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1 STANDARDS OF LEARNING CONTENT REVIEW NOTES ALGEBRA I 3 rd Nine Weeks, 2018-2019
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Page 1: STANDARDS OF LEARNING CONTENT REVIEW NOTESstar.spsk12.net/math/Algebra I/AlgebraICRN3NW.pdf6 Example 7: Γ‘5𝑝7 Γ‘4𝑝 J5 L7 J4 L = J5βˆ’4 L7βˆ’1= J L6 Example 8: ( Γ”2 Γ• Γ–5 Γ”2

1

STANDARDS OF LEARNING

CONTENT REVIEW NOTES

ALGEBRA I

3rd Nine Weeks, 2018-2019

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OVERVIEW

Algebra I Content Review Notes are designed by the High School Mathematics Steering Committee as a resource

for students and parents. Each nine weeks’ Standards of Learning (SOLs) have been identified and a detailed

explanation of the specific SOL is provided. Specific notes have also been included in this document to assist

students in understanding the concepts. Sample problems allow the students to see step-by-step models for

solving various types of problems. A β€œ ” section has also been developed to provide students with the

opportunity to solve similar problems and check their answers. The answers to the β€œ ” problems are

found at the end of the document.

The document is a compilation of information found in the Virginia Department of Education (VDOE)

Curriculum Framework, Enhanced Scope and Sequence, and Released Test items. In addition to VDOE

information, Prentice Hall textbook series and resources have been used. Finally, information from various

websites is included. The websites are listed with the information as it appears in the document.

Supplemental online information can be accessed by scanning QR codes throughout the document. These will

take students to video tutorials and online resources. In addition, a self-assessment is available at the end of the

document to allow students to check their readiness for the nine-weeks test.

The Algebra I Blueprint Summary Table is listed below as a snapshot of the reporting categories, the number of

questions per reporting category, and the corresponding SOLs.

Algebra I Blueprint Summary Table

Reporting Categories No. of Items SOL

Expressions & Operations 12 A.1

A.2a – c

A.3

Equations & Inequalities 18 A.4a – f

A.5a – d

A.6a – b

Functions & Statistics 20 A.7a – f

A.8

A.9

A.10

A.11

Total Number of Operational Items 50

Field-Test Items* 10

Total Number of Items 60

* These field-test items will not be used to compute the students’ scores on the test.

It is the Mathematics Instructors’ desire that students and parents will use this document as a tool toward the

students’ success on the end-of-year assessment.

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Laws of Exponents & Polynomial Operations A.2 The student will perform operations on polynomials, including

a) applying the laws of exponents to perform operations on expressions;

Monomial is a single term. It could refer to a number, a variable, or a product of a number and one or more variables. Some examples of monomials include:

14π‘Žπ‘Β² βˆ’ 6𝑑 1 1

2π‘₯²𝑦𝑧²

When you multiply monomials that have a common base, you add the exponents.

Example 1: Multiply 𝑏2 βˆ™ 𝑏5

𝑏2 βˆ™ 𝑏5 = 𝑏2+5 = 𝑏7

This works because when you raise a number or variable to a power, it is like multiplying it by itself that many times. When you then multiply this by another power of the same number or variable, you are just multiplying it by itself that many more times.

Example 2: π‘…π‘’π‘€π‘Ÿπ‘–π‘‘π‘’ 43 π‘Žπ‘›π‘‘ 42π‘Žπ‘  π‘šπ‘’π‘™π‘‘π‘–π‘π‘™π‘–π‘π‘Žπ‘‘π‘–π‘œπ‘› π‘π‘Ÿπ‘œπ‘π‘™π‘’π‘šπ‘ . π‘ˆπ‘ π‘’ π‘‘β„Žπ‘–π‘  π‘‘π‘œ π‘ π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 43 βˆ™ 42.

43 = 4 βˆ™ 4 βˆ™ 4 42 = 4 βˆ™ 4

43 βˆ™ 42 = 4 βˆ™ 4 βˆ™ 4 βˆ™ 4 βˆ™ 4 = 45

Example 3: Simplify 1

2π‘₯𝑦3𝑧5 βˆ™ 14π‘₯3𝑧 βˆ™ 𝑦6𝑧2

(1

2 βˆ™ 14) (π‘₯ βˆ™ π‘₯3)(𝑦3 βˆ™ 𝑦6)(𝑧5 βˆ™ 𝑧 βˆ™ 𝑧2)

7π‘₯4𝑦9𝑧8 When you raise a power to a power, you multiply the exponents.

(32)4

32 βˆ™ 32 βˆ™ 32 βˆ™ 32

32+2+2+2

38

Example 4: Simplify (π‘Ž2𝑏)4

(π‘Ž2𝑏)4 = π‘Ž2βˆ™4𝑏1βˆ™4

π‘Ž8𝑏4

This means 3Β² times itself 4 times!

Scan this QR code to go to a video tutorial on

multiplying monomials!

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5

Example 5: Simplify (5𝑑5𝑒7𝑓2)3

53𝑑5βˆ™3𝑒7βˆ™3𝑓2βˆ™3

125𝑑15𝑒21𝑓6

Often, you will be asked to multiply monomials and raise powers to a power. Make sure that you follow the ORDER OF OPERATIONS! Raise to powers first, then multiply.

Example 6: Simplify (βˆ’2π‘₯³𝑦𝑧²)Β³(2π‘₯³𝑦𝑧²)⁴

(βˆ’8π‘₯9𝑦3𝑧6) βˆ™ (16π‘₯12𝑦4𝑧8)

βˆ’128π‘₯21𝑦7𝑧14

Laws of Exponents Simplify each expression

1. 6π‘Ž3𝑏5(βˆ’π‘Žπ‘2)

2. (βˆ’5π‘₯2𝑦𝑧3)2

3. [(π‘šπ‘›6𝑝3)2]4

4. (π‘Ž4𝑏2𝑐7)4 (13π‘Ž2𝑐)3 When you divide monomials with like bases, you will subtract the exponents.

Anything raised to the zero power is equal to ONE!

π‘₯0 = 1 150 = 1 (βˆ’235𝑦7)0 = 1 To find the power of a quotient, raise both the numerator and the denominator to the power. (Remember to follow the order of operations!)

(π‘Ž

𝑏)

5

= π‘Ž5

𝑏5

Page 6: STANDARDS OF LEARNING CONTENT REVIEW NOTESstar.spsk12.net/math/Algebra I/AlgebraICRN3NW.pdf6 Example 7: Γ‘5𝑝7 Γ‘4𝑝 J5 L7 J4 L = J5βˆ’4 L7βˆ’1= J L6 Example 8: ( Γ”2 Γ• Γ–5 Γ”2

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Example 7: 𝑛5𝑝7

𝑛4𝑝

𝑛5𝑝7

𝑛4𝑝= 𝑛5βˆ’4 𝑝7βˆ’1 = 𝑛 𝑝6

Example 8: (π‘Ž2𝑏𝑐5

π‘Ž2𝑐2 )3

(π‘Ž2𝑏𝑐5

π‘Ž2𝑐2)

3

= (π‘Ž2βˆ’2𝑏𝑐5βˆ’2)3 = (π‘Ž0𝑏𝑐3)3 = 13𝑏3𝑐9 = 𝑏3𝑐9

You will also see negative exponents in monomials. When you have a negative exponent, you will reciprocate that variable (move it to the other side of the fraction bar) and the exponent will become positive.

As an example: 2βˆ’3 = .125 = 1

8 π‘œπ‘Ÿ 2βˆ’3 =

1

23

When simplifying monomials with negative exponents, you can start by β€˜flipping over’ all of the negative exponents to make them positive. Then, simplify.

Example 9: π‘Žβˆ’3𝑏2π‘βˆ’5

π‘Žβˆ’7𝑏𝑐10

π‘Žβˆ’3𝑏2π‘βˆ’5

π‘Žβˆ’7𝑏𝑐10

π‘Ž7𝑏2

π‘Ž3𝑏𝑐10𝑐5

π‘Ž4𝑏

𝑐15

Example 10: (2π‘₯2π‘¦βˆ’4

3π‘₯𝑦5 )βˆ’3

(2π‘₯2π‘¦βˆ’4

3π‘₯𝑦5)

βˆ’3

= (3π‘₯𝑦5

2π‘₯2π‘¦βˆ’4)

3

= (3π‘₯𝑦5𝑦4

2π‘₯2)

3

= (3𝑦9

2π‘₯)

3

= 27𝑦27

8π‘₯3

Remember that anything to the zero power equals 1!

Scan this QR code to go to a video tutorial on

dividing monomials!

Scan this QR code to go to a video tutorial on

simplifying monomials

with negative exponents!

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Exponents Laws of Exponents

Simplify each expression

5. 33π‘Ž5𝑏9

11π‘Ž6𝑏2

6. (π‘₯4𝑦2

2π‘₯𝑦)

3

7. (2π‘₯π‘¦βˆ’3)5

8. 𝑛4𝑛2π‘šβˆ’5

𝑛6π‘šβˆ’2𝑝0

9. 15π‘Žπ‘βˆ’2𝑐5

9π‘Žβˆ’4𝑏2𝑐2

10. (6π‘₯4𝑦2π‘§βˆ’3

9π‘₯βˆ’2𝑦0π‘§βˆ’1 )

βˆ’2

Polynomials A.2 The student will perform operations on polynomials, including

b) adding, subtracting, and multiplying polynomials.

Adding and subtracting polynomials is the same as COMBINING LIKE TERMS. In order for two terms to be like terms, they must have the same variables and the same exponents.

Like Terms NOT Like Terms

5π‘Žπ‘2, βˆ’3π‘Žπ‘2,2

3π‘Žπ‘2 5π‘Žπ‘2, βˆ’3π‘Ž2𝑏,

2

3π‘Žπ‘

Each of these terms contain an β€˜π‘Žπ‘2 β€˜, therefore they are like terms.

Although these terms have the same variables, corresponding variables do not

have the same exponents. Therefore, these are NOT like terms.

Example 1: (2π‘₯2𝑦 + 5π‘₯𝑦 βˆ’ 7𝑦2) + (4π‘₯2𝑦 βˆ’ 10π‘₯𝑦 + 3𝑦2)

(2π‘₯2𝑦 + 4π‘₯2𝑦) + (5π‘₯𝑦 βˆ’ 10π‘₯𝑦) + (βˆ’7𝑦2 + 3𝑦2)

6π‘₯2𝑦 βˆ’ 5π‘₯𝑦 βˆ’ 4𝑦2

Like terms are underlined here. Remember that each term takes the sign in front of it!

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8

Remember that if you are subtracting a polynomial, you are subtracting all of the terms (Therefore, you must distribute the negative to each term first!)

Example 2: (βˆ’3π‘Žπ‘ βˆ’ 5π‘Ž4𝑏2 + 𝑏) βˆ’ (4π‘Žπ‘ βˆ’ 6𝑏)

βˆ’3π‘Žπ‘ βˆ’ 5π‘Ž4𝑏2 + 𝑏 βˆ’ 4π‘Žπ‘ + 6𝑏

βˆ’7π‘Žπ‘ βˆ’ 5π‘Ž4𝑏2 + 7𝑏

Polynomials Simplify each expression

1. (π‘šπ‘›π‘2 βˆ’ 7𝑛𝑝2 + 12π‘šπ‘›) + (4π‘šπ‘›π‘2 βˆ’ 3π‘šπ‘›)

2. (4π‘Ž + 9𝑏 βˆ’ 3𝑐 + 2𝑑) + (2π‘Ž βˆ’ 𝑏 βˆ’ 5𝑐 + 3𝑑2)

3. (12π‘₯2 βˆ’ 6π‘₯𝑦 + 9𝑦2) βˆ’ (3π‘₯2 + π‘₯𝑦 βˆ’ 4𝑦2)

4. (32π‘Žπ‘2 + 5π‘Ž2𝑏 βˆ’ 21𝑏2) βˆ’ (π‘Žπ‘ + 14𝑏2 βˆ’ 5π‘Ž2𝑏) 5. (2𝑑𝑒 βˆ’ 8𝑒 + 7𝑑) + (βˆ’π‘‘π‘’ βˆ’ 4𝑑) βˆ’ (3𝑒 + 𝑑 βˆ’ 5𝑑𝑒) To multiply a polynomial by a monomial, simply distribute the monomial to each term in the polynomial. You will use the rules of exponents to simplify each term.

Example 3: 5π‘₯ (3π‘₯2 βˆ’ 6π‘₯𝑦 + 2𝑦2)

(5π‘₯ βˆ™ 3π‘₯2) βˆ’ (5π‘₯ βˆ™ 6π‘₯𝑦) + (5π‘₯ βˆ™ 2𝑦2)

15π‘₯3 βˆ’ 30π‘₯2𝑦 + 10π‘₯𝑦2

Distribute the negative to everything in the second set of parentheses!

Then, COMBINE LIKE TERMS!

Scan this QR code to go to a video tutorial on

adding and subtracting polynomials.

Distribute the 5π‘₯ to each term.

Then, simplify each term

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Example 4: βˆ’2π‘Ž2𝑏 (5π‘Žπ‘3 βˆ’ 6π‘Ž2𝑏5 + π‘Ž2𝑏 βˆ’ π‘Žπ‘3 ) (βˆ’2π‘Ž2𝑏 βˆ™ 5π‘Žπ‘3) + (βˆ’2π‘Ž2𝑏 βˆ™ βˆ’6π‘Ž2𝑏5 ) + (βˆ’2π‘Ž2𝑏 βˆ™ π‘Ž2𝑏 ) + (βˆ’2π‘Ž2𝑏 βˆ™ βˆ’π‘Žπ‘3)

βˆ’10π‘Ž3𝑏4 + 12π‘Ž4𝑏6 βˆ’ 2π‘Ž4𝑏2 + 2π‘Ž3𝑏4

βˆ’8π‘Ž3𝑏4 + 12π‘Ž4𝑏6 βˆ’ 2π‘Ž4𝑏2 To multiply two polynomials together, distribute each term in the first polynomial to each term in the second polynomial. When you are multiplying two binomials together this may be called FOIL. FOIL stands for:

F – First – multiply the first term in each binomial together O – Outer – multiply the outermost term in each binomial together I – Inner – multiply the innermost term in each binomial together L – Last – multiply the last term in each binomial together (This is the exact same as distributing the first term, then distributing the second term) Don’t forget to combine like terms when possible. Example 5: (2π‘₯ + 5)(3π‘₯ βˆ’ 2) First Outer Inner Last

(2π‘₯ βˆ™ 3π‘₯) + (2π‘₯ βˆ™ βˆ’2) + (5 βˆ™ 3π‘₯) + (5 βˆ™ βˆ’2)

6π‘₯2 βˆ’ 4π‘₯ + 15π‘₯ βˆ’ 10

6π‘₯2 + 11π‘₯ βˆ’ 10

Example 6: (π‘Ž2 + 2𝑏2)(4π‘Ž βˆ’ 3π‘Žπ‘ + 6𝑏)

(π‘Ž2)(4π‘Ž βˆ’ 3π‘Žπ‘ + 6𝑏) + (2𝑏2)(4π‘Ž βˆ’ 3π‘Žπ‘ + 6𝑏)

4π‘Ž3 βˆ’ 3π‘Ž3𝑏 + 6π‘Ž2𝑏 + 8π‘Žπ‘2 βˆ’ 6π‘Žπ‘3 + 12𝑏3

Don’t forget to check for like terms!

Scan this QR code to go to a video tutorial on

multiplying monomials and polynomials.

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10

Example 7: (4𝑦 βˆ’ 3)2 (4𝑦 βˆ’ 3)(4𝑦 βˆ’ 3) (4𝑦 βˆ™ 4𝑦) + (4𝑦 βˆ™ βˆ’3) + (βˆ’3 βˆ™ 4𝑦) + (βˆ’3 βˆ™ βˆ’3)

16𝑦2 βˆ’ 12𝑦 βˆ’ 12𝑦 + 9

16𝑦2 βˆ’ 24𝑦 + 9

Polynomials

6. βˆ’2π‘šπ‘›2(4π‘š2𝑛 βˆ’ 3π‘šπ‘›) 7. 2π‘Žπ‘2𝑐 (4π‘Ž + 𝑏 βˆ’ 3𝑐) + 5π‘Žπ‘2𝑐 (6π‘Ž βˆ’ 3𝑏 βˆ’ 5𝑐) 8. (6π‘₯ + 5)(6π‘₯ βˆ’ 5)

9. (5π‘Ž2𝑏 βˆ’ 2𝑏2)(4π‘Žπ‘ + 𝑏)

10. (3π‘₯𝑦 βˆ’ 5π‘₯)2 Factoring A.2 The student will perform operations on polynomials, including

c) factoring completely first- and second-degree binomials and trinomials in one variable.

The prime factorization of a number or monomial is that number or monomial broken

down into the product of its prime factors.

Example 1: Write the prime factorization of 18π‘₯3

18π‘₯3

18π‘₯3𝑦2 = 2 βˆ™ 3 βˆ™ 3 βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ or 2 βˆ™ 32 βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯

To find the greatest common factor (GCF) of two or more monomials, break each down

into its prime factorization. The GCF is the product of all of the shared factors.

Remember that to square something means to multiply it by itself!

9 2

3 3

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11

Example 2: What is the greatest common factor of 9π‘Ž, 15π‘Ž, π‘Žπ‘›π‘‘ 6π‘Ž2

𝐺𝐢𝐹 = 3π‘Ž

You can use the GCF to help you rewrite (factor) polynomials. If all of the terms in the

polynomial have common factors you can pull these factors out from the terms to factor

the polynomial.

Example 3: Factor 8π‘₯2 + 20π‘₯

8π‘₯2 = 2 βˆ™ 2 βˆ™ 2 βˆ™ π‘₯ βˆ™ π‘₯

20π‘₯ = 2 βˆ™ 2 βˆ™ 5 βˆ™ π‘₯

4π‘₯ (2π‘₯ + 5)

Example 4: Factor 15π‘Ž3 βˆ’ 15π‘Ž2

15π‘Ž3 = 3 βˆ™ 5 βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™

βˆ’30π‘Ž2 = βˆ’1 βˆ™ 2 βˆ™ 3 βˆ™ 5 βˆ™ π‘Ž βˆ™ π‘Ž βˆ™

15π‘Ž2 = 3 βˆ™ 5 βˆ™ π‘Ž βˆ™ π‘Ž βˆ™

15π‘Ž2 (π‘Ž βˆ’ 1)

Factoring

1. Write the prime factorization of 180π‘Ž2

2. Find the greatest common factor of 15π‘₯2 π‘Žπ‘›π‘‘ 42π‘₯

3. Factor 8𝑐3 + 14𝑐 βˆ’ 12

Circle each factor that they ALL have in common!

GCF = 2 βˆ™ 2 βˆ™ π‘₯ = 4π‘₯

Pull the GCF out from each term and rewrite. Check your work by distributing.

GCF = 3 βˆ™ 5 βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ 𝑏 = 15π‘Ž2

Pull the GCF out from each term and rewrite. Check your work by distributing.

Scan this QR code to go to a video tutorial on

greatest common factors.

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Simplifying Radicals

A.3 The student will simplify a) square roots of whole numbers and monomial algebraic expressions;

To simplify a radical, you will pull out any perfect square factors (i.e. 4, 9, 16, 25, etc.)

√18 = √9 βˆ™ 2

The square root of 9 is equal to 3, so you can pull the square root of 9 from underneath

the radical sign to find the simplified answer 3 √2 , which means 3 times the square

root of 2. You can check this simplification in your calculator by verifying that

√18 = 3√2 .

Another way to simplify radicals, if you don’t know the factors of a number, is to create

a factor tree and break the number down to its prime factors. When you have broken

the number down to all of its prime factors you can pull out pairs of factors for square

roots, which will multiply together to make perfect squares.

Example 5: Simplify √128 √2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2

2 βˆ™ 2 βˆ™ 2 √2 = 8√2

Example 6: Simplify 3√32π‘₯3𝑦

To simplify a root of a higher index, you pull out factors that occur the same number of

times as the index of the radical. As an example, if you are simplifying √64π‘Ž75 , you

would only pull out factors that occurred 5 times, since 5 is the index of the root.

2 64

8 8

4 2 2 4

2 2 2 2

16 2 x x x y

4 4

2 2 2 2

3√2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ π‘₯ βˆ™ 𝑦

3 βˆ™ 2 βˆ™ 2 βˆ™ π‘₯√2π‘₯𝑦 = 12π‘₯√2π‘₯𝑦

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Example 7: Simplify √64π‘Ž73

√2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ 2 βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž βˆ™ π‘Ž3

2 βˆ™ 2 βˆ™ π‘Ž βˆ™ π‘Ž βˆšπ‘Ž3

4π‘Ž2 βˆšπ‘Ž3

Simplifying Radicals Simplify the following radicals.

4. √4π‘₯4𝑦3

5. 6π‘Žβˆš15π‘Žπ‘4𝑐3

6. √48𝑐4𝑑23

Factoring Special Cases A.2 The student will perform operations on polynomials, including

c) factoring completely first- and second-degree binomials and trinomials in one variable.

A.3 The student will simplify

a) square roots of whole numbers and monomial algebraic expression b) cube roots of integers

Factoring Trinomials

To factor a trinomial of the form π‘₯2 + 𝑏π‘₯ + 𝑐, first find two integers whose sum is equal

to 𝑏, and whose product is equal to π‘Ž βˆ™ 𝑐 . You can start by listing all of the factors of π‘Ž βˆ™ 𝑐, and then see which two factors add up to the coefficient of 𝑏. Once you have determined which factors to use, you can put all of your terms β€œin a box” and factor the rows and columns.

Scan this QR code to go to a video tutorial on

simplifying radicals.

Because this is a cube root, I pulled out things that occurred 3 times.

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Example 1: Factor π‘₯2 + 6π‘₯ + 8 π‘Ž βˆ™ 𝑐 = 1 βˆ™ 8 = 8 So, we are looking for factors of 8 that add up to 6! Factors of 8 Sum of factors 1, 8 9 2, 4 6 Put terms β€œin a box”

Sometimes you will not be able to find factors of π‘Ž βˆ™ 𝑐 that sum to b. When this happens, the trinomial is PRIME.

Example 2: Factor 2π‘₯2 + 5π‘₯ βˆ’ 2 π‘Ž βˆ™ 𝑐 = 2 βˆ™ βˆ’2 = βˆ’4 So, we are looking for factors of -4 that add up to 5! Factors of 8 Sum of factors 1, -4 -3 -1, 4 3 -2, 2 0 Nothing works, therefore this trinomial is PRIME When factoring, anytime the 𝑏 term is negative and the 𝑐 term is positive, your answer will have two minus signs!

First Term

(π‘Žπ‘₯2)

One Factor (__π‘₯)

Other Factor (__π‘₯)

Last Term

(c)

π‘₯2 2π‘₯

4π‘₯ 8

Find the greatest common factor in each row and each column. These will give you your two binomials!

π‘₯ 4

π‘₯ 2

(π‘₯ + 4) (π‘₯ + 2)

Check your answer by FOIL-ing!

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15

Example 3: Factor 8π‘₯2 βˆ’ 21π‘₯ + 10 π‘Ž βˆ™ 𝑐 = 8 βˆ™ 10 = 80 So, we are looking for factors of 80 that add up to βˆ’21! Factors of 80 Sum of factors -4, -20 -24 -5, -16 -21

Example 4: Factor 3π‘₯2 + 24π‘₯ + 45

Pull out a GCF first!! 3(π‘₯2 + 8π‘₯ + 15) π‘Ž βˆ™ 𝑐 = 1 βˆ™ 15 = 15 So, we are looking for factors of 15 that add up to 8!

Factoring Special Cases Factor each of the trinomials below

1. π‘₯2 + 7π‘₯ + 12

2. 2π‘₯2 βˆ’ 14π‘₯𝑦 βˆ’ 36𝑦2

3. 6π‘₯2 + 17π‘₯ + 5

4. π‘₯2 βˆ’ 9π‘₯ + 1

8π‘₯2 βˆ’5π‘₯

βˆ’16π‘₯ 10

π‘₯2 3π‘₯

5π‘₯ 15

Find the greatest common factor in each row and each column. These will give you your two binomials!

π‘₯

βˆ’2

8π‘₯ βˆ’5

(8π‘₯ βˆ’ 5) (π‘₯ βˆ’ 2)

Check your answer by FOIL-ing!

Scan this QR code to go to a video tutorial on

factoring trinomials.

Find the greatest common factor in each row and each column. These will give you your two binomials!

π‘₯

5

π‘₯ 3

3(π‘₯ + 5) (π‘₯ + 3)

Check your answer by FOIL-ing! Don’t forget your GCF in the front.

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16

To solve a quadratic equation (i.e. find its solutions, roots, or zeros), set one side equal

to zero (put the quadratic in standard form), then factor. Set each factor equal to zero

to find the values for π‘₯ that are the solutions to the quadratic.

Example 5: Find the zeros of π‘₯2 βˆ’ 18 = 7π‘₯

Start by getting one side equal to zero and write in standard form.

π‘₯2 βˆ’ 18 = 7π‘₯

βˆ’7π‘₯ βˆ’ 7π‘₯

π‘₯2 βˆ’ 7π‘₯ βˆ’ 18 = 0 Now factor the trinomial.

We are looking for factors of βˆ’18 that add up to βˆ’7. βˆ’9 and 2 work!

(x + 2)(x βˆ’ 9) = 0 Set both factors equal to zero!

π‘₯ + 2 = 0 π‘Žπ‘›π‘‘ π‘₯ βˆ’ 9 = 0

π‘₯ = βˆ’2 π‘Žπ‘›π‘‘ 9 or {βˆ’2, 9}

You can check your answer in your calculator by graphing the quadratic. The solutions

are the x-intercepts, so this graph should cross the x-axis at -2 and 9.

Factoring Special Cases Find the solution to each trinomial

5. π‘₯2 + 9π‘₯ + 20 = 0

6. 2π‘₯2 + 6 = 7π‘₯

7. 2π‘₯3 + 10π‘₯2 βˆ’ 10π‘₯ = 2π‘₯

π‘₯2 βˆ’9π‘₯

2π‘₯ βˆ’18

π‘₯

2

π‘₯ βˆ’9

-2 9

Scan this QR code to go to a video tutorial on solving

trinomials by factoring.

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Special Cases

A perfect square trinomial can be factored to two binomials that are the same, so you

can write it as the binomial squared.

π‘Ž2 + 2π‘Žπ‘ + 𝑏2 = (π‘Ž + 𝑏)2 π‘Ž2 βˆ’ 2π‘Žπ‘ + 𝑏2 = (π‘Ž βˆ’ 𝑏)2

Example 6: Factor 4π‘₯2 βˆ’ 24π‘₯ + 36

If your first and last terms are perfect squares you can check for a perfect square

trinomial. Take the square root of the first and last number and see if the product

of those is equal to Β½ of the middle number.

√4 = 2 π‘Žπ‘›π‘‘ √36 = 6 6 βˆ™ 2 = 12 , π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 1

2 π‘œπ‘“ 24

Now that we know this case works, you can write the binomial factor squared

(2π‘₯ βˆ’ 6)2

Remember to check your answer by FOIL-ing the binomials back out!

Another special case is if the quadratic is represented as the difference of two perfect

squares (i.e. 4π‘₯2 βˆ’ 16). If both the first and last term are perfect squares, and the two

terms are being subtracted their factorization can be written as (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏). As an

example 4π‘₯2 βˆ’ 16 = (2π‘₯ + 4)(2π‘₯ βˆ’ 4). Remember that you can check your work by

FOIL-ing.

Example 7: Factor completely 3π‘₯2 βˆ’ 27

To begin, you should factor out a GCF. In this case it would be 3.

3(π‘₯2 βˆ’ 9) Now you are left with a difference of squares!

3(π‘₯ + 3)(π‘₯ βˆ’ 3)

Factoring Special Cases

8. πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ 4π‘₯2 βˆ’ 9𝑦2

9. 𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘Ÿπ‘œπ‘œπ‘‘(𝑠) π‘₯2 + 12π‘₯ + 36 = 0

10. πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ π‘π‘œπ‘šπ‘π‘™π‘’π‘‘π‘’π‘™π‘¦ 8π‘₯3 βˆ’ 56π‘₯2 + 98π‘₯

Scan this QR code to go to a video tutorial on factoring

special cases.

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18

Solving Quadratic Equations A.4 The student will solve multistep linear and quadratic equations in two variables

b) solving quadratic equations algebraically and graphically;

Graphing a quadratic equation

Standard form for a quadratic function is: 𝑓(π‘₯) = π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 , π‘Ž β‰  0

The graph of a quadratic equation will be a parabola.

If π‘Ž > 0, then the parabola opens upward. If π‘Ž < 0, then the parabola opens downward.

The axis of symmetry is the line = βˆ’π‘

2π‘Ž .

The x-coordinate of the vertex is βˆ’π‘

2π‘Ž . The y-coordinate of the vertex is found by

plugging that x value into the equation and solving for 𝑓(π‘₯).

The y-intercept is (0, 𝑐).

To graph a quadratic:

1. Identify a, b, and c.

2. Find the axis of symmetry (π‘₯ = βˆ’π‘

2π‘Ž ), and lightly sketch.

3. Find the vertex. The x-coordinate is βˆ’π‘

2π‘Ž . Use this to find the y-coordinate.

4. Plot the y-intercept (c), and its reflection across the axis of symmetry.

5. Draw a smooth curve through your points.

The vertex of a parabola is its turning point, or the β€˜tip’ of the parabola. In this picture, the turning point is at (2, 0).

Example 1: Graph 𝑦 = 2π‘₯2 βˆ’ 4π‘₯ + 3

Step 1: Identify a, b, and c. π‘Ž = 2, 𝑏 = βˆ’4, π‘Žπ‘›π‘‘ 𝑐 = 3

Step 2: Find and sketch the axis of symmetry.

π‘₯ = βˆ’π‘

2π‘Ž π‘₯ =

βˆ’(βˆ’4)

2(2) π‘₯ =

4

4 π‘₯ = 1

Step 3: Find the vertex.

Scan this QR code to go to a video tutorial on graphing and

solving quadratic equations.

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The x-coordinate is 1. Plug this in to find y.

𝑦 = 2(1)2 βˆ’ 4(1) + 3 𝑦 = 2 βˆ’ 4 + 3 𝑦 = 1

The vertex is (1, 1).

Step 4: Plot the y-intercept and its reflection.

Because c = 3, the y-intercept is (0, 3). Reflecting this point across x = 1

gives the point (2, 3).

Step 5: Draw a smooth curve.

Remember to check your graphs in your calculator!

You might be asked to find the solutions of a quadratic equation by graphing it. The

solutions to a quadratic equation are the points where it crosses the x-axis.

A quadratic can have two solutions, only one solution, or no solutions at all.

Sometimes you will need to find the solution to a quadratic that cannot be factored. In

that case, you can use the quadratic formula: π‘₯ =βˆ’π‘Β±βˆšπ‘2βˆ’4π‘Žπ‘

2π‘Ž

You just substitute the values for a, b, and c into the quadratic formula and simplify.

Two Solutions (the parabola

crosses the x-axis twice)

One Solution (the parabola

crosses the x-axis one time)

No Solutions (the parabola does

not cross the x-axis)

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Example 2: Solve 5π‘₯2 βˆ’ 2π‘₯ βˆ’ 9 = 0

π‘Ž = 5 𝑏 = βˆ’2 𝑐 = βˆ’9 Plug these values into the quadratic formula

π‘₯ =βˆ’π‘Β±βˆšπ‘2βˆ’4π‘Žπ‘

2π‘Ž π‘₯ =

2±√(βˆ’2)2βˆ’4(5)(βˆ’9)

2(5) π‘₯ =

2±√4+180

10 π‘₯ =

2±√184

10

Your two solutions are π‘₯ =2+√184

10=

2+2√46

10=

𝟏+βˆšπŸ’πŸ”

πŸ“ and π‘₯ =

2βˆ’βˆš184

10=

2+2√46

10=

𝟏+βˆšπŸ’πŸ”

πŸ“

Solving Quadratic Equations

1. Sketch the graph of 𝑦 = π‘₯2 + 4

2. Sketch the graph of 𝑦 = βˆ’2π‘₯2 + 6π‘₯

3. Find the solution(s) by graphing 𝑦 = π‘₯2 βˆ’ 16

4. Find the solution(s) by graphing 𝑦 = π‘₯2 βˆ’ 10π‘₯ + 25

5. Find the solution(s), use the quadratic formula 3π‘₯2 + 6π‘₯ βˆ’ 5 = 0

6. Find the zero(s) of the quadratic, use any method you like. 2π‘₯2 = 4π‘₯ βˆ’ 9

Scan this QR code to go to a video tutorial on using the

quadratic formula.

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Answers to the

problems: Laws of Exponents

1. βˆ’6π‘Ž4𝑏7

2. 25π‘₯4𝑦2𝑧6

3. π‘š8𝑛48𝑝24

4. 2197π‘Ž22𝑏8𝑐31

5. 3𝑏7

π‘Ž

6. π‘₯9𝑦3

8

7. 32π‘₯5

𝑦15

8. 1

π‘š3

9. 5π‘Ž5𝑐3

3𝑏4

10. 9𝑧4

4π‘₯12𝑦4

Polynomials

1. 5π‘šπ‘›π‘2 βˆ’ 7𝑛𝑝2 + 9π‘šπ‘›

2. 6π‘Ž + 8𝑏 βˆ’ 8𝑐 + 2𝑑 + 3𝑑2

3. 9π‘₯2 βˆ’ 7π‘₯𝑦 + 13𝑦2

4. 32π‘Žπ‘2 + 10π‘Ž2𝑏 βˆ’ 35𝑏2 βˆ’ π‘Žπ‘

5. 6𝑑𝑒 βˆ’ 11𝑒 + 2𝑑

6. βˆ’8π‘š3𝑛3 + 6π‘š2𝑛3

7. 38π‘Ž2𝑏2𝑐 βˆ’ 13π‘Žπ‘3𝑐 βˆ’ 31π‘Žπ‘2𝑐2

8. 36π‘₯2 βˆ’ 25

9. 20π‘Ž3𝑏2 + 5π‘Ž2𝑏2 βˆ’ 8π‘Žπ‘3 βˆ’ 2𝑏3

10. 9π‘₯2𝑦2 βˆ’ 30π‘₯2𝑦 + 25π‘₯2 Factoring & Simplifying Radicals

1. 2 βˆ™ 2 βˆ™ 3 βˆ™ 3 βˆ™ 5 βˆ™ π‘Ž βˆ™ π‘Ž

2. 3π‘₯

3. 2(4𝑐3 + 7𝑐 βˆ’ 6)

4. 2π‘₯2π‘¦βˆšπ‘¦

5. 6π‘Žπ‘2π‘βˆš15π‘Žπ‘

6. 2𝑐 √6𝑐𝑑23

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Factoring Special Cases

1. (π‘₯ + 4)(π‘₯ + 3)

2. 2(π‘₯ + 2𝑦)(π‘₯ βˆ’ 9𝑦)

3. (2π‘₯ + 5)(3π‘₯ + 1)

4. Prime

5. π‘₯ = βˆ’5, βˆ’4 or {βˆ’5, βˆ’4}

6. π‘₯ = 2,3

2 or {

3

2, 2}

7. π‘₯ = βˆ’6, 0, 1 or {βˆ’6, 0, 1}

8. (2π‘₯ + 3𝑦)(2π‘₯ βˆ’ 3𝑦)

9. π‘₯ = βˆ’6 or {βˆ’6}

10. 2π‘₯ (2π‘₯ βˆ’ 7)2

Solving Quadratic Equations 1.

2.

3. π‘₯ = 4, βˆ’4 or {βˆ’4, 4}

4. π‘₯ = 5 or {5}

5. βˆ’3 Β±2 √6

3

6. No Solution


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