10

Mathematics
Quarter 1 – Module 11:
Illustrates Polynomial Equation
(M10ALIi-1)
Mathematics– Grade 10
Self-Learning Module (SLM)
Quarter 1 – Module 11: Illustrates Polynomial Equation
First Edition, 2020

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Mathematics
Quarter 1 – Module 11:
Illustrates Polynomial Equation
(M10ALIi-1)
Introductory Message
For the facilitator:

Welcome to the (Mathematics 10 Self-Learning Module (SLM) Illustration of

Polynomial Equation!
This module was collaboratively designed, developed and reviewed by educators both
from public and private institutions to assist you, the teacher or facilitator in helping
the learners meet the standards set by the K to 12 Curriculum while overcoming
their personal, social, and economic constraints in schooling.

This learning resource hopes to engage the learners into guided and independent
learning activities at their own pace and time. Furthermore, this also aims to help
learners acquire the needed 21st century skills while taking into consideration their
needs and circumstances.

In addition to the material in the main text, you will also see this box in the body of
the module:

Notes to the Teacher

This contains helpful tips or strategies that
will help you in guiding the learners.

As a facilitator you are expected to orient the learners on how to use this module.
You also need to keep track of the learners' progress while allowing them to manage
their own learning. Furthermore, you are expected to encourage and assist the
learners as they do the tasks included in the module.

2
For the learner:

Welcome to the Mathematics 10 Self-Learning Module (SLM) on Illustration of

Polynomial Equation!

The hand is one of the most symbolized part of the human body. It is often used to
depict skill, action and purpose. Through our hands we may learn, create and
accomplish. Hence, the hand in this learning resource signifies that you as a learner
is capable and empowered to successfully achieve the relevant competencies and
skills at your own pace and time. Your academic success lies in your own hands!

This module was designed to provide you with fun and meaningful opportunities for
guided and independent learning at your own pace and time. You will be enabled to
process the contents of the learning resource while being an active learner.

This module has the following parts and corresponding icons:

What I Need to Know This will give you an idea of the skills or
competencies you are expected to learn in the
module.

What I Know This part includes an activity that aims to

check what you already know about the
lesson to take. If you get all the answers
correct (100%), you may decide to skip this
module.

What’s In This is a brief drill or review to help you link

the current lesson with the previous one.

What’s New In this portion, the new lesson will be

introduced to you in various ways such as a
story, a song, a poem, a problem opener, an
activity or a situation.

What is It This section provides a brief discussion of the

lesson. This aims to help you discover and
understand new concepts and skills.

What’s More This comprises activities for independent

practice to solidify your understanding and
skills of the topic. You may check the
answers to the exercises using the Answer
Key at the end of the module.

What I Have Learned This includes questions or blank

sentence/paragraph to be filled in to process
what you learned from the lesson.

What I Can Do This section provides an activity which will

help you transfer your new knowledge or skill
into real life situations or concerns.

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 Assessment This is a task which aims to evaluate your
level of mastery in achieving the learning
competency.

Additional Activities In this portion, another activity will be given

to you to enrich your knowledge or skill of the
lesson learned. This also tends retention of
learned concepts.

Answer Key This contains answers to all activities in the

module.

At the end of this module you will also find:

References This is a list of all sources used in developing

this module.

The following are some reminders in using this module:

1. Use the module with care. Do not put unnecessary mark/s on any part of the
module. Use a separate sheet of paper in answering the exercises.
2. Don’t forget to answer What I Know before moving on to the other activities
included in the module.
3. Read the instruction carefully before doing each task.
4. Observe honesty and integrity in doing the tasks and checking your answers.
5. Finish the task at hand before proceeding to the next.
6. Return this module to your teacher/facilitator once you are through with it.
If you encounter any difficulty in answering the tasks in this module, do not
hesitate to consult your teacher or facilitator. Always bear in mind that you are
not alone.

We hope that through this material, you will experience meaningful learning and
gain deep understanding of the relevant competencies. You can do it!

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 What I Need to Know

This module was designed and written with you in mind. It is here to help you master
the Illustration of Polynomial Equation. The scope of this module permits it to be
used in many different learning situations. The language used recognizes the diverse
vocabulary level of students. The lessons are arranged to follow the standard
sequence of the course. But the order in which you read them can be changed to
correspond with the textbook you are now using.

The module is divided into four lessons, namely:

 Lesson 1 – Definition and concept of polynomial equation;
 Lesson 2 – Types of polynomial equation;
 Lesson 3 – Finding the roots using Zero product Property and Factor Theorem
 Lesson 4 – Creating polynomial equation given the roots

After going through this module, you are expected to:

1. define and understand the concept of polynomial equation;
2. identify the types of polynomial equation;
3. apply the zero-product property and factor theorem in finding zeros;
4. create polynomial equation given the roots.

What I Know

This module was designed and crafted based on your needs. It is here to help
you understand and know more about polynomial equations. The content of this
module allows everyone especially the learners to use it in may learning situations.
The language used was securely understood by the learners. The lesson were also
arranged following the standard sequence of the course. But to remind you, the order
in which you read them can be changed to correspond with the textbook you are now
using.

1. 𝑥 − 2 = 0
a. 1 b. 0 c. -1 d. 2

2. 𝑥 + 3 = 0
a. 3 b. -3 c. -2 d. 0

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 3. 𝑥(𝑥 − 4) = 0
a. 0,4 b. 1,4 c. 0,-4 d. 1,-4

4. (𝑥 + 1)(𝑥 − 3) = 0
a. -1,-3 b. 1,3 c. -1,3 d. 1,-3

5. 𝑥 2 + 𝑥 − 2 = 0
a. 2,-1 b. 2,1 c. -2, -1 d. -2, 1

6. 𝑥 2 (𝑥 − 9)(2𝑥 + 1) = 0
a. 0,-9,1/2 b. 0,9,-1/2 c. 0,-9,-1/2 d. 0,9,1/2

7. (𝑥 + 4)(𝑥 2 − 𝑥 − 2) = 0
a. -4, 1, 2 b. -4,-1, 2 c. 4,-1,-2 d. -4, 1,-2

8. 2𝑥(𝑥 2 − 36) = 0
a. 0, 6, -6 b. 0, 4, 9 c. 0, -4, 9 d. 0, 4, -9

9. (𝑥 + 8)(𝑥 − 7)(𝑥 2 − 5𝑥 + 6) = 0
a. -8, 7, -3, -2 b. 8, -7, 3, 2 c. -8, -7, 3, 2 d. -8, 7,
3, 2

10. (3𝑥 + 1)2 (𝑥 + 7)(𝑥 − 2) = 0

a. -1/3, 1/3,7,2 b. 1/3,1/3,7,2 c. -1/3,-1/3,-7,2 d. 1/3,-
1/3,7,2

Very Good! You did a good job. You’re now ready for the next set of activities.

Lesson
Illustration of Polynomial
11 Equation
Great Job! You are now ready to start with a new lesson. Do you know that
polynomial equations play a vital role in Mathematics. In this module you are going
to learn how to illustrate polynomial equations.

What’s In

Let us remember that in your previous lesson, you have learned how to find
roots using Factor Theorem and Rational Root Theorem. Let us have a short review,
are you ready?

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 The Factor Theorem states that a polynomial P(x) has a factor (x - r) if and
only P(r) = 0. It is a special case of the Remainder Theorem where the remainder P(r)
= 0. It is used as a linking factor and zeros of the polynomial.

Example: Given that (x – 2) is a factor of the polynomial,𝑥 3 − 𝑘𝑥 2 − 24𝑥 + 28 find

k.

Solution:
Factor Theorem Synthetic Division

P(2) = 23 − 𝑘(2)2 − 24(2) + 28 = 0 2 1 3 -24 28

8 − 4𝑘 − 48 + 28 = 0 −4𝑘 = 2 10 -28
12 𝑘 = −3 1 5 -14 0

Therefore, the value of k in the given polynomial 𝑥 3 − 𝑘𝑥 2 − 24𝑥 + 28 is -3 and its

remainder is 0.

What’s New

This module focuses on illustrating polynomials equation and finding the

roots. Some polynomial equations are given below. Complete the table by identifying
the degree and the number of real roots. The first one is done for you.

Polynomial Equation Degree No. of Real

Roots
1. (𝑥 + 1)2 (𝑥 − 5) = 0 3 3
2. 𝑥−8=0
3. (𝑥 + 2)(𝑥 − 2) = 0
4. (𝑥 − 3)(𝑥 + 1)(𝑥 − 1) = 0
5. 𝑥(𝑥 − 4)(𝑥 + 5)(𝑥 − 1) = 0
6. (𝑥 − 1)(𝑥 − 3)3 = 0
7. (𝑥 2 − 4𝑥 + 13)(𝑥 − 5)3 = 0
8. (𝑥 + 1)5 (𝑥 − 1)2
9. (𝑥 2 + 4)(𝑥 − 3)3 = 0
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10. (𝑥 − √2)6 (𝑥 + √2) = 0

Nice one! The activity shows the relationship between the number of roots and
the degree of a polynomial equation.
You are now ready to discover more about Polynomial Equation!

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 What is It

Below are some important matters that we need to discuss in order for you to
understand polynomial equation. Read carefully and understand all salient point
written on this part of the module.

POLYNOMIAL EQUATION

A polynomial equation is an equation that has multiple terms made up of

numbers and variables. Usually, it is expressed in the form 𝑎𝑛 (𝑥 𝑛 ), where 𝑎 is the
coefficient, 𝑥 is the variable, and 𝑛 is the exponent. The value of the exponent should
always be a positive integer.

If we expand the polynomial equation we will get the general expression,

𝑓(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + 𝑎𝑛−2 𝑥 𝑛−2 + ⋯ + 𝑎1 𝑥 + 𝑎0

Examples of Polynomial Equation

1. 3𝑥 + 4 = 0
2. 2𝑥 2 + 3𝑥 + 1 = 0
3. 𝑥 3 + 4𝑥 2 + 2𝑥 + 6 = 0

In the example number 3, 𝑥 3 + 4𝑥 2 + 2𝑥 + 6 = 0, the degree is 3, the leading

coefficient is 1, and the constant is 6. To understand these terms let’s define
them one-by-one.

 Degree of the Polynomial – is the highest among the degrees (exponents) on

the equation.
 Leading Coefficient – is the number written in front of the variable with the
largest exponent.
 Constant – is the term of degree 0; it is the term in which the variable does
not appear.

TYPES OF POLYNOMIALS

1. Monomials – Monomials are the algebraic expressions with one term.

It is an expression that contains any count of like terms.

Examples:

5x2 8x 6xy -3y 4x+2x

Note: 4x + 2x is considered as monomial because we can combine like terms,

therefore, it is considered as 6x.

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 2. Binomials – Binomials are the algebraic expressions with two unlike
terms separated by addition or subtraction.

Examples:

2x2 + 3, x + y, x3 – 2x, x – 5, 4ab + 2ac

3. Trinomials – Trinomials are the algebraic expressions with three

unlike terms.

Examples:

2x2 + 5x – 3, 2x2 + 4xy + 2y2, -3x3 – 7x2 + 8

TYPES OF POLYNOMIAL EQUATION/TYPES OF ALGEBRAIC EQUATIONS

1. Linear equation. This type of equation is in the form ax + b = 0,

where a is not equal to 0. This is always in the first degree or let’s just say
the highest exponent is 1.

Examples:

5x + 2=0, -2y – 4=0, 5x = 0, 4a – 2= 6

2. Quadratic equation. This type of equation has a general form of ax2

+ bx + c = 0, where a, b and c are numbers and a is never zero. The other
two letters, b and c, can be zero.

Examples:

2x2 + 5x – 4 =0, y2 + 2y – 3 = 0

3. Cubic equation, which has the general form of ax3+bx2 + cx + d = 0,

where a, b, c and d are numbers but a cannot be zero. The way to identify
these types of equations is to look for the x3. The 3 should be your highest
exponent or the degree of the polynomial.

Examples:

x3 + 4x2 -3x +2=0, y3 – 5y2 + 6y – 8 = 0

FINDING THE ROOTS OF POLYNOMIAL EQUATION USING ZERO-PRODUCT

PROPERTY and FACTOR THEOREM

The Zero Product Property says that, if for example, we have ab = 0, then
either a = 0, b =0, or both a and b are equal to 0.

So here:

1. (x+3) (x-2) (x+1) (x-1)=0

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 Either (x+3), (x-2), (x+1), or (x-1) = 0
Then for x + 3 = 0, we transpose the 3 to the other side, so x = -3.
In the same way, we solve x - 2 =0, x + 1 = 0, x - 1 = 0.
So the answer here is x = -3, 2, -1, 1.

2. (x+5) (x-5) (x+5) (x-1) =0

Similarly solving, x + 5 =0, x - 5 = 0, x+5 = 0, and x - 1 =0

x = -5, 5, -5, 1
Note that we write both -5, even if it happened two times.

3. (x+4)2 (x-3)3 = 0

So here, when there are exponents:

(x+4)2 = 0, this is the same as (x+4)(x+4) = 0, right?
So we also have -4 twice.
Similarly, we also have (x-3) three times, so we have 3 three times.
x = -4, -4, 3, 3, 3

Multiplicity – states that the equations having 2 or more the same roots can be
written as one provided mentioning how many times it occurs. (ex. Roots are
1,2,2,3 = 1, 3, and 2(2times)).

Factor Theorem is the reverse of the Remainder Theorem: If you synthetic-

divide a polynomial by x = a and get a zero remainder, then, not only is x = a a zero
of the polynomial (courtesy of the Remainder Theorem), but x – a is also a factor of
the polynomial (courtesy of the Factor Theorem).

Just as with the Remainder Theorem, the point here is not to do the long
division of a given polynomial by a given factor. This Theorem isn't repeating what
you already know, but is instead trying to make your life simpler. When faced with a
Factor Theorem exercise, you will apply synthetic division and then check for a zero
remainder.

 Use the Factor Theorem to determine whether x – 1 is a factor of

f (x) = 2x4 + 3x2 – 5x + 7.
For x – 1 to be a factor of f (x) = 2x4 + 3x2 – 5x + 7, the Factor Theorem says
that x = 1 must be a zero of f (x). To test whether x – 1 is a factor, I will first
set x – 1 equal to zero and solve to find the proposed zero, x = 1. Then I will use
synthetic division to divide f (x) by x = 1. Since there is no cubed term, I will be
careful to remember to insert a "0" into the first line of the synthetic division to
represent the omitted power of x in 2x4 + 3x2 – 5x + 7:

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 1 2 0 3 -5 7
2 2 5 0
2 2 5 0 7

Since the remainder is not zero, then the Factor Theorem says that:

x – 1 is not a factor of f (x).

 Using the Factor Theorem, verify that x +4 is a factor of

f (x) = 5x4 + 16x3 – 15x2 + 8x + 16.
If x + 4 is a factor, then (setting this factor equal to zero and solving) x = –4 is a
root. To do the required verification, I need to check that, when I use synthetic
division on f (x), with x = –4, I get a zero remainder:

-4 5 16 -15 8 16
-6 16 -4 -16
5 -4 1 4 0
The remainder is zero, so the Factor Theorem says that:
x + 4 is a factor of 5x4 + 16x3 – 15x2 + 8x + 16.

In practice, the Factor Theorem is used when factoring polynomials "completely".

Rather than trying various factors by using long division, you will use synthetic
division and the Factor Theorem. Any time you divide by a number (being a potential
root of the polynomial) and get a zero remainder in the synthetic division, this means
that the number is indeed a root, and thus "x minus the number" is a factor. Then
you will continue the division with the resulting smaller polynomial, continuing until
you arrive at a linear factor (so you've found all the factors) or a quadratic (to which
you can apply the Quadratic Formula).

WRITING POLYNOMIAL EQUATION GIVEN THE ROOTS

The roots of an equation are the values that make it equal zero. If this is a
regular polynomial, then that means there are as many factors (at least) as there are
roots. So the equation is the product of three factors if there are three roots. Each
root corresponds to one of the factors equaling zero, so you can deal with them
individually. Think of each of the roots as a separate function if you like:

f(x)g(x)h(x) = 0, so f(x) = 0 or g(x) = 0 or h(x) = 0

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 If you start with the equation x3 - 4x2 - 7x + 10 = 0 you can factor it and get (x
- 1)(x + 2)(x - 5) = 0 and thus the roots are x = 1, x = -2 and x = 5. To solve your
problems you apply this process backwards.

Example:

Find a polynomial with roots 1, -2 and 5. Start with the roots x = 1, x = -2 and
x = 5 and construct the polynomial (x - 1)(x + 2)(x - 5) = 0. You can then expand this
expression if you wish and getx3 - 4x2 - 7x + 10 = 0

What’s More

At this point, you already know some important ideas about polynomial
equations. Now, using what you have learned earlier about polynomial equation,
complete the table below. The rational roots are already given, answer only what is
missing.

Polynomial Equation Leading Constant Roots

Coefficient Term
11𝑥 − 6 = 0 (1) (2) 1,2,3
𝑥 3 − 𝑥 2 − 10𝑥 − 8 = 0 (3) (4) -2,-1,4
𝑥 3 + 2𝑥 2 − 23𝑥 − 60 = 0 (5) (6) -4,-3,5
4𝑥 2 + 3𝑥 + 2 = 0 (7) (8) -1/2, 1, 1, 2
16𝑥 3 + 21𝑥 2 + 4𝑥 − 12 = 0 (9) (10) -2/3,1,2,3

What I Have Learned

Here is another activity that lets you apply what you have learned about types
of polynomials.

Name each polynomial by its degree and number of terms.

1. 3𝑥 2 + 2𝑥 + 8 = 0
Degree: ____________________
Term: _____________________

2. 17𝑎2 + 2𝑎 − 7 = 0
Degree: ____________________
Term: _____________________

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 3. 10𝑎 + 12 = 0
Degree: ____________________
Term: _____________________

4. −16𝑥 2 + 8𝑥 + 52 = 0
Degree: ____________________
Term: _____________________

5. −5x − 6 = 0
Degree: ____________________
Term: ______________________

Excellent! Now you are up to the final test of this module.

What I Can Do

Here are other activities that let you apply what you learned about Polynomial
Equations.

For each item below, give or create a polynomial equation with integer
coefficients that has the following roots.

1.) -1,2,-6 3.) 0,-4,-5,1 5.) -2, 3,2,-3

2.) 2, -7 4.) -2,3,5

Nice work! Now you’re up for the final challenge of this module

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 Assessment

I hope you had a good time going over this module. For you to determine how
much you’ve learned, please perform the following exercises.

I. Find the roots of the following polynomial equation using Zero-Product

Property.

1. (𝑥 + 3)(𝑥 − 2)(𝑥 + 1)(𝑥 − 1) = 0

2. (𝑥 + 5)(𝑥 − 5)(𝑥 + 5)(𝑥 − 1) = 0
3. (𝑥 + 4)2 (𝑥 − 3)3 = 0
4. 𝑥(𝑥 − 3)4 (𝑥 + 6)2 = 0
5. 𝑥 2 (𝑥 − 9) = 0

II. Identify the roots of each equation. State the multiplicity of each root if
there is.

1. −𝑥 4 + 2𝑥 3 + 8𝑥 2 − 48 = 0 ______________________________________

2. 𝑥 3 + 3𝑥 2 + 3𝑥 + 1 = 0 _____________________________________

3. 𝑥 3 + 5𝑥 2 + 𝑥𝑥 − 48 = 0 ______________________________________

4. 𝑥 3 + 10𝑥 2 + 17𝑥 = 28 ______________________________________

5. 3𝑥 3 + 10𝑥 2 − 27 = 10 ______________________________________
Good Job! You did well on this module!

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 15
Assessment What I have What’s What’s What I
I. 1.) -3,2,-1,1 learned more New Know
2.) -5,1,-5,5 1. Quadratic 1. 11 1. 3 1. d
3.) -4,-4,3,3,3 Trinomial 2. -6 2. 1 2. b
4.) 2. Quadratic 3. 1 3. 2 3. a
0,3,3,3,3,6,6 Trinomial 4. -8 4. 3 4. c
5.) 0,0,-9 3. Linear 5. 1 5. 4 5. d
Binomial 6. -60 6. 4 6. b
II. 1.) -2,0,4 4. Quadratic 7. 4 7. 5 7. b
2.) -1, 3 (3 Trinomial 8. 2 8. 7 8. a
times) 5. Linear 9. 16 9. 5 9. a
3.) 3, -4 (2 Binomial 10. -12 10. 12 10. c
times)
4.) -7,-4,1
5.) -5, -1/3, 2
Answer Key
and 3 examples of polynomial equations with a relatively long list of possible roots.
Give 3 examples of polynomial equations with a relatively short list of possible roots,
Additional Activities
References

Grade 10 Learner’s Material

https://www.purplemath.com/modules/factrthm.htm
https://www.shsu.edu/~kws006/Precalculus/2.3_Zeroes_of_Polynomials_files/S%
26Z%203.2.pdf
https://www.onlinemathlearning.com/remainder-theorem.html
https://www.shmoop.com/polynomial-equations/remainder-theorem-
examples.html
https://cdn.kutasoftware.com/Worksheets/Alg2/The%20Remainder%20Theorem.
pdf

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 Disclaimer
This Self-learning Module (SLM) was developed by DepEd SOCCSKSARGEN
with the primary objective of preparing for and addressing the new normal.
Contents of this module were based on DepEd’s Most Essential Learning
Competencies (MELC). This is a supplementary material to be used by all
learners of Region XII in all public schools beginning SY 2020-2021. The
process of LR development was observed in the production of this module.
This is version 1.0. We highly encourage feedback, comments, and
recommendations.

For inquiries or feedback, please write or call:

Department of Education – SOCCSKSARGEN

Learning Resource Management System (LRMS)

Regional Center, Brgy. Carpenter Hill, City of Koronadal

Telefax No.: (083) 2288825/ (083) 2281893

Email Address: region12@deped.gov.ph