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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 “ ” section has also been developed to provide students with the

opportunity to solve similar problems and check their answers. The answers to the “ ” 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:

14𝑎𝑏² − 6𝑑 1 1

2𝑥²𝑦𝑧²

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

Example 1: Multiply 𝑏2 ∙ 𝑏5

𝑏2 ∙ 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: 𝑅𝑒𝑤𝑟𝑖𝑡𝑒 43 𝑎𝑛𝑑 42𝑎𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑜𝑏𝑙𝑒𝑚𝑠. 𝑈𝑠𝑒 𝑡ℎ𝑖𝑠 𝑡𝑜 𝑠𝑖𝑚𝑝𝑙𝑖𝑓𝑦 43 ∙ 42.

43 = 4 ∙ 4 ∙ 4 42 = 4 ∙ 4

43 ∙ 42 = 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 = 45

Example 3: Simplify 1

2𝑥𝑦3𝑧5 ∙ 14𝑥3𝑧 ∙ 𝑦6𝑧2

(1

2 ∙ 14) (𝑥 ∙ 𝑥3)(𝑦3 ∙ 𝑦6)(𝑧5 ∙ 𝑧 ∙ 𝑧2)

7𝑥4𝑦9𝑧8 When you raise a power to a power, you multiply the exponents.

(32)4

32 ∙ 32 ∙ 32 ∙ 32

32+2+2+2

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Example 4: Simplify (𝑎2𝑏)4

(𝑎2𝑏)4 = 𝑎2∙4𝑏1∙4

𝑎8𝑏4

This means 3² times itself 4 times!

Scan this QR code to go to a video tutorial on

multiplying monomials!

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Example 5: Simplify (5𝑑5𝑒7𝑓2)3

53𝑑5∙3𝑒7∙3𝑓2∙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 (−2𝑥³𝑦𝑧²)³(2𝑥³𝑦𝑧²)⁴

(−8𝑥9𝑦3𝑧6) ∙ (16𝑥12𝑦4𝑧8)

−128𝑥21𝑦7𝑧14

Laws of Exponents Simplify each expression

1. 6𝑎3𝑏5(−𝑎𝑏2)

2. (−5𝑥2𝑦𝑧3)2

3. [(𝑚𝑛6𝑝3)2]4

4. (𝑎4𝑏2𝑐7)4 (13𝑎2𝑐)3 When you divide monomials with like bases, you will subtract the exponents.

Anything raised to the zero power is equal to ONE!

𝑥0 = 1 150 = 1 (−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!)

(𝑎

𝑏)

5

= 𝑎5

𝑏5

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Example 7: 𝑛5𝑝7

𝑛4𝑝

𝑛5𝑝7

𝑛4𝑝= 𝑛5−4 𝑝7−1 = 𝑛 𝑝6

Example 8: (𝑎2𝑏𝑐5

𝑎2𝑐2 )3

(𝑎2𝑏𝑐5

𝑎2𝑐2)

3

= (𝑎2−2𝑏𝑐5−2)3 = (𝑎0𝑏𝑐3)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: 2−3 = .125 = 1

8 𝑜𝑟 2−3 =

1

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

Example 9: 𝑎−3𝑏2𝑐−5

𝑎−7𝑏𝑐10

𝑎−3𝑏2𝑐−5

𝑎−7𝑏𝑐10

𝑎7𝑏2

𝑎3𝑏𝑐10𝑐5

𝑎4𝑏

𝑐15

Example 10: (2𝑥2𝑦−4

3𝑥𝑦5 )−3

(2𝑥2𝑦−4

3𝑥𝑦5)

−3

= (3𝑥𝑦5

2𝑥2𝑦−4)

3

= (3𝑥𝑦5𝑦4

2𝑥2)

3

= (3𝑦9

2𝑥)

3

= 27𝑦27

8𝑥3

Remember that anything to the zero power equals 1!


dividing monomials!


simplifying monomials

with negative exponents!

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Exponents Laws of Exponents

Simplify each expression

5. 33𝑎5𝑏9

11𝑎6𝑏2

6. (𝑥4𝑦2

2𝑥𝑦)

3

7. (2𝑥𝑦−3)5

8. 𝑛4𝑛2𝑚−5

𝑛6𝑚−2𝑝0

9. 15𝑎𝑏−2𝑐5

9𝑎−4𝑏2𝑐2

10. (6𝑥4𝑦2𝑧−3

9𝑥−2𝑦0𝑧−1 )

−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

5𝑎𝑏2, −3𝑎𝑏2,2

3𝑎𝑏2 5𝑎𝑏2, −3𝑎2𝑏,

2

3𝑎𝑏

Each of these terms contain an ‘𝑎𝑏2 ‘, 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: (2𝑥2𝑦 + 5𝑥𝑦 − 7𝑦2) + (4𝑥2𝑦 − 10𝑥𝑦 + 3𝑦2)

(2𝑥2𝑦 + 4𝑥2𝑦) + (5𝑥𝑦 − 10𝑥𝑦) + (−7𝑦2 + 3𝑦2)

6𝑥2𝑦 − 5𝑥𝑦 − 4𝑦2

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

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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: (−3𝑎𝑏 − 5𝑎4𝑏2 + 𝑏) − (4𝑎𝑏 − 6𝑏)

−3𝑎𝑏 − 5𝑎4𝑏2 + 𝑏 − 4𝑎𝑏 + 6𝑏

−7𝑎𝑏 − 5𝑎4𝑏2 + 7𝑏

Polynomials Simplify each expression

1. (𝑚𝑛𝑝2 − 7𝑛𝑝2 + 12𝑚𝑛) + (4𝑚𝑛𝑝2 − 3𝑚𝑛)

2. (4𝑎 + 9𝑏 − 3𝑐 + 2𝑑) + (2𝑎 − 𝑏 − 5𝑐 + 3𝑑2)

3. (12𝑥2 − 6𝑥𝑦 + 9𝑦2) − (3𝑥2 + 𝑥𝑦 − 4𝑦2)

4. (32𝑎𝑏2 + 5𝑎2𝑏 − 21𝑏2) − (𝑎𝑏 + 14𝑏2 − 5𝑎2𝑏) 5. (2𝑡𝑢 − 8𝑢 + 7𝑡) + (−𝑡𝑢 − 4𝑡) − (3𝑢 + 𝑡 − 5𝑡𝑢) 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: 5𝑥 (3𝑥2 − 6𝑥𝑦 + 2𝑦2)

(5𝑥 ∙ 3𝑥2) − (5𝑥 ∙ 6𝑥𝑦) + (5𝑥 ∙ 2𝑦2)

15𝑥3 − 30𝑥2𝑦 + 10𝑥𝑦2

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

Then, COMBINE LIKE TERMS!


adding and subtracting polynomials.

Distribute the 5𝑥 to each term.

Then, simplify each term

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Example 4: −2𝑎2𝑏 (5𝑎𝑏3 − 6𝑎2𝑏5 + 𝑎2𝑏 − 𝑎𝑏3 ) (−2𝑎2𝑏 ∙ 5𝑎𝑏3) + (−2𝑎2𝑏 ∙ −6𝑎2𝑏5 ) + (−2𝑎2𝑏 ∙ 𝑎2𝑏 ) + (−2𝑎2𝑏 ∙ −𝑎𝑏3)

−10𝑎3𝑏4 + 12𝑎4𝑏6 − 2𝑎4𝑏2 + 2𝑎3𝑏4

−8𝑎3𝑏4 + 12𝑎4𝑏6 − 2𝑎4𝑏2 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: (2𝑥 + 5)(3𝑥 − 2) First Outer Inner Last

(2𝑥 ∙ 3𝑥) + (2𝑥 ∙ −2) + (5 ∙ 3𝑥) + (5 ∙ −2)

6𝑥2 − 4𝑥 + 15𝑥 − 10

6𝑥2 + 11𝑥 − 10

Example 6: (𝑎2 + 2𝑏2)(4𝑎 − 3𝑎𝑏 + 6𝑏)

(𝑎2)(4𝑎 − 3𝑎𝑏 + 6𝑏) + (2𝑏2)(4𝑎 − 3𝑎𝑏 + 6𝑏)

4𝑎3 − 3𝑎3𝑏 + 6𝑎2𝑏 + 8𝑎𝑏2 − 6𝑎𝑏3 + 12𝑏3

Don’t forget to check for like terms!


multiplying monomials and polynomials.

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Example 7: (4𝑦 − 3)2 (4𝑦 − 3)(4𝑦 − 3) (4𝑦 ∙ 4𝑦) + (4𝑦 ∙ −3) + (−3 ∙ 4𝑦) + (−3 ∙ −3)

16𝑦2 − 12𝑦 − 12𝑦 + 9

16𝑦2 − 24𝑦 + 9

Polynomials

6. −2𝑚𝑛2(4𝑚2𝑛 − 3𝑚𝑛) 7. 2𝑎𝑏2𝑐 (4𝑎 + 𝑏 − 3𝑐) + 5𝑎𝑏2𝑐 (6𝑎 − 3𝑏 − 5𝑐) 8. (6𝑥 + 5)(6𝑥 − 5)

9. (5𝑎2𝑏 − 2𝑏2)(4𝑎𝑏 + 𝑏)

10. (3𝑥𝑦 − 5𝑥)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 18𝑥3

18𝑥3

18𝑥3𝑦2 = 2 ∙ 3 ∙ 3 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 or 2 ∙ 32 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥

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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Example 2: What is the greatest common factor of 9𝑎, 15𝑎, 𝑎𝑛𝑑 6𝑎2

𝐺𝐶𝐹 = 3𝑎

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 8𝑥2 + 20𝑥

8𝑥2 = 2 ∙ 2 ∙ 2 ∙ 𝑥 ∙ 𝑥

20𝑥 = 2 ∙ 2 ∙ 5 ∙ 𝑥

4𝑥 (2𝑥 + 5)

Example 4: Factor 15𝑎3 − 15𝑎2

15𝑎3 = 3 ∙ 5 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙

−30𝑎2 = −1 ∙ 2 ∙ 3 ∙ 5 ∙ 𝑎 ∙ 𝑎 ∙

15𝑎2 = 3 ∙ 5 ∙ 𝑎 ∙ 𝑎 ∙

15𝑎2 (𝑎 − 1)

Factoring

1. Write the prime factorization of 180𝑎2

2. Find the greatest common factor of 15𝑥2 𝑎𝑛𝑑 42𝑥

3. Factor 8𝑐3 + 14𝑐 − 12

Circle each factor that they ALL have in common!

GCF = 2 ∙ 2 ∙ 𝑥 = 4𝑥

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

GCF = 3 ∙ 5 ∙ 𝑎 ∙ 𝑎 ∙ 𝑏 = 15𝑎2

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


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 ∙ 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 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2

2 ∙ 2 ∙ 2 √2 = 8√2

Example 6: Simplify 3√32𝑥3𝑦

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 √64𝑎75 , 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 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑦

3 ∙ 2 ∙ 2 ∙ 𝑥√2𝑥𝑦 = 12𝑥√2𝑥𝑦

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Example 7: Simplify √64𝑎73

√2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎3

2 ∙ 2 ∙ 𝑎 ∙ 𝑎 √𝑎3

4𝑎2 √𝑎3

Simplifying Radicals Simplify the following radicals.

4. √4𝑥4𝑦3

5. 6𝑎√15𝑎𝑏4𝑐3

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 𝑥2 + 𝑏𝑥 + 𝑐, first find two integers whose sum is equal

to 𝑏, and whose product is equal to 𝑎 ∙ 𝑐 . You can start by listing all of the factors of 𝑎 ∙ 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 “in a box” and factor the rows and columns.


simplifying radicals.

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

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Example 1: Factor 𝑥2 + 6𝑥 + 8 𝑎 ∙ 𝑐 = 1 ∙ 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 “in a box”

Sometimes you will not be able to find factors of 𝑎 ∙ 𝑐 that sum to b. When this happens, the trinomial is PRIME.

Example 2: Factor 2𝑥2 + 5𝑥 − 2 𝑎 ∙ 𝑐 = 2 ∙ −2 = −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

(𝑎𝑥2)

One Factor (__𝑥)

Other Factor (__𝑥)

Last Term

(c)

𝑥2 2𝑥

4𝑥 8

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

𝑥 4

𝑥 2

(𝑥 + 4) (𝑥 + 2)

Check your answer by FOIL-ing!

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Example 3: Factor 8𝑥2 − 21𝑥 + 10 𝑎 ∙ 𝑐 = 8 ∙ 10 = 80 So, we are looking for factors of 80 that add up to −21! Factors of 80 Sum of factors -4, -20 -24 -5, -16 -21

Example 4: Factor 3𝑥2 + 24𝑥 + 45

Pull out a GCF first!! 3(𝑥2 + 8𝑥 + 15) 𝑎 ∙ 𝑐 = 1 ∙ 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. 𝑥2 + 7𝑥 + 12

2. 2𝑥2 − 14𝑥𝑦 − 36𝑦2

3. 6𝑥2 + 17𝑥 + 5

4. 𝑥2 − 9𝑥 + 1

8𝑥2 −5𝑥

−16𝑥 10

𝑥2 3𝑥

5𝑥 15


𝑥

−2

8𝑥 −5

(8𝑥 − 5) (𝑥 − 2)

Check your answer by FOIL-ing!


factoring trinomials.


𝑥

5

𝑥 3

3(𝑥 + 5) (𝑥 + 3)

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

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 𝑥 that are the solutions to the quadratic.

Example 5: Find the zeros of 𝑥2 − 18 = 7𝑥

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

𝑥2 − 18 = 7𝑥

−7𝑥 − 7𝑥

𝑥2 − 7𝑥 − 18 = 0 Now factor the trinomial.

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

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

𝑥 + 2 = 0 𝑎𝑛𝑑 𝑥 − 9 = 0

𝑥 = −2 𝑎𝑛𝑑 9 or {−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. 𝑥2 + 9𝑥 + 20 = 0

6. 2𝑥2 + 6 = 7𝑥

7. 2𝑥3 + 10𝑥2 − 10𝑥 = 2𝑥

𝑥2 −9𝑥

2𝑥 −18

𝑥

2

𝑥 −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.

𝑎2 + 2𝑎𝑏 + 𝑏2 = (𝑎 + 𝑏)2 𝑎2 − 2𝑎𝑏 + 𝑏2 = (𝑎 − 𝑏)2

Example 6: Factor 4𝑥2 − 24𝑥 + 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 ½ of the middle number.

√4 = 2 𝑎𝑛𝑑 √36 = 6 6 ∙ 2 = 12 , 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 1

2 𝑜𝑓 24

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

(2𝑥 − 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. 4𝑥2 − 16). If both the first and last term are perfect squares, and the two

terms are being subtracted their factorization can be written as (𝑎 + 𝑏)(𝑎 − 𝑏). As an

example 4𝑥2 − 16 = (2𝑥 + 4)(2𝑥 − 4). Remember that you can check your work by

FOIL-ing.

Example 7: Factor completely 3𝑥2 − 27

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

3(𝑥2 − 9) Now you are left with a difference of squares!

3(𝑥 + 3)(𝑥 − 3)

Factoring Special Cases

8. 𝐹𝑎𝑐𝑡𝑜𝑟 4𝑥2 − 9𝑦2

9. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡(𝑠) 𝑥2 + 12𝑥 + 36 = 0

10. 𝐹𝑎𝑐𝑡𝑜𝑟 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑙𝑦 8𝑥3 − 56𝑥2 + 98𝑥

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

special cases.

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(𝑥) = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 , 𝑎 ≠ 0

The graph of a quadratic equation will be a parabola.

If 𝑎 > 0, then the parabola opens upward. If 𝑎 < 0, then the parabola opens downward.

The axis of symmetry is the line = −𝑏

2𝑎 .

The x-coordinate of the vertex is −𝑏

2𝑎 . The y-coordinate of the vertex is found by

plugging that x value into the equation and solving for 𝑓(𝑥).

The y-intercept is (0, 𝑐).

To graph a quadratic:

1. Identify a, b, and c.

2. Find the axis of symmetry (𝑥 = −𝑏

2𝑎 ), and lightly sketch.

3. Find the vertex. The x-coordinate is −𝑏

2𝑎 . 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 ‘tip’ of the parabola. In this picture, the turning point is at (2, 0).

Example 1: Graph 𝑦 = 2𝑥2 − 4𝑥 + 3

Step 1: Identify a, b, and c. 𝑎 = 2, 𝑏 = −4, 𝑎𝑛𝑑 𝑐 = 3

Step 2: Find and sketch the axis of symmetry.

𝑥 = −𝑏

2𝑎 𝑥 =

−(−4)

2(2) 𝑥 =

4

4 𝑥 = 1

Step 3: Find the vertex.

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

solving quadratic equations.

19

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

𝑦 = 2(1)2 − 4(1) + 3 𝑦 = 2 − 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: 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎

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)

20

Example 2: Solve 5𝑥2 − 2𝑥 − 9 = 0

𝑎 = 5 𝑏 = −2 𝑐 = −9 Plug these values into the quadratic formula

𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎 𝑥 =

2±√(−2)2−4(5)(−9)

2(5) 𝑥 =

2±√4+180

10 𝑥 =

2±√184

10

Your two solutions are 𝑥 =2+√184

10=

2+2√46

10=

𝟏+√𝟒𝟔

𝟓 and 𝑥 =

2−√184

10=

2+2√46

10=

𝟏+√𝟒𝟔

𝟓

Solving Quadratic Equations

1. Sketch the graph of 𝑦 = 𝑥2 + 4

2. Sketch the graph of 𝑦 = −2𝑥2 + 6𝑥

3. Find the solution(s) by graphing 𝑦 = 𝑥2 − 16

4. Find the solution(s) by graphing 𝑦 = 𝑥2 − 10𝑥 + 25

5. Find the solution(s), use the quadratic formula 3𝑥2 + 6𝑥 − 5 = 0

6. Find the zero(s) of the quadratic, use any method you like. 2𝑥2 = 4𝑥 − 9

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

quadratic formula.

21

Answers to the

problems: Laws of Exponents

1. −6𝑎4𝑏7

2. 25𝑥4𝑦2𝑧6

3. 𝑚8𝑛48𝑝24

4. 2197𝑎22𝑏8𝑐31

5. 3𝑏7

𝑎

6. 𝑥9𝑦3

8

7. 32𝑥5

𝑦15

8. 1

𝑚3

9. 5𝑎5𝑐3

3𝑏4

10. 9𝑧4

4𝑥12𝑦4

Polynomials

1. 5𝑚𝑛𝑝2 − 7𝑛𝑝2 + 9𝑚𝑛

2. 6𝑎 + 8𝑏 − 8𝑐 + 2𝑑 + 3𝑑2

3. 9𝑥2 − 7𝑥𝑦 + 13𝑦2

4. 32𝑎𝑏2 + 10𝑎2𝑏 − 35𝑏2 − 𝑎𝑏

5. 6𝑡𝑢 − 11𝑢 + 2𝑡

6. −8𝑚3𝑛3 + 6𝑚2𝑛3

7. 38𝑎2𝑏2𝑐 − 13𝑎𝑏3𝑐 − 31𝑎𝑏2𝑐2

8. 36𝑥2 − 25

9. 20𝑎3𝑏2 + 5𝑎2𝑏2 − 8𝑎𝑏3 − 2𝑏3

10. 9𝑥2𝑦2 − 30𝑥2𝑦 + 25𝑥2 Factoring & Simplifying Radicals

1. 2 ∙ 2 ∙ 3 ∙ 3 ∙ 5 ∙ 𝑎 ∙ 𝑎

2. 3𝑥

3. 2(4𝑐3 + 7𝑐 − 6)

4. 2𝑥2𝑦√𝑦

5. 6𝑎𝑏2𝑐√15𝑎𝑐

6. 2𝑐 √6𝑐𝑑23

22

Factoring Special Cases

1. (𝑥 + 4)(𝑥 + 3)

2. 2(𝑥 + 2𝑦)(𝑥 − 9𝑦)

3. (2𝑥 + 5)(3𝑥 + 1)

4. Prime

5. 𝑥 = −5, −4 or {−5, −4}

6. 𝑥 = 2,3

2 or {

3

2, 2}

7. 𝑥 = −6, 0, 1 or {−6, 0, 1}

8. (2𝑥 + 3𝑦)(2𝑥 − 3𝑦)

9. 𝑥 = −6 or {−6}

10. 2𝑥 (2𝑥 − 7)2

Solving Quadratic Equations 1.

2.

3. 𝑥 = 4, −4 or {−4, 4}

4. 𝑥 = 5 or {5}

5. −3 ±2 √6

3

6. No Solution

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