Chapter 4

Factoring and Quadratic Equations

Lesson 1: Factoring by GCF, DOTS, and Case I

Lesson 2: Factoring by Grouping & Case II

Lesson 3: Factoring by Sum and Difference of Perfect Cubes

Lesson 4: Solving Quadratic Equations by Factoring

Lesson 5: Solving Quadratic Inequalities

This assignment is a teacher-modified version of Algebra 2 Common Core Copyright (c) 2016 eMath

Instruction, LLC used by permission.

Chapter 4

Lesson 1

GCF, DOTS, and Case I Factoring

Factoring:

When we factor it is important to remember that __________________________; we are simply rewrite

it in an equivalent form.

GCF or “Greatest Common Factor”:

Factoring by GCF means that we “__________________” what terms have in _______________. This

can be a combination of numbers, variables, or both.

Exercise #1: Factor out by GCF.

(a) 3x2 + 6x (b) 20x – 5x2

(c) 10x2y3 – 25xy4 (d) 15x – 30x3

(e) 2x2 + 8x + 10 (f) 8x3 – 12x2 + 20x

Exercise #2: Rewritten in factored form 20x2 – 36x is equivalent to

(1) 2x(10x – 15) (3) 5x(4x + 7)

(2) 4x(5x – 9) (4) 9x(x – 4)

Conjugate Multiplication Pattern

“DOTS” Factoring:

Another type of factoring stems from _______________________ multiplication. This type of factoring

is known as Difference of Two Squares or “DOTS” factoring. When we factor these types of expressions,

we _____________ conjugate multiplication.

Exercise #3: Write each of the following binomials as the product of a conjugate pair.

(a) x2 – 9 (b) 25 – 4x2

(c)

(d)

Exercise #4: Factor each expression completely.

Factoring Completely: When we factor completely, it means that we factor until we ________________

______________________________. It is important to always look for a GCF first.

(a) 28x2 – 7 (b) 3x3 – 48x

(c) 16x4 – 81 (d) 2x4 – 162

Trinomial Case I Factoring:

Another type of factoring is trinomial factoring. This is when we have a trinomial. In Case I factoring, the

leading coefficient is _________.

To factor these, it is helpful to look at the ______________ term and the ___________________ term.

Exercise #5: Write each of the following trinomials in factored form.

(a)x2 – 7x – 18 (b) x2 + 14x + 24

(c) x2 + x – 12 (d) x2 – 5x + 6

(e) x2 – 15x + 44 (f) x2 – 6x – 16

Exercise #6: Factor each expression completely.

(a) 4x2 + 8x + 4 (b) 5x2 - 25x – 30

(c) 2x2 + 8x – 64 (d) 3x2 + 18x + 27

Chapter 4

Lesson 1 HOMEWORK

GCF, DOTS, and Case I Factoring

Fluency:

1. Factor the following completely:

(a) 230 35x x (b) 3 220 5 15x x x (c) 3 214 35 7x x x

(d) 327 12x x (e) 28 512x (f) 2 125

9x

(h) (i) (k) 2 17 30x x

2. If one factor of 56x4y3 – 42x2y6 is 14x2y3, what is the other factor?

(1) 4x2 – 3y3 (3) 4x2y – 3xy3

(2) 4x2 – 3y2 (4) 4x2y – 3xy2

3. When factored completely, x3 – 13x2 – 30x is

(1) x(x + 3)(x – 10) (3) x(x + 2)(x – 15)

(2) x(x – 3)(x – 10) (4) x(x – 2)(x + 15)

Applications

4. The area of any rectangular shape is given by the product of its width and length. If the area of a

particular rectangular garden is given by 215 35A x x and its width is given by 5x , then find an

expression for the garden’s length. Justify your response.

5. The volume of a particular rectangular box is given by the equation 350 2V x x . The height and

length of the box are shown on the diagram below. Find the width of the box in terms of x. Recall that

V L W H for a rectangular box.

2x 5x

?

Chapter 4

Lesson 2

Grouping & Case II Factoring

Factoring by Grouping:

A new type of factoring is factoring by grouping. This type of factoring requires us to see structure in expressions. We usually factor by grouping when we have a polynomial that has four or more terms.

Example Steps

x3 + 2x2 + 3x + 6 1. ________________ terms together that have a _____________ factor.

2. Factor each group

3. Factor by __________

4. Distribute to check (if needed)

Exercise #1: Factor the expression 2x3 – 6x2 + 5x – 15. Justify each step with one of the three major properties of real numbers, i.e. the commutative, associative, or distributive.

Exercise #2: Use the method of factoring by grouping to completely factor the following expressions.

(a) 3x3 + 2x2 - 27x - 18 (b) 18x3 + 9x2 - 2x - 1

(c) x5 + 4x3 + 2x2 + 8 (d) 5x3 + 10x2 + 20x + 40

Exercise #3: Write the expression (x + 3)(x - 4) + 5(x + 3) as the equivalent product of binomials.

Exercise #4: Consider the expression x2 + ab - ax - bx.

(a) How can you rewrite the expression so that the first two terms share a common factor (other than 1)?

(b) Write this expression as an equivalent product of binomials.

Exercise #5: Louis factored the expression 2x3 + 10x2 + 7x + 21 below. Is he correct? Explain.

2x3 + 10x2 + 7x + 21 = 2x2(x+5) +7(x+3)

= (2x2 + 7)(x + 5 + x + 3)

= (2x2 + 7)(2x + 8)

Case II Trinomial Factoring:

Case II Trinomial factoring is used when the leading coefficient is not 1, even after any GCF was taken out.

Example Steps

2x2 – 7x + 6 1. See if you can factor out a GCF.

2. Multiply the coefficient of the first tem with the last term.

3. Split the middle term. You must determine the signs and coefficients for the two terms.

4.Factor by grouping.

Exercise #6: Factor the following trinomials.

(a) 3x2 + 19x - 40 (b) 2x2 - 15x + 18

(c) 15x2 + 13x + 2 (d) 10x2 + 13x – 30

(e) 12x2 + 8x - 15 (f) 36x2 - 35x + 6

Exercise #7: Factor the following trinomials completely.

(a) 10x2 + 55x - 105  (b) 12x2 + 57x - 15

(c) 2x2 + 20x + 50  (d) 12x2 + 29x - 8

Chapter 4

Lesson 2 HOMEWORK

Grouping & Case II Factoring

Fluency:

1. Rewrite each of the following as a product of binomials.

(a) 10 3 5 3x x x x (b) 3 7 5 5 2 4x x x x

(c) 210 6 35 21x x x (d)

212 3 20 5x x x

(e) 22 29 15x x (f)

218 25 8x x

2. Which of the following is the correct factorization of the trinomial 212 23 10x x ?

(1) 6 1 3 10x x (3) 4 5 3 2x x

(2) 6 2 2 5x x (4) 4 5 3 2x x

3. Factor each expression completely.

(a) 215 110 120x x (b)

3 210 26 12x x x

Reasoning:

4. Consider the expression: 3 25 9 45x x x . Enter this expression on your calculator and find its

zeroes. Provide evidence. Then, factor it completely. Do you see the relationship between the factors

and the zeroes?

Chapter 4

Lesson 3

Factoring Sum & Difference of Cubes

Two special factoring formulas are the sum and difference of perfect cubes. These formulas are:

a3 + b3 = ______________________________________________

a3 – b3 = ______________________________________________

Yes, these formulas have to be memorized!!! The quadratic portion usually cannot be factored.

Examples:

When you have a pair of cubes, carefully apply the appropriate rule. By “carefully,” I mean “using

parentheses to keep track of everything, especially the negative signs.” Here are some typical problems:

Factor:

1.) x3 – 8 Hint: It helps to write the example in cubed form. For this

example, we would write this as: (x)3 – (2)3. By doing this, we

can clearly see which is our “a” value and which is our “b” value.

Now, finish factoring by using the rules.

Exercise #2: Factor each of the following.

(a) a3 + 27 (b) x3 – 64

(c) 8b3 + 216c6 (d) x9 – 125

Excerise #3: Factor each of the following completely.

(a) 2x3 + 128 (b) x9 – 512

(c) x6 – 729 (d) 2 – 686y3

Exercise #4: Daniel incorrectly factored x6 – 125 as (x3 + 5)(x3 – 5). Where did Daniel go wrong? What is

the correct factorization?

Exercise #5: y6 – 64 can be first factored as a difference of cubes or as a difference of squares.

(a) Factor y6 – 64 first as a difference of cubes, and then factor completely.

(b) Factor y6 - 64 first as a difference of squares and then factor completely.

(c) Based on your answers to part (a) and part (b), if a polynomial can be factored as a difference

of cubes or a difference of squares which one should you do first? Why?

Chapter 4

Lesson 3 Homework

Factoring Sum & Difference of Cubes

Fluency

Exercise #1: Each of the following expressions is either a difference of perfect cubes or a sum of perfect

cubes. Factor appropriately.

(a) 216z3 – w3 (b) 27r3 + 1000s3

(c) 250t3 + 16s3 (d) 8k6 – 27q3

(e) a6 – 8b3 (f) 64y6 – 1

Exercise #2: Louis incorrectly factored x3 – 216 as (x2 + 6)(x – 36). Where did Louis go wrong? What is the

correct factorization?

Exercise #3: Factor 216s3 + 27t3.

(1) (6s + 3t)(36s2 – 18st + 9t2) (3) (6s – 3t)(36s2 + 18st + 9t2)

(2) (9s + 3t)(36s2 + 9t2) (4) (6s + 3t)(36s2 + 18st + 9t2)

Chapter 4

Lesson 4

Solving Quadratic Equations

Solving Quadratic Equations

The Zero Product Law:

If the product of multiple factors is equal to zero then at least one of the factors must be equal to zero.

The Zero Product Law can be used to solve any quadratic equation that is ________________ (not

prime). To utilize this technique, we must first set the equation ______________ to ___________ and

then factor the non-zero side.

Exercise #1: Solve each of the following quadratic equations using the Zero Product Law.

(a) x2 + 3x – 14 = -2x + 10 (b) 3x2 + 12x – 7 = x2 + 3x – 2

Exercise #2: Consider the system of equations shown below consisting of a parabola and a line.

y = 3x2 - 8x + 5 and y = 4x + 5

(a) Find the intersection points of these curves algebraically.

(b) Using your calculator, sketch the graph of this system on the axes on the axes below. Be sure to label

the curves with equations, the intersection points, and the window.

(c)Verify your answer to part (a) by using the Intersect command on your calculator.

Exercise #3: The parabola shown below has the equation y = x2 – 2x – 3.

(a) Write the coordinates of the two x-intercepts of the

graph.

(b) Algebraically find the x-intercepts of the parabola.

Exercise #4: Algebraically find the set of x-intercepts (zeros) for each parabola given below.

(a) y = 4x2 – 1 (b) y = 3x2 + 13x – 10

(c) y = 4x2 – 10x (c) y = x2 + 13x – 14

Exercise #5: A quadratic function of the form y = x2 + bx + c is shown below.

(a) What are the x-intercepts of this parabola?

(b) Based on your answer in part (a), write the equation

of this quadratic function first in factored form and then

in trinomial form.

Chapter 4

Lesson 4 Homework

Solving Quadratic Equations

Fluency:

1. Solve each of the following equations for the value of x.

(a) 212 8 0x x (b)

2 4 40 10 15x x x

(c) 2 26 15 2 2 10 4x x x x (d)

24 3 11 3 2x x x

Applications:

2. Consider the system of equations shown below consisting of one linear and one quadratic equation.

24 5 and 2 5 10y x y x x

(a) Find the intersection points of this system algebraically.

(b) Using your calculator, sketch a graph of this system to the right. Be sure to label the curves with

equations, the intersection points, and the window.

(c) Verify with the Intersect function on your calculator.

y

x

3. Algebraically, find the zeroes (x-intercepts) of each quadratic function given below.

(a) 212 18y x x (b) 22 6 8y x x

Reasoning

4. A quadratic function of the form 2y x bx c is shown on the graph below.

(a) What are the x-intercepts of this parabola?

(b) Based on your answer to part (a), write the equation of this quadratic

function first in factored form and then in trinomial form.

y

x

Chapter 4

Lesson 5

Solving Quadratic Inequalities

Quadratic Inequalities in One Variable:

Quadratic inequalities are similar to quadratic equations. When solving these types of problems, it is

important to first find the zeros of the quadratic equation. Then, we put these zeros on a number line

and test each interval to see where we must shade. Finally, we write our answer as an inequality.

< or >: we need to put an open circle on our number line.

< or >: we need to put a close circle on our number line.

Exercise #1: Solve each of the following quadratic inequalities. Graph your solutions on a number line

and write your final answer in set-builder notation.

(a) x2 – 5x – 36 < 0 (b) x2 – x – 12 > 0

(c) 2x2 – 13x – 7 > 0 (d) 5x2 + 28x – 12 < 0

(e) 2x2 – 4x – 8 > 10x – 8 (f) x2 + 14x – 6 < 14x + 19

Exercise #2: The number line graph is the solution to which of the following

inequalities?

(1) x2 – 2x – 8 > 0 (3) x2 – 2x – 8 > 0

(2) x2 + 2x – 8 < 0 (4) x2 + 2x – 8 < 0

Exercise #3: Which of the following represents the solution set of the inequality -2x2 + 7x – 3 > 0?

(1)

(3)

(2)

(4)

Chapter 4

Lesson 5 Homework

Solving Quadratic Inequalities

Fluency

1. Which of the following values of x is in the solution set of the inequality2 2 0x x ?

(1) 1 (3) 0

(2) 2 (4) 4

2. The solution set of the inequality 2 25x is which of the following?

(1) 5, (3) , 5 5,

(2) 5, 5 (4) , 5

3. The solution to the inequality 2 9 0x can be expressed graphically as

(1) (3)

(2) (4)

-5 0 5 -5 0 5

-5 0 5 -5 0 5

4. Find the solution set to each of the quadratic inequalities shown below. Represent your solution set

using any acceptable notation and graphically on a number line.

(a) 2 5 6x x (b)

2 10 24x x

(c) 28 50 5 10 5x x x (d)

2 27 4 3 3 4 4x x x x