term: The term in a Polynomial with the highest degree
Example: From above: 5π‘2 is the leading term, it has the highest degree
Descending Order: Writing terms from highest to lowest degree
Example: From Above: Is in descending order, degree goes 2, 1, 0
Polynomial Degree: The highest degree on a term, becomes the degree of the polynomial
Example: From Above: 5π‘2 is the leading term with degree 2, so the
Polynomial is of degree 2
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Combining like Terms
Combining like terms is doing exactly that
When we have a long list of terms written as a Polynomial, we can combine any that are
Like Terms: same variables, same exponent
Example: 3π2 β 4 + 2π β 2π2 + 4π β 8
Like Terms are: 3π2 πππ β2π2 , β 4 πππ β 8, 2π πππ 4π
So, 3π2 β 2π2 = π2
2π + 4π = 6π
β4 β 8 = β12
And Descending Order: ππ + ππ β ππ
Example: Combine the Like Terms and leave the simplified expression in Descending Order
5π₯π¦ + 5π₯2 + 2π₯ β 6 β 4π¦π₯ + 2π₯ + 6 β 3π₯2
Like Terms are:
+5π₯2πππ β3π₯2 so 5π₯2 β 3π₯2 = 2π₯2 π·πππππ ππ 2
5π₯π¦ πππ β 4π¦π₯ so 5π₯π¦ β 4π₯π¦ = π₯π¦ π·πππππ ππ 2
+2π₯ πππ + 2π₯ so 2π₯ + 2π₯ = 4π₯ π·πππππ ππ 1
β6 πππ + 6 so β6 + 6 = 0 π·πππππ ππ 0
Since π₯2πππ π₯π¦ are both degree 2, which one goes first?
We list them ALPHABETICALLY, π₯2 = π₯π₯ πππ π₯π₯ πππππ ππππππ π₯π¦
πππ + ππ + ππ
Degree 2 Degree 1
Degree 0
π₯π¦ and π¦π₯ are the same, in
multiplication order doesnβt matter,
π₯π¦ = π¦π₯
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Section 6.1 β Practice Questions
Identify the number of terms, what are they, and their degrees?
1. 3π₯ β 4π₯2 β 5
2. 4π₯π¦π§
3. β2π₯π¦π§ β 5π₯π¦ + 4
4. 5π₯3π¦ + 4π₯π¦3 β 6π₯π¦π§ 5. 5
6. 3π₯ + 4π¦ + 5π§ β π₯2
Put the following Polynomials in DESCENDING ORDER
7. 3 + 4π₯2 β 5π₯ 8. β2π‘ + 4π‘3 β 2π‘ β 3π‘2
9. 2 β π₯ + 5π₯2
10. π§2 β 4π§ + 5 11. π₯ + π₯π¦ + π₯π§ β π¦
12. β5π₯π¦ β π¦ + 2π₯ + π₯2
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Simplify the following, put your answer in DESCENDING ORDER
13. 3π‘ + 4 β 6π‘ + 2π‘2 + 4 β 3π‘2 14. 7π§3 + 2π§ β 4π§2 + 1 β 4π§2 β 5 + 3π§2 + 3π§ 15. 5π₯π¦ + 3 β 5π¦π₯ + 2 16. β4π + 5π2 β 7 β 5π2 + 4π + 7
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17. 1
3π2 + 2π β
1
6π2 + 4 β 9
18. β4.9π₯ β 3.2π¦ β 1.3π₯ + 4.2π¦ + 1
19. 11
5π₯ +
2
3π¦ β
3
5π₯ β
1
3π¦ + 10
20. 1
4π2 β π β
1
2π +
3
8π2 +
5
16π2
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Section 6.2 β Addition and Subtraction of Polynomials
All we are doing here is grouping like terms, however it involves a couple extra steps
Letβs start with Addition
Addition of Polynomials
Example: (π₯2 + 4π₯ β 7) + (2π₯2 β 3π₯ + 4)
We have 2 Polynomials, shown in brackets, and we are adding the second Polynomial to the first.
Step 1: In addition just drop the Brackets, keep the sign on the first term of the second
Polynomial, since itβs positive, nothing changes
π₯2 + 4π₯ β 7 + 2π₯2 β 3π₯ + 4
Step 2: Group the Like Terms
π₯2 + 4π₯ β 7 + 2π₯2 β 3π₯ + 4
= πππ + π β π
Make sure your answer is in DESCENDING ORDER!
We canβt SOLVE for the unknown yet, this is as far as we will go in this class.
If you make the Polynomial equal to something, then we can solve: We do this in Grade 10.
Example: (β4π2 + 3 β 2π) + (2 β 3π2 + 7π)
β4π2 + 3 β 2π + 2 β 3π2 + 7π
β4π2 β 3π2 β 2π + 7π + 3 + 2
βπππ + ππ + π
Drop the brackets, leave the sign on
the 1st term of the second Polynomial
Rearrange the terms so like terms are
together, descending order right
away is a bonus
Leave the solution in Descending Order
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Example: (5π₯π¦ + 3 β 2π₯) + (β3 + 2π₯ β 5π₯π¦)
5π₯π¦ + 3 β 2π₯ β 3 + 2π₯ β 5π₯π¦
5π₯π¦ β 5π₯π¦ β 2π₯ + 2π₯ + 3 β 3
π
Example: (4 + 3π‘2 β 7π₯) + (6π₯ β 2π‘2 β 12)
4 + 3π‘2 β 7π₯ + 6π₯ β 2π‘2 β 12
3π‘2 β 2π‘2 β 7π₯ + 6π₯ + 4 β 12
ππ β π β π
Subtraction of Polynomials
There is 1 very important concept to understand with subtraction
Consider this: (π₯2 + 5π₯ β 4) β (2π₯2 β 5π₯ β 4)
We are subtracting this from the 1st one. The subtraction symbol MUST affect each term.
Think about WATERBOMBING in the negative symbol
The signs change
(π₯2 + 5π₯ β 4) β (2π₯2 β 5π₯ β 4)
After you WATERBOMB in the negative you can change the signs and DROP the BRACKETS
Remember:
o πππππ‘ππ£π β πππππ‘ππ£π = πππ ππ‘ππ£π
o πππππ‘ππ£π β πππ ππ‘ππ£π = πππππ‘ππ£π
π₯2 + 5π₯ β 4 β 2π₯2 + 5π₯ + 4
Drop the brackets, leave the sign on
the 1st term of the second Polynomial
Rearrange the terms so like terms are
together, descending order right
away is a bonus
Leave the solution in Descending Order,
if everything cancels out, zero is a valid
answer!
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Now we GROUP the LIKE TERMS and weβre done
π₯2 β 2π₯2 = βπ₯2
5π₯ + 5π₯ = 10π₯
β4 + 4 = 0
So, in Descending Order,
βππ + πππ
Example: (3π₯2 β 4π₯ + 2) β (6π₯2 + 5π₯ β 12)
Step 1: Drop the Brackets 3π₯2 β 4π₯ + 2 β (6π₯2 + 5π₯ β 12)
of 1st Polynomial
Step 2: Waterbomb in the 3π₯2 β 4π₯ + 2 β 6π₯2 β 5π₯ + 12
(β) to the second one
and Drop the Brackets
Step 3: Group the LIKE TERMS βπππ β ππ + ππ
and put the result in
DESCENDING ORDER
Example: (9π2 + 4π + 5) β (β3π2 β 4π + 5)
9π2 + 4π + 5 + 3π2 + 4π β 5
9π2 + 3π2 + 4π + 4π + 5 β 5
ππππ + ππ
Example: β(2π‘2 + 4π‘ β 6) β (8π‘2 β 5π‘ + 2)
β2π‘2 β 4π‘ + 6 β 8π‘2 + 5π‘ β 2
β2π‘2 β 8π‘2 β 4π‘ + 5π‘ + 6 β 2
βππππ + π + π
Waterbomb in the negative sign
Group the LIKE TERMS
Waterbomb in the negative sign Waterbomb in the negative
Group the LIKE TERMS
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Section 6.2 β Practice Questions
Add the following Polynomials, leave answer in DESCENDING order.
1. (π₯ + 4) + (π₯ β 7)
2. (2π₯2 β 4π₯ β 7) + (3π₯2 β 7 + 4π₯)
3. (3π₯π¦ + 4π₯3 + 4) + (2π₯π¦ β 4π₯3 β 4) 4. (10 + 4π‘2 + 4π‘) + (2π‘ β 7π‘2 β 8)
5. (π3 + 2π2 + π + 4) + (3π3 β 2π2 β 7π + 15)
6. (4 + 6π₯ β 2π₯2) + (βπ₯2 β 2π₯) 7. (π‘2 + 4) + (βπ‘2 β 4)
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8. (βπ₯2 + 2 β 3π₯) + (β4π₯2 + π₯ β 5) 9. (β2π₯ + π₯2 β 2π¦2) + (βπ¦2 β π₯ + 2π₯2)
10. (4π₯ β 2π₯2) + (β5 + π₯2) 11. (β3 + 4π₯2 + 4π₯) + (5π₯ β 2π₯2 + 4)
12. (3π₯ β 2π₯π¦ + 2π¦) + (π₯π¦ β 3π¦) + (β3π¦ β π₯)
13. (β2π¦ + 3π₯ + π₯π¦) + (2π₯π¦ β π₯ β π¦) + (βπ₯ β 4π₯π¦)
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Subtract the Polynomials, leave answer in DESCENDING order.
14. (3π₯2 + 4π₯ β 7) β (β2π₯2 + 4π₯ + 9) 15. (π‘3 β 5π‘ + 4π‘2) β (π‘2 β 7π‘ β 2π‘2)
16. (π§ β 4) β (3π§ β 7) 17. (π€ β 7) β (2π€ + 4)
18. (π + 6) β (β2π β 2) 19. (π + 14) β (β5π + 7)
20. (2π2 + π β 7π) β (3π2 β π β 4π) β (6π2 β 8π + 7π)
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21. (5 β π‘) β (β7 + π‘) β (12 + π‘2) 22. (β2π₯ β 3π¦) β (4π₯ + 2π¦) β (π₯ β 3π¦) 23. (β5π₯ β 2π¦ + 3π§) β (β2π₯ + 9π¦) β (βπ₯ + π¦ β 2π§)
Perform the Combined Operations
24. (2π π‘ β π β π‘) β (β3π π‘ + π‘) + (βπ + 2π‘) 25. (β3π₯ + 4π¦) + (6π₯ β 5π¦) β (2π₯ + 11π¦ β 5π§) 26. (β2π₯π¦ + 9π§) + (4π₯2 β 11π§) β (6π₯2 + 8π₯π¦ β 11π§)
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Section 6.3 β Multiply, Divide, Combined Operations, and Tiles
Multiplication of Polynomials
Multiplication is awesome, all we do is WATERBOMB (DISTRIBUTIVITY) and use our
Exponent Laws for the Variables
Remember: When we multiply a COMMON BASE we ADD the exponents!
Also, the Order of Multiplication does not matter!
2 β 3 ππ π‘βπ π πππ ππ 3 β 2
π₯π¦π§ ππ π‘βπ π πππ ππ π¦π₯π§, π§π¦π₯, ππ π§π₯π¦
It makes no difference, keep this in mind
Example: 3(π₯ + 4)
3(π₯ + 4) Waterbomb in the 3
3 π‘ππππ π₯ ππ 3
3 π‘ππππ 4 ππ 12
So,
π(π + π) = ππ + ππ
Example: β4π₯(π₯ + 6)
β4π₯ β π₯ + β4π₯ β 6
βπππ β πππ
Example: π₯(π₯ + π¦)
π₯ β π₯ + π₯ β π¦
ππ + ππ
Example: 4π(3ππ β 2π)
4π β 3ππ + 4π β β2π
πππππ β πππ
Example: π‘2(6π β 4π‘)
π‘2 β 6π + π‘2 β β4π‘
6π‘2π β 4π‘3
βπππ + ππππ
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Example: 14π(2π2 β 4π)
14π β 2π2 + 14π β β4π
ππππ β ππππ
Division of Polynomials
Division of Polynomials is just fractions and exponent laws
Consider this:
2 + 3
7=
2
7+
3
7
So,
4π + 2
2=
4π
2+
2
2 β 2π + 1
Example:
π‘2 + 7π‘
π‘=
π‘2
π‘+
7π‘
π‘ β π‘ + 7
Example:
4π§3 β 2π§2 + 12π§
2π§
4π§3
2π§+
β2π§2
2π§+
12π§
2π§
πππ β π + π
Remember this?
We can break Polynomials down the same way.
π‘2
π‘= π‘
7π‘
π‘= 7
Exponent Laws
Anything divided by itself is 1
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Example:
8π₯π¦π§ + 4π¦π§ + 2π§
2π§
8π₯π¦π§
2π§+
4π¦π§
2π§+
2π§
2π§
πππ + ππ + π
Combined Operations
It is very rare that you only have to add, subtract, multiply, or divide only.
More often than not it involves a combination of steps
Example:
3π(π + 4) β 2π(4π + 6)
3π2 + 12π β 8π2 β 12π
β5π2
So itβll take a few steps, multiply first, add/subtract, then combine the terms and leave
your answer in Descending Order
Example:
4π‘(π‘2 + 5)
π‘
4π‘3 + 20π‘
π‘
4π‘3
π‘+
20π‘
π‘ = πππ + ππ
Waterbomb to remove the BRACKETS
Combine LIKE TERMS and SIMPLIFY
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Example:
3π₯(π₯ + 4)
π₯+
5π₯(3π₯ β 12)
3π₯
3π₯2 + 12π₯
π₯+
15π₯2 β 60π₯
3π₯
3π₯2
π₯+
12π₯
π₯+
15π₯2
3π₯+
β60π₯
3π₯
3π₯ + 12 + 5π₯ β 20
ππ β π
Example:
2π2(π β 4)
πβ
6(π2 + 2π)
2
2π3 β 8π2
πβ (
6π2 + 12π
2)
2π3
πβ
8π2
πβ (
6π2
2+
12π
2)
2π2 β 8πβ(3π2 + 6π)
2π2 β 8π β 3π2 β 6π
2π2 β 3π2 β 8π β 6π
βππ β πππ
Waterbomb to remove brackets
Divide each term by the denominator
Group LIKE TERMS and SIMPLIFY
Waterbomb to remove brackets
Since youβre subtracting, put brackets
around the second Polynomial so you
donβt forget to subtract each term
Divide each term by its denominator
Waterbomb in the NEGATIVE symbol
Group LIKE TERMS
Simplify the final solution
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Algebra Tiles: The Visual Representation of Polynomials
If the Tiles are Shaded In, they are the POSTIVE representation, non-shaded are NEGATIVE
That means that the Shaded and Non-Shaded CANCEL OUT
So here are a couple Examples:
Add the following:
π₯
π₯ 1
π₯
1 1
1 π‘πππ π₯ π‘πππ π₯2 π‘πππ
πππ
πππ
πππ ππππππ ππ’π‘
ππππππ ππ’π‘
ππππππ ππ’π‘
+ =
βπ₯2 β 2π₯ + 2 π₯2 + π₯ β 3 βπ₯ β 1
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Subtract the following (this is tricky):
Multiply the following:
β
Remember when we
illustrated INTEGERS
We need a NEGATIVE to
take away and we donβt
have one so we bring in β0β
Now we TAKE AWAY
=
π₯2 + π₯ β 1 βπ₯2 + π₯ β 1 2π₯2
π₯(βπ₯ β 2) = βπ₯2 β 2π₯
βπ₯ β 2
π₯
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Section 6.3 β Practice Questions
Multiply the following. Leave answer in DESCENDING order.
1. β3(π₯ β 7)
2. β(π‘2 β 7π‘ + 4)
3. 4π‘π(β2π‘2 + 3π)
4. 4π2(π2 + 7π β 2)
5. βπ§(π§ + 4)
6. 2π₯(2π¦ + π₯ β 3π§)
7. π₯π¦(π₯π¦π§ + π§ β π₯π¦) 8. 2π π‘(β3π + 4π‘ β π π‘)
9. β2π₯2(3π₯2 β 2π¦2 + 4π§2)
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Divide the following. Leave answer in DESCENDING order.
10. 3π₯+12
3 11.
π‘2+4π‘
π‘
12. 3π₯2β9π₯+6
3
13. 5π3+10π2β5π
5π 14.
β4π‘2+2π‘
2π‘
15. βπ2ππ β ππ2π + πππ2
βπππ
16. 18π§4β6π§3+3π§2
β3π§2 17. 4π12+6π3β8π2
β2πβ2
18. βπ2π2π+ππ2π2β π2π2π2
ππ2π
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Perform the Combined Operations. Answer in DESCENDING order.
19. β3(π₯2 + 4π₯) + 5π₯(π₯ β 6)
20. 2π‘(π‘2β4π‘)
π‘ β3π‘(4π‘ β 5)
21. 7π(3π2+4π)
7+
9π(6π2β3π)
3
22. β3π§3(π§β3)
3β
4π§2(3π§+6π§2)
3
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23. Add
24. Multiply
+ βπ₯2 π₯2 βπ₯ βπ₯
β1 +1
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Extra Work Space
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Answer Key
Section 6.1
1. πππππ : 3 3π₯, β4π₯2, β5
π·πππππ: 1, 2, 0
2. πππππ : 1 4π₯π¦π§
π·πππππ: 3
3. πππππ : 3 β2π₯π¦π§, β5π₯π¦, 4 π·πππππ: 3, 2, 0
4. πππππ : 3 5π₯3π¦, 4π₯π¦3, β6π₯π¦π§
π·πππππ: 4, 4, 3
5. πππππ : 1 5
π·πππππ: 0
6. πππππ : 4 3π₯, 4π¦, 5π§, βπ₯2
π·πππππ: 1, 1, 1, 2
7. 4π₯2 β 5π₯ + 3 8. 4π‘3 β 3π‘2 β 4π‘
9. 5π₯2 β π₯ + 2 10. π§2 β 4π§ + 5 11. π₯π¦ + π₯π§ + π₯ β π¦ 12. π₯2 β 5π₯π¦ + 2π₯ β π¦
13. βπ‘2 β 3π‘ + 8 14. 7π§3 β 5π§2 + 5π§ β 4 15. 5 16. 0
17. 1
6π2 + 2π β 5 18. β6.2π₯ + π¦ + 1 19.
8
5π₯ +
1
3π¦ + 10 20.
15
16π2 β
3
2π
Section 6.2
1. 2π₯ β 3 2. 5π₯2 β 14 3. 5π₯π¦ 4. β3π‘2 + 6π‘ + 2
5. 4π3 β 6π + 19 6. β3π₯2 + 4π₯ + 4 7. 0 8. β5π₯2 β 2π₯ β 3
9. 3π₯2 β 3π¦2 β 3π₯ 10. βπ₯2 + 4π₯ β 5 11. 2π₯2 + 9π₯ + 1 12. βπ₯π¦ + 2π₯ β 4π¦
13. βπ₯π¦ + π₯ β 3π¦ 14. 5π₯2 β 16 15. π‘3 + 5π‘2 + 2π‘ 16. β2π§ + 3
17. βπ€ β 11 18. 3π + 8 19. 6π + 7 20. β7π2
21. βπ‘2 β 2π‘ 22. β7π₯ β 2π¦ 23. β2π₯ β 12π¦ + 5π§ 24. 5π π‘ β 2π
25. π₯ β 12π¦ + 5π§ 26. β2π₯2 β 10π₯π¦ + 9π§
Section 6.3
1. β3π₯ + 21 2. βπ‘2 + 7π‘ β 4 3. β8π‘3π + 12π‘π2 4. 4π4 + 28π3 β 8π2
5. βπ§2 β 4π§ 6. 2π₯2 + 4π₯π¦ β 6π₯π§ 7. π₯2π¦2π§ β π₯2π¦2 + π₯π¦π§ 8. β2π 2π‘2 β 6π 2π‘ + 8π π‘2
9. β6π₯4 + 4π₯2π¦2 β 8π₯2π§2 10. π₯ + 4 11. π‘ + 4 12. π₯2 β 3π₯ + 2
13. π2 + 2π β 1 14. β2π‘ + 1 15. π + π β π 16. β6π§2 + 2π§ β 1
17. β2π14 β 3π5 + 4π4 18. βππ β π + π 19. 2π₯2 β 42π₯ 20. β10π‘2 + 7π‘
21. 21π3 β 5π2 22. β9π§4 β π§3 23. β4π₯ β 1 24. 2π₯