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Math and Science Resource Center

Skill Blaster for Math 102/103: College Algebra

The worksheets in this packet are intended to help you develop the fundamental math skills necessary to succeed in your college algebra course. If you find that you struggle to complete a worksheet, weekly

help sessions are available through the Math and Science Resource Center (MSRC).

Intended Schedule

Week Lecture Topic Skill Blaster Area of Focus 1 Solving Linear Equations No Meeting 2 Graphing Plotting Points on a Graph 3 Linear Functions and Slope Intercept Form Using Slope Intercept to Graph 4 Point Slope Form & Linear Inequalities Solving Equations in Standard Form for Y 5 Factoring Laws of Exponents 6 Polynomials Identifying Like Terms, Add/Subtract 7 Factoring Continued Distributing and Foiling 8 Factoring & Solving Quadratic Equations Add/Subtract Fractions 9 Simplify/Add/Subtract/Multiply/Divide

Rational Functions No Meeting (Spring Break) – Simplifying Fractions with Variables

10 Complex Fractions & Solving Rational Functions

Fractions with Variable Expressions in the Denominator

11 Radicals Simplifying Perfect Squares and Cubes 12 Radicals Continued Simplify Non-Perfect Squares and Cubes 13 Add/Subtract/Multiply Radicals Identify Prime Quadratic Expressions 14 Quadratic Formula & Completing the Square Factoring Plan 15 Final Preparation No Meeting


Skill Blaster for College Algebra MATH103

Week Two

Topic: Plotting Points on a Graph

Questions to reflect upon: Do you know how to take a set of points and plot them on a graph? Do you know how to look at a graph and determine the ordered pair associated with a point?

Problem:

Plot the points (1, 2), (−4,−5), (−2, 1) and (1,−2)

Demonstration of problem:

(1, 2) Understand that an ordered pair is a description

of a point given as (x, y).

(−4,−5) → 𝑥𝑥 = −4,𝑦𝑦 = −5 (−2, 1) → 𝑥𝑥 = −2,𝑦𝑦 = 1

(1,−2) → 𝑥𝑥 = 1,𝑦𝑦 = −2

Understand the structure of a graph. The x axis is represented as the horizontal axis while the y axis is represented as the vertical axis. The point where the two axes cross is the origin where both x and y are zero. The left side of the horizontal axis becomes more negative, while the right side becomes more positive. The bottom portion of the vertical axis becomes more negative, while the upper portion becomes more positive.

Now plot the points. Start by moving horizontally to the value of x, then up or down to the value of corresponding y.

STEP 1

STEP 2

-5

0

5

-5 0 5

-5

0

5

-5 0 5

STEP 3

x y

-

-

+

+

Evaluate: Complete these three problems PRIOR to the Skill Blaster Session.

Write out any questions for your Skill Blaster Instructor in the Notes section. If you encountered any difficulties or uncertainties with these problems, be sure to record where you encountered them and why you feel you struggled.

1. Plot the following points: (2,4), (−1, 3), (3,4), (5,−1), (0,−1), (−2,0)

2. Plot the following points: X Y -2 -2 -1 -3 -4 5 2 3 2 -2

3. Identify the points in the graph as ordered pairs.

Notes:

-5

0

5

-5 0 5

-5

0

5

-5 0 5

-5

0

5

-5 0 5



Week Three

Topic: Using Slope Intercept to Graph

Questions to reflect upon: Do you know how to plot a line on a graph? Do you know what slope means? Do you know what an intercept means? Do you know how to use the slope and the y-intercept of a line to graph? Do you know how to put an equation in slope-intercept form?

Problem:

Given 𝑦𝑦 = −2𝑥𝑥 + 1, identify the slope, the y − intercept, and graph the line.


𝑦𝑦 = −2𝑥𝑥 + 1 Identify if the equation is in slope intercept form. Slope intercept form is y = mx + b, where m is the value for slope and b is the y value of the y intercept.

m = 2, b = 1 Slope is 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

. In this case, because m = -2, the

slope is −2 (𝑑𝑑𝑑𝑑𝑑𝑑𝑟𝑟)1(𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡)

, which means that for every 2

units the line falls, it moves right 1. In this case, because b = 1, the y-intercept is (0,1).

When graphing, plot the y-intercept first.

• Two points are enough to plot the line. Use

the 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟

technique starting from point

(0,1). From 𝑦𝑦 = 1, go down 2 units to 𝑦𝑦 = −1. From 𝑥𝑥 = 0, move 1 unit to the right to 𝑥𝑥 = 1.

• Then connect the two points with a line that passes through them.

*Remember, your line needs to have arrows on BOTH sides. Without them, you are only graphing a segment of the line, which is incorrect.

STEP 1

STEP 2

-5

0

5

-5 0 5

STEP 3

-5

0

5

-5 0 5


Write out any questions for your Skill Blaster Instructor in the Notes section. If you encountered any difficulties or uncertainties with these problems, be sure to record where you encountered them and why you feel you struggled.

1. Given 𝑦𝑦 = 4𝑥𝑥 − 2, find:

Slope: m =

Y-intercept: ( , )

A second point: ( , )

And Graph

2. Given 𝑦𝑦 = −3𝑥𝑥, find:

Slope: m =

Y-intercept: ( , )


And Graph

3. Given 𝑦𝑦 = 32𝑥𝑥 + 3, find:

Slope: m =

Y-intercept: ( , )


And Graph

Notes:

-5

0

5

-5 0 5

-5

0

5

-5 0 5

-5

0

5

-5 0 5



Week Four

Topic: Solving Equations in Standard Form for Y

Questions to reflect upon: Do you know how to solve an equation for a specific variable? Do you know the difference in standard form and point slope form?

Problem:

Solve 3𝑥𝑥 + 4𝑦𝑦 = 2 for y.


3𝑥𝑥 + 4𝑦𝑦 = 2 Recognize the difference of standard form and slope-intercept form.

Standard form: ax + by = c.

Slope-intercept: y = mx + b.

The given equation is in standard form, and solving the equation for y will put it in slope-intercept form.

4𝑦𝑦 = −3𝑥𝑥 + 2 Because we want to isolate y, we start by moving the x term to the other side. Because the x term was positive, we subtracted it from both sides.

𝑦𝑦 = −34𝑥𝑥 + 2

4 Now we can get y by itself by dividing both sides

by 4.

*Remember, when we multiply or divide by a number for the whole equation, EACH term must get the same treatment.

𝑦𝑦 = −34𝑥𝑥 + 1

2 *Always check if any part of your answer can

reduce!

STEP 1

STEP 2

STEP 3

STEP 4


Write out any questions for your Skill Blaster Instructor in the Notes section below. If you encountered any difficulties or uncertainties with these problems, be sure to record where you encountered them and why you feel you struggled.

Solve the following equations for y:

1. −2𝑥𝑥 + 32𝑦𝑦 = 3

2. 3𝑦𝑦 + 12𝑥𝑥 = 2

3. 32𝑥𝑥 − 𝑦𝑦 = 1

4

Notes:



Week Five

Topic: Laws of Exponents

Questions to reflect upon: Do you know the laws of exponents? Do you know how to apply these rules?

Problem:

Simplify (𝑥𝑥1𝑦𝑦4𝑧𝑧−2)3

𝑥𝑥−3𝑦𝑦2𝑧𝑧0 using the laws of exponents.


(𝑥𝑥1𝑦𝑦4𝑧𝑧−2)3

𝑥𝑥−3𝑦𝑦2𝑧𝑧0= (𝑥𝑥1𝑦𝑦4𝑧𝑧−2)3

𝑥𝑥−3𝑦𝑦2(1) 𝑎𝑎0 = 1

= (𝑥𝑥1𝑦𝑦4𝑧𝑧−2)3

𝑥𝑥−3𝑦𝑦2

(𝑥𝑥1𝑦𝑦4𝑧𝑧−2)3

𝑥𝑥−3𝑦𝑦2= 𝑥𝑥

1∗3𝑦𝑦4∗3𝑧𝑧−2∗3

𝑥𝑥−3𝑦𝑦2 (𝑎𝑎𝑥𝑥)𝑦𝑦 = 𝑎𝑎𝑥𝑥∗𝑦𝑦

= 𝑥𝑥3𝑦𝑦12𝑧𝑧−6

𝑥𝑥−3𝑦𝑦2

𝑥𝑥3𝑦𝑦12𝑧𝑧−6

𝑥𝑥−3𝑦𝑦2= 𝑥𝑥

3𝑥𝑥3𝑦𝑦12

𝑦𝑦2𝑧𝑧6 𝑎𝑎−𝑥𝑥 = 1

𝑎𝑎𝑥𝑥

𝑥𝑥3𝑥𝑥3𝑦𝑦12

𝑦𝑦2𝑧𝑧6= 𝑥𝑥6𝑦𝑦12

𝑦𝑦2𝑧𝑧6 𝑎𝑎𝑥𝑥𝑎𝑎𝑦𝑦 = 𝑎𝑎𝑥𝑥+𝑦𝑦

𝑥𝑥6𝑦𝑦12

𝑦𝑦2𝑧𝑧6= 𝑥𝑥

6𝑦𝑦10

𝑧𝑧6 𝑎𝑎𝑥𝑥

𝑎𝑎𝑦𝑦= 𝑎𝑎𝑥𝑥−𝑦𝑦

𝑥𝑥6𝑦𝑦10

𝑧𝑧6 Make sure you cannot simplify anymore!

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

STEP 6



1. Simplify (𝑥𝑥3𝑦𝑦2𝑧𝑧1)2

𝑥𝑥2𝑦𝑦4𝑧𝑧1

2. Simplify 𝑥𝑥−2𝑦𝑦−2

(𝑥𝑥−3𝑦𝑦−2𝑧𝑧4)4

3. Simplify (𝑥𝑥−4𝑦𝑦−3)−2

(𝑦𝑦−2𝑧𝑧1)−1

Notes:



Week Six

Topic: Identifying Like Terms, Adding and Subtracting Like Terms

Questions to reflect upon: Do you know how to identify like-terms? Do you know how to combine like terms through addition or subtraction?

Problem:

𝑥𝑥𝑦𝑦3 − 3𝑥𝑥2 + 4𝑦𝑦3 − 8𝑥𝑥𝑦𝑦3 + 10𝑥𝑥 − 5𝑦𝑦3 − 3𝑥𝑥


𝑥𝑥𝑦𝑦3 − 3𝑥𝑥2 + 4𝑦𝑦3 − 8𝑥𝑥𝑦𝑦3 + 10𝑥𝑥 − 5𝑦𝑦3 − 3𝑥𝑥

Identify the different terms you have in your equation, and determine which are like-terms. In order to be like terms, they have to have the same factors.

𝑥𝑥𝑦𝑦3 − 8𝑥𝑥𝑦𝑦3 + 4𝑦𝑦3 − 5𝑦𝑦3 − 3𝑥𝑥2 + 10𝑥𝑥 − 3𝑥𝑥

You may find it easier to combine these terms by reorganizing them so they are next to one another. This way you can better see the quantity of terms.

−7𝑥𝑥𝑦𝑦3 − 𝑦𝑦3 − 3𝑥𝑥2 + 7𝑥𝑥

Combine the terms through the operation given between them.

−7𝑥𝑥𝑦𝑦3 − 𝑦𝑦3 − 3𝑥𝑥2 + 7𝑥𝑥

STEP 1

STEP 2

STEP 3



Simplify:

1. −4𝑎𝑎2𝑏𝑏2 + 𝑎𝑎2 + 2𝑏𝑏3 − 8𝑎𝑎2 + 10𝑎𝑎2𝑏𝑏2 − 5𝑏𝑏3 − 3𝑎𝑎2

2. −𝑟𝑟3 − 3𝑟𝑟2 − 4𝑟𝑟3 − 8𝑟𝑟3 − 10𝑟𝑟 − 5𝑟𝑟3

3. 𝑢𝑢𝑢𝑢3 − 3𝑢𝑢𝑢𝑢2 + 4𝑢𝑢3 − (8𝑢𝑢𝑢𝑢3 + 10𝑢𝑢 − 5𝑢𝑢3 − 3𝑢𝑢)

Notes:



Week Seven

Topic: Distributing and FOILing

Questions to reflect upon: Do you know how and when to distribute? Do you know how to FOIL?

Problem:

Simplify − 4(𝑥𝑥𝑦𝑦)2 + 2(𝑥𝑥 + 𝑦𝑦)2


−4(𝑥𝑥𝑦𝑦)2 + 2(𝑥𝑥 + 𝑦𝑦)2 Remember order of operations. In this case, the first item we can take care of are the exponents.

We have two cases of exponents:

−4(𝑥𝑥𝑦𝑦)2 and 2(𝑥𝑥 + 𝑦𝑦)2 The difference in these two cases is the number of terms. −4(𝑥𝑥𝑦𝑦)2 is 1 term. 2(𝑥𝑥 + 𝑦𝑦)2 is 2 terms. You CANNOT distribute and exponents across more than one term. The first case distributes the exponent, the second case uses FOIL.

−4(𝑥𝑥𝑦𝑦)2 = −4(𝑥𝑥2𝑦𝑦2)

2(𝑥𝑥 + 𝑦𝑦)2 = 2(𝑥𝑥 + 𝑦𝑦)(𝑥𝑥 + 𝑦𝑦)

= −2(𝑥𝑥 + 𝑦𝑦)(𝑥𝑥 + 𝑦𝑦)

= −2(𝑥𝑥2 + 𝑥𝑥𝑦𝑦 + 𝑥𝑥𝑦𝑦 + 𝑦𝑦2) = 2(𝑥𝑥2 + 2𝑥𝑥𝑦𝑦 + 𝑦𝑦2)

−4(𝑥𝑥2𝑦𝑦2) + 2(𝑥𝑥2 + 2𝑥𝑥𝑦𝑦 + 𝑦𝑦2) Now distribute the multiple to EACH term.

−4𝑥𝑥2𝑦𝑦2 + 2𝑥𝑥2 + 4𝑥𝑥𝑦𝑦 + 2𝑦𝑦2 Check if you can reduce further by combining like terms.

STEP 1

STEP 2

STEP 3



Simplify:

1. −(𝑎𝑎𝑏𝑏𝑎𝑎)3 + 2(𝑎𝑎 + 𝑏𝑏)2

2. (𝑟𝑟 − 𝑡𝑡)2 + 4(𝑟𝑟𝑡𝑡)2 − 6𝑟𝑟2

3. −2(2𝑥𝑥 + 3𝑦𝑦)2 − (𝑥𝑥3𝑦𝑦2)2

Notes:



Week Eight

Topic: Add/Subtract Fractions

Questions to reflect upon: Do you know what PEMDAS stands for? Do you fully understand fundamental mathematical operations of addition, subtraction, multiplication and division? Do you know how to add or subtract fractions?

Problem:

−12

− 73

+ 94


−12

− 73

+ 94 Find the Lowest Common Denominator.

LCD = 12

−12

− 73

+ 94

−1(6)2(6)

− 7(4)3(4)

+ 9(3)4(3)

Multiply EACH fraction by the multiple it is

missing. *Remember, if you change the bottom of the fraction, you have to change the top as well.

− 612

− 2812

+ 2712

Now that all the fractions have the same

denominator, you can combine them into one fraction.

−6−28+2712

Use PEMDAS to simplify the numerator. −34+27

12

−712

*Always check if your answer can reduce!

STEP 1

STEP 2

2 x 6 = 122 3 x 4 = 122 4 x 3 = 122

STEP 3

STEP 4

STEP 5



1. 104

− 43

+ 14

2. −87

+ 12− 4

14

3. 12

+ 26

+ 64− 1

2

Notes:



Week Nine

Topic: Simplifying Fractions with Variables

Questions to reflect upon: Do you know how to find the LCD of fractions? Do you know how to handles fractions with variables in the numerator?

Problem:

Simplify the expression 2𝑥𝑥3

+1 + 𝑥𝑥

2−

34

into one fraction.


2𝑥𝑥3

+ 1+𝑥𝑥2− 3

4 Find the Lowest Common Denominator.

LCD = 12

2𝑥𝑥(4)3(4)

+ (1+𝑥𝑥)(6)2(6) − 3(3)

4(3) Multiply EACH fraction by the multiple it is

missing. Pay attention to the number of terms in the numerator, and distribute as needed!

8𝑥𝑥12

+ 6+6𝑥𝑥12

− 912



8𝑥𝑥+6+6𝑥𝑥−912

Use PEMDAS to simplify the numerator.

*Remember, you cannot combine unlike terms. 14𝑥𝑥−312

14𝑥𝑥−312

*Always check if your answer can reduce! In

order to reduce, every term in the fraction must have the same factor.

STEP 2

3 x 4 = 122 2 x 6 = 122 4 x 3 = 122

STEP 3

STEP 4

STEP 1

STEP 5



Simplify the expression into one fraction:

1. 𝑥𝑥2

+ 1+𝑥𝑥2− 1

4

2. 2𝑧𝑧5

+ 𝑧𝑧3− 𝑧𝑧−3

5+ 1

15

3. 𝑦𝑦4

+ 3𝑥𝑥+45

+ 5𝑦𝑦2− 𝑦𝑦

4

Notes:



Week Ten

Topic: Fractions with Variable Expressions in the Denominator

Questions to reflect upon: Do you know how to find an LCD involving variables?

Problem:

Simplify the expression 2

3𝑥𝑥+

1𝑥𝑥 + 1

−33

into one fraction.


23𝑥𝑥

+ 1𝑥𝑥+1

− 53 Find the Lowest Common Denominator.

LCD = 3x(x+1) *Remember, the LCD is all the factors that make up the denominators!

2(𝑥𝑥+1)3𝑥𝑥(𝑥𝑥+1)

+ 1(3𝑥𝑥)(𝑥𝑥+1)(3𝑥𝑥)

− 5(𝑥𝑥(𝑥𝑥+1))3(𝑥𝑥(𝑥𝑥+1))

Multiply EACH fraction by the multiple it is

missing. Pay attention to the number of terms in the denominators, and distribute as needed!

2𝑥𝑥+23𝑥𝑥(𝑥𝑥+1)

+ 3𝑥𝑥3𝑥𝑥(𝑥𝑥+1)

− 5(𝑥𝑥2+𝑥𝑥)3𝑥𝑥(𝑥𝑥+1)



2𝑥𝑥+2+3𝑥𝑥−5(𝑥𝑥2+𝑥𝑥)3𝑥𝑥(𝑥𝑥+1)

Use PEMDAS to simplify the numerator.

*Remember, you cannot combine unlike terms. 2𝑥𝑥+2+3𝑥𝑥−5𝑥𝑥2−5𝑥𝑥

3𝑥𝑥(𝑥𝑥+1)

−3𝑥𝑥2+23𝑥𝑥(𝑥𝑥+1)

*Always check if your answer can reduce! For a

fraction with MORE THAN ONE term in the numerator, every term must be able to reduce by the same factor!

−3𝑥𝑥2+23𝑥𝑥(𝑥𝑥+1)

STEP 2

STEP 3

STEP 4

STEP 1

STEP 5



Simplify the expression into one fraction:

1. 1

𝑥𝑥+1+ 2

2− 4

𝑥𝑥

2. 32𝑥𝑥− 1

3𝑥𝑥− 3

𝑥𝑥−4

3. 𝑥𝑥2𝑥𝑥

+ 112

+ 23𝑥𝑥− 2𝑥𝑥

2𝑥𝑥−1

Notes:



Week Eleven

Topic: Simplifying Perfect Squares and Cubes

Questions to reflect upon: Do you know how to identify and simplify perfect squares or perfect cubes?

Problem:

Simplify �8𝑥𝑥3𝑦𝑦6𝑧𝑧93


�8𝑥𝑥3𝑦𝑦6𝑧𝑧93 Recognize the kind of root, and what the root means. Here we have a cube root, so in order to pull anything out the radical, it must be some quantity cubed.

�8𝑥𝑥3𝑦𝑦6𝑧𝑧93 You may find it easier to break down each factor into some quantity cubed. This way, you can visualize the factor that will be removed from the root. Use a factor tree for the constant.

�23𝑥𝑥3𝑦𝑦3𝑦𝑦3𝑧𝑧3𝑧𝑧3𝑧𝑧33 For EACH cubed term, we can move ONE of that base outside the radical. Because this is a perfect cube, meaning every factor can be written as some quantity cubed, we will move everything outside.

2𝑥𝑥𝑦𝑦𝑦𝑦𝑧𝑧𝑧𝑧𝑧𝑧 *Always check if any part of your answer can reduce!

2𝑥𝑥𝑦𝑦2𝑧𝑧3

2𝑥𝑥𝑦𝑦2𝑧𝑧3

STEP 1

STEP 2

STEP 3

STEP 4

8

4 2

2 2



Simplify:

1. √27𝑟𝑟12𝑡𝑡33

2. √81𝑢𝑢6𝑢𝑢8

3. �34𝑥𝑥12𝑦𝑦4𝑧𝑧204

Notes:



Week Twelve

Topic: Simplifying Non-Perfect Squares and Cubes

Questions to reflect upon: Do you know how to identify and simplify non-perfect squares or cubes?

Simplify �16𝑥𝑥4𝑦𝑦5𝑧𝑧63


�16𝑥𝑥4𝑦𝑦5𝑧𝑧63 Recognize the kind of root, and what the root means. Here we have a cube root, so in order to pull any factor out the radical, it must be some quantity cubed.

�16𝑥𝑥4𝑦𝑦5𝑧𝑧63 You may find it easier to break down each factor into some quantity cubed. This way, you can visualize the factor that will be removed from the root. Use a factor tree for the constant.

�2321𝑥𝑥3𝑥𝑥1𝑦𝑦3𝑦𝑦2𝑧𝑧3𝑧𝑧33 For EACH cubed term, we can move ONE of that base outside the radical. Because this is NOT a perfect cube, there will still be factors left inside the radical.

2𝑥𝑥𝑦𝑦𝑧𝑧𝑧𝑧�21𝑥𝑥1𝑦𝑦23 *Always check if any part of your answer can reduce! For radicals, if the value of an exponent inside the radical is higher than the index, you haven’t reduced enough!

2𝑥𝑥𝑦𝑦𝑧𝑧2�2𝑥𝑥𝑦𝑦23

2𝑥𝑥𝑦𝑦𝑧𝑧2�2𝑥𝑥𝑦𝑦23

STEP 1

STEP 2

STEP 3

STEP 4

16

4 4

2 2 2 2



Simplify:

1. �24𝑦𝑦7𝑧𝑧23

2. √75𝑡𝑡2𝑠𝑠13

3. �75𝑥𝑥10𝑦𝑦8𝑧𝑧134

Notes:



Week Thirteen

Topic: Identifying Prime Expressions

Questions to reflect upon: Do you know what Prime means? Can you identify when an expression is Prime?

Problem:

Determine if the following are Prime:

𝑎𝑎. 8 − 𝑥𝑥3, b. 4𝑥𝑥2 + 32, 𝑎𝑎. 𝑥𝑥2 + 4, 𝑑𝑑. 3𝑥𝑥3 − 8𝑦𝑦3


𝑎𝑎. 8 − 𝑥𝑥3 No GCF Find GCF if possible.

b. 4𝑥𝑥2 + 32 GCF = 4

𝑎𝑎. 𝑥𝑥2 + 4 No GCF

𝑑𝑑. 3𝑥𝑥3 − 8𝑦𝑦3 No GCF

IF YOU HAVE FOUND A GCF IN STEP 1, THE PROBLEM IS NOT PRIME. Even if the result of step 1 is something that is no longer factorable, finding a GCF means the problem was NOT prime.

Follow the next step of the factoring plan: identify the number of terms and whether or not the expression fits that role.

𝑎𝑎. 8 − 𝑥𝑥3 NOT PRIME, Factors to: (2 − 𝑥𝑥)(4 + 2𝑥𝑥 + 𝑥𝑥2)

Not prime, because it fits the difference of cube formula.

b. 4(𝑥𝑥2 + 8) NOT PRIME

Not prime, because we were able to factor a GCF in the previous step. This is the factored form.

𝑎𝑎. 𝑥𝑥2 + 4 PRIME

Prime, because we do not have a formula for the addition of squares.

𝑑𝑑. 3𝑥𝑥3 − 8𝑦𝑦3 PRIME

Prime, because, although it looks like the difference of cubes, there is not a valid “a” value.

STEP 1

STEP 2 𝑎𝑎. 8 − 𝑥𝑥3 No GCF

b. 4(𝑥𝑥2 + 8) NOT PRIME

𝑎𝑎. 𝑥𝑥2 + 4 No GCF

𝑑𝑑. 3𝑥𝑥3 − 8𝑦𝑦3 No GCF

Evaluate: Complete these three problems PRIOR to the Skill Blaster Session. Write out any questions for your Skill Blaster Instructor in the Notes section below. If you encountered any difficulties or uncertainties with these problems, be sure to record where you encountered them and why you feel you struggled.

Determine if the following are Prime:

1. 27 − 9𝑥𝑥2

2. −𝑥𝑥2 + 4𝑦𝑦4

3. 62 + 8𝑦𝑦3

Notes:



Week Fourteen

Topic: Factoring Plan

Questions to reflect upon: Do you know the steps to factor? Can you identify when to use special factoring rules?

Problem:

Factor 18 − 2𝑥𝑥2


The factoring plan:

1. Factor out a GCF if possible. 2. Identify the number of terms

a. 2 terms – Use a Special Factoring Rule b. 3 terms – Use Trial and Error or the AC method c. 4 terms – Use Grouping

18 − 2𝑥𝑥2 The GCF is 2, so factor 2 from each term.

= 2(9 − 𝑥𝑥2)

2(9 − 𝑥𝑥2) There are two terms, which means we need to use a special factoring rule. In this case, we can use the difference of squares formula: (𝑎𝑎2 − 𝑏𝑏2) = (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)

2(3 + 𝑥𝑥)(3− 𝑥𝑥)

2(3 + 𝑥𝑥)(3− 𝑥𝑥) Always check to make sure your answer can not be factored or reduced any further!

STEP 1

STEP 2

STEP 3

a = 3

b = x



Factor:

1. 𝑦𝑦3 + 27

2. 27 − 3𝑥𝑥4

3. 𝑥𝑥2 − 3𝑥𝑥 − 10

Notes:

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