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1
STANDARDS OF LEARNING
CONTENT REVIEW NOTES
ALGEBRA I
3rd Nine Weeks, 2018-2019
2
OVERVIEW
Algebra I Content Review Notes are designed by the High School Mathematics Steering Committee as a resource
for students and parents. Each nine weeksβ 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 β section has also been developed to provide students with the
opportunity to solve similar problems and check their answers. The answers to the β β 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 β c
A.3
Equations & Inequalities 18 A.4a β f
A.5a β d
A.6a β b
Functions & Statistics 20 A.7a β 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 studentsβ scores on the test.
It is the Mathematics Instructorsβ desire that students and parents will use this document as a tool toward the
studentsβ success on the end-of-year assessment.
3
4
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:
14ππΒ² β 6π 1 1
2π₯Β²π¦π§Β²
When you multiply monomials that have a common base, you add the exponents.
Example 1: Multiply π2 β π5
π2 β π5 = π2+5 = π7
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 ππ€πππ‘π 43 πππ 42ππ ππ’ππ‘πππππππ‘πππ ππππππππ . ππ π π‘βππ π‘π π πππππππ¦ 43 β 42.
43 = 4 β 4 β 4 42 = 4 β 4
43 β 42 = 4 β 4 β 4 β 4 β 4 = 45
Example 3: Simplify 1
2π₯π¦3π§5 β 14π₯3π§ β π¦6π§2
(1
2 β 14) (π₯ β π₯3)(π¦3 β π¦6)(π§5 β π§ β π§2)
7π₯4π¦9π§8 When you raise a power to a power, you multiply the exponents.
(32)4
32 β 32 β 32 β 32
32+2+2+2
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Example 4: Simplify (π2π)4
(π2π)4 = π2β4π1β4
π8π4
This means 3Β² times itself 4 times!
Scan this QR code to go to a video tutorial on
multiplying monomials!
5
Example 5: Simplify (5π5π7π2)3
53π5β3π7β3π2β3
125π15π21π6
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 (β2π₯Β³π¦π§Β²)Β³(2π₯Β³π¦π§Β²)β΄
(β8π₯9π¦3π§6) β (16π₯12π¦4π§8)
β128π₯21π¦7π§14
Laws of Exponents Simplify each expression
1. 6π3π5(βππ2)
2. (β5π₯2π¦π§3)2
3. [(ππ6π3)2]4
4. (π4π2π7)4 (13π2π)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 (β235π¦7)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
π)
5
= π5
π5
6
Example 7: π5π7
π4π
π5π7
π4π= π5β4 π7β1 = π π6
Example 8: (π2ππ5
π2π2 )3
(π2ππ5
π2π2)
3
= (π2β2ππ5β2)3 = (π0ππ3)3 = 13π3π9 = π3π9
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: 2β3 = .125 = 1
8 ππ 2β3 =
1
23
When simplifying monomials with negative exponents, you can start by βflipping overβ all of the negative exponents to make them positive. Then, simplify.
Example 9: πβ3π2πβ5
πβ7ππ10
πβ3π2πβ5
πβ7ππ10
π7π2
π3ππ10π5
π4π
π15
Example 10: (2π₯2π¦β4
3π₯π¦5 )β3
(2π₯2π¦β4
3π₯π¦5)
β3
= (3π₯π¦5
2π₯2π¦β4)
3
= (3π₯π¦5π¦4
2π₯2)
3
= (3π¦9
2π₯)
3
= 27π¦27
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!
7
Exponents Laws of Exponents
Simplify each expression
5. 33π5π9
11π6π2
6. (π₯4π¦2
2π₯π¦)
3
7. (2π₯π¦β3)5
8. π4π2πβ5
π6πβ2π0
9. 15ππβ2π5
9πβ4π2π2
10. (6π₯4π¦2π§β3
9π₯β2π¦0π§β1 )
β2
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
5ππ2, β3ππ2,2
3ππ2 5ππ2, β3π2π,
2
3ππ
Each of these terms contain an βππ2 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₯2π¦ + 5π₯π¦ β 7π¦2) + (4π₯2π¦ β 10π₯π¦ + 3π¦2)
(2π₯2π¦ + 4π₯2π¦) + (5π₯π¦ β 10π₯π¦) + (β7π¦2 + 3π¦2)
6π₯2π¦ β 5π₯π¦ β 4π¦2
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: (β3ππ β 5π4π2 + π) β (4ππ β 6π)
β3ππ β 5π4π2 + π β 4ππ + 6π
β7ππ β 5π4π2 + 7π
Polynomials Simplify each expression
1. (πππ2 β 7ππ2 + 12ππ) + (4πππ2 β 3ππ)
2. (4π + 9π β 3π + 2π) + (2π β π β 5π + 3π2)
3. (12π₯2 β 6π₯π¦ + 9π¦2) β (3π₯2 + π₯π¦ β 4π¦2)
4. (32ππ2 + 5π2π β 21π2) β (ππ + 14π2 β 5π2π) 5. (2π‘π’ β 8π’ + 7π‘) + (βπ‘π’ β 4π‘) β (3π’ + π‘ β 5π‘π’) 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 β 6π₯π¦ + 2π¦2)
(5π₯ β 3π₯2) β (5π₯ β 6π₯π¦) + (5π₯ β 2π¦2)
15π₯3 β 30π₯2π¦ + 10π₯π¦2
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: β2π2π (5ππ3 β 6π2π5 + π2π β ππ3 ) (β2π2π β 5ππ3) + (β2π2π β β6π2π5 ) + (β2π2π β π2π ) + (β2π2π β βππ3)
β10π3π4 + 12π4π6 β 2π4π2 + 2π3π4
β8π3π4 + 12π4π6 β 2π4π2 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 β First β multiply the first term in each binomial together O β Outer β multiply the outermost term in each binomial together I β Inner β multiply the innermost term in each binomial together L β Last β multiply the last term in each binomial together (This is the exact same as distributing the first term, then distributing the second term) Donβt forget to combine like terms when possible. Example 5: (2π₯ + 5)(3π₯ β 2) First Outer Inner Last
(2π₯ β 3π₯) + (2π₯ β β2) + (5 β 3π₯) + (5 β β2)
6π₯2 β 4π₯ + 15π₯ β 10
6π₯2 + 11π₯ β 10
Example 6: (π2 + 2π2)(4π β 3ππ + 6π)
(π2)(4π β 3ππ + 6π) + (2π2)(4π β 3ππ + 6π)
4π3 β 3π3π + 6π2π + 8ππ2 β 6ππ3 + 12π3
Donβt 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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Example 7: (4π¦ β 3)2 (4π¦ β 3)(4π¦ β 3) (4π¦ β 4π¦) + (4π¦ β β3) + (β3 β 4π¦) + (β3 β β3)
16π¦2 β 12π¦ β 12π¦ + 9
16π¦2 β 24π¦ + 9
Polynomials
6. β2ππ2(4π2π β 3ππ) 7. 2ππ2π (4π + π β 3π) + 5ππ2π (6π β 3π β 5π) 8. (6π₯ + 5)(6π₯ β 5)
9. (5π2π β 2π2)(4ππ + π)
10. (3π₯π¦ β 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₯3π¦2 = 2 β 3 β 3 β π₯ β π₯ β π₯ or 2 β 32 β π₯ β π₯ β 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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Example 2: What is the greatest common factor of 9π, 15π, πππ 6π2
πΊπΆπΉ = 3π
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 β 2 β 2 β π₯ β π₯
20π₯ = 2 β 2 β 5 β π₯
4π₯ (2π₯ + 5)
Example 4: Factor 15π3 β 15π2
15π3 = 3 β 5 β π β π β π β
β30π2 = β1 β 2 β 3 β 5 β π β π β
15π2 = 3 β 5 β π β π β
15π2 (π β 1)
Factoring
1. Write the prime factorization of 180π2
2. Find the greatest common factor of 15π₯2 πππ 42π₯
3. Factor 8π3 + 14π β 12
Circle each factor that they ALL have in common!
GCF = 2 β 2 β π₯ = 4π₯
Pull the GCF out from each term and rewrite. Check your work by distributing.
GCF = 3 β 5 β π β π β π = 15π2
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.)
β18 = β9 β 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 β2 , which means 3 times the square
root of 2. You can check this simplification in your calculator by verifying that
β18 = 3β2 .
Another way to simplify radicals, if you donβt 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 β128 β2 β 2 β 2 β 2 β 2 β 2 β 2
2 β 2 β 2 β2 = 8β2
Example 6: Simplify 3β32π₯3π¦
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 β64π75 , 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
3β2 β 2 β 2 β 2 β 2 β π₯ β π₯ β π₯ β π¦
3 β 2 β 2 β π₯β2π₯π¦ = 12π₯β2π₯π¦
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Example 7: Simplify β64π73
β2 β 2 β 2 β 2 β 2 β 2 β π β π β π β π β π β π β π3
2 β 2 β π β π βπ3
4π2 βπ3
Simplifying Radicals Simplify the following radicals.
4. β4π₯4π¦3
5. 6πβ15ππ4π3
6. β48π4π23
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 + ππ₯ + π, first find two integers whose sum is equal
to π, and whose product is equal to π β π . You can start by listing all of the factors of π β π, and then see which two factors add up to the coefficient of π. Once you have determined which factors to use, you can put all of your terms βin a boxβ 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 β π = 1 β 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 βin a boxβ
Sometimes you will not be able to find factors of π β π that sum to b. When this happens, the trinomial is PRIME.
Example 2: Factor 2π₯2 + 5π₯ β 2 π β π = 2 β β2 = β4 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 π term is negative and the π term is positive, your answer will have two minus signs!
First Term
(ππ₯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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Example 3: Factor 8π₯2 β 21π₯ + 10 π β π = 8 β 10 = 80 So, we are looking for factors of 80 that add up to β21! 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 β π = 1 β 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 β 14π₯π¦ β 36π¦2
3. 6π₯2 + 17π₯ + 5
4. π₯2 β 9π₯ + 1
8π₯2 β5π₯
β16π₯ 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₯
β2
8π₯ β5
(8π₯ β 5) (π₯ β 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! Donβt 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 π₯ that are the solutions to the quadratic.
Example 5: Find the zeros of π₯2 β 18 = 7π₯
Start by getting one side equal to zero and write in standard form.
π₯2 β 18 = 7π₯
β7π₯ β 7π₯
π₯2 β 7π₯ β 18 = 0 Now factor the trinomial.
We are looking for factors of β18 that add up to β7. β9 and 2 work!
(x + 2)(x β 9) = 0 Set both factors equal to zero!
π₯ + 2 = 0 πππ π₯ β 9 = 0
π₯ = β2 πππ 9 or {β2, 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 β 10π₯ = 2π₯
π₯2 β9π₯
2π₯ β18
π₯
2
π₯ β9
-2 9
Scan this QR code to go to a video tutorial on solving
trinomials by factoring.
17
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.
π2 + 2ππ + π2 = (π + π)2 π2 β 2ππ + π2 = (π β π)2
Example 6: Factor 4π₯2 β 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.
β4 = 2 πππ β36 = 6 6 β 2 = 12 , π€βππβ ππ 1
2 ππ 24
Now that we know this case works, you can write the binomial factor squared
(2π₯ β 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 β 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 + π)(π β π). As an
example 4π₯2 β 16 = (2π₯ + 4)(2π₯ β 4). Remember that you can check your work by
FOIL-ing.
Example 7: Factor completely 3π₯2 β 27
To begin, you should factor out a GCF. In this case it would be 3.
3(π₯2 β 9) Now you are left with a difference of squares!
3(π₯ + 3)(π₯ β 3)
Factoring Special Cases
8. πΉπππ‘ππ 4π₯2 β 9π¦2
9. πΉπππ π‘βπ ππππ‘(π ) π₯2 + 12π₯ + 36 = 0
10. πΉπππ‘ππ πππππππ‘πππ¦ 8π₯3 β 56π₯2 + 98π₯
Scan this QR code to go to a video tutorial on factoring
special cases.
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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: π(π₯) = ππ₯2 + ππ₯ + π , π β 0
The graph of a quadratic equation will be a parabola.
If π > 0, then the parabola opens upward. If π < 0, then the parabola opens downward.
The axis of symmetry is the line = βπ
2π .
The x-coordinate of the vertex is βπ
2π . The y-coordinate of the vertex is found by
plugging that x value into the equation and solving for π(π₯).
The y-intercept is (0, π).
To graph a quadratic:
1. Identify a, b, and c.
2. Find the axis of symmetry (π₯ = βπ
2π ), and lightly sketch.
3. Find the vertex. The x-coordinate is βπ
2π . 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 βtipβ of the parabola. In this picture, the turning point is at (2, 0).
Example 1: Graph π¦ = 2π₯2 β 4π₯ + 3
Step 1: Identify a, b, and c. π = 2, π = β4, πππ π = 3
Step 2: Find and sketch the axis of symmetry.
π₯ = βπ
2π π₯ =
β(β4)
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.
19
The x-coordinate is 1. Plug this in to find y.
π¦ = 2(1)2 β 4(1) + 3 π¦ = 2 β 4 + 3 π¦ = 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₯ =βπΒ±βπ2β4ππ
2π
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 5π₯2 β 2π₯ β 9 = 0
π = 5 π = β2 π = β9 Plug these values into the quadratic formula
π₯ =βπΒ±βπ2β4ππ
2π π₯ =
2Β±β(β2)2β4(5)(β9)
2(5) π₯ =
2Β±β4+180
10 π₯ =
2Β±β184
10
Your two solutions are π₯ =2+β184
10=
2+2β46
10=
π+βππ
π and π₯ =
2ββ184
10=
2+2β46
10=
π+βππ
π
Solving Quadratic Equations
1. Sketch the graph of π¦ = π₯2 + 4
2. Sketch the graph of π¦ = β2π₯2 + 6π₯
3. Find the solution(s) by graphing π¦ = π₯2 β 16
4. Find the solution(s) by graphing π¦ = π₯2 β 10π₯ + 25
5. Find the solution(s), use the quadratic formula 3π₯2 + 6π₯ β 5 = 0
6. Find the zero(s) of the quadratic, use any method you like. 2π₯2 = 4π₯ β 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. β6π4π7
2. 25π₯4π¦2π§6
3. π8π48π24
4. 2197π22π8π31
5. 3π7
π
6. π₯9π¦3
8
7. 32π₯5
π¦15
8. 1
π3
9. 5π5π3
3π4
10. 9π§4
4π₯12π¦4
Polynomials
1. 5πππ2 β 7ππ2 + 9ππ
2. 6π + 8π β 8π + 2π + 3π2
3. 9π₯2 β 7π₯π¦ + 13π¦2
4. 32ππ2 + 10π2π β 35π2 β ππ
5. 6π‘π’ β 11π’ + 2π‘
6. β8π3π3 + 6π2π3
7. 38π2π2π β 13ππ3π β 31ππ2π2
8. 36π₯2 β 25
9. 20π3π2 + 5π2π2 β 8ππ3 β 2π3
10. 9π₯2π¦2 β 30π₯2π¦ + 25π₯2 Factoring & Simplifying Radicals
1. 2 β 2 β 3 β 3 β 5 β π β π
2. 3π₯
3. 2(4π3 + 7π β 6)
4. 2π₯2π¦βπ¦
5. 6ππ2πβ15ππ
6. 2π β6ππ23
22
Factoring Special Cases
1. (π₯ + 4)(π₯ + 3)
2. 2(π₯ + 2π¦)(π₯ β 9π¦)
3. (2π₯ + 5)(3π₯ + 1)
4. Prime
5. π₯ = β5, β4 or {β5, β4}
6. π₯ = 2,3
2 or {
3
2, 2}
7. π₯ = β6, 0, 1 or {β6, 0, 1}
8. (2π₯ + 3π¦)(2π₯ β 3π¦)
9. π₯ = β6 or {β6}
10. 2π₯ (2π₯ β 7)2
Solving Quadratic Equations 1.
2.
3. π₯ = 4, β4 or {β4, 4}
4. π₯ = 5 or {5}
5. β3 Β±2 β6
3
6. No Solution
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