10
SCHOOLS DIVISION OFFICE DAGUPAN CITY
GENESIS ADVANCED INTECH ACADEMY
Dagupan City
MATHEMATICS
Quarter 1
Name: ________________________________________
Section:________________________________________
Table of Contents
What I Know ................................................................................1
Module 1 ......................................................................................2
Module 2 ......................................................................................6
Module 3 ......................................................................................11
Module 4 ......................................................................................14
Module 5 ......................................................................................21
Module 6 ......................................................................................24
Module 7 ......................................................................................31
Module 8 ......................................................................................33
Module 9 ......................................................................................37
Assessment ..................................................................................41
Answer Key ..................................................................................42
References… ................................................................................46
DIRECTIONS: Read the questions carefully. Choose the letter of the correct answer.
1. Find the missing term of the sequence: 15, 11, 7, 3, __, -5.
a. -1
b. 0
c. 1
d. 2
2. Given the nth term, find the 7th term of the sequence: an=4n+9.
a. -23
b. -37
c. 23
d. 37
3. What is the 100th term of the Arithmetic Sequence 3, 7, 11, 15...?
a. 396
b. 399
c. 400
d. 410
4. If five arithmetic means are inserted between 11 and 65, find the second
arithmetic mean.
a. 20
b. 29
c. 38
d. 47
5. What is the sum of all the odd integers between 6 and 16?
a. 53
b. 54
c. 55
d. 56
6. Given -1, 1, -1, 1, -1..., find the 99th term.
a. -1
b. 0
c. 1
d. 2
7. In a theater, there are 15 chairs in the first row, but each row has 4 more
chairs than the previous one. How many chairs are there in the 14th row?
a. 59
b. 63
c. 67
d. 71
8. If (x3 – x2 + 3x – 3) is divided by (x² + 3), then the possible degree of quotient is
.
a. 0
b. 1
c. 2
d. 3
9. What is the remainder when (x3 + 9x2 + 13x +1) is divided by (x + 2)?
a. -1
b. -3
c. 1
d. 3
10. The area of the rectangular field is (x3 + 5x2 – x – 5) square meters. If the width
measures (x + 5) meters, what is the measure of its length?
a. (x² – 1) meters
c. (x – 1) meters
b. (x2 + 1) meters
d. (x + 1) meters
Module 1
This module was designed and written with you in mind. It is here to help you
master on how to generate patterns, to illustrate an arithmetic sequence, and to
determine nth term of an arithmetic sequence. 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 one lesson, namely:
Lesson 1 – Sequences and Arithmetic Sequence
After going through this module, you are expected to:
1. Generate patterns (M10AL-la-1)
2. Illustrate an arithmetic sequence (M10AL-lb-1)
3. Determine the nth term of an arithmetic sequence
Lesson
1
Sequences and Arithmetic
Sequence
Sequence is a function whose domain is the finite set {1, 2, 3,…, n} or the infinite
set {1, 2, 3,…}.
Each number in the sequence is called a term.
The finite sequence 4, 11, 18, 25, 32 has 5 terms,
where 4 is the first term (a1), 11 is the second term
2
3
4
5
(a2), 18 is the third term (a3), 25 is the fourth term
(a4), and 32 is the fifth term (a5). The notation a1,
11 18 25 32 a2, a3,…, an is used to denote the different terms
in a sequence.
Example of Finite Sequence:
n
an
1
4
The sequence 6, 8, 10, 12,… is an example of
infinite sequence which can be calculated by
substituting the values of n in the equation
2
3
4
…
an=2n+4. The “…” at the end indicates that the
sequence has no end. The expression an refers to
8 10 12 …
the general rule or the nth term of the sequence.
Example of Infinite Sequence:
n
an
1
6
An arithmetic sequence is a sequence where every term after the first is obtained
by adding a constant called the common difference (d). The common difference is
obtained by subtracting the two consecutive terms. Examples:
Ex 1:
13, 19,
25,…
d= 6
7,
6
6
Ex 2: -8, -4, 0, 4, 8, 12
d= 4 4 4 4 4
Ex 3: 12, 7, 2, -3, -8
Ex 4: 5, 3, 1, -1, -3,…
d = -5 -5 -5 -5
d = -2 -2 -2 -2
If every term is greater than the previous term in an arithmetic sequence, then the
common difference is positive. However, if every term is lesser than the previous
term, then the common difference is negative.
In finding any term in the arithmetic sequence, use the rule:
𝒂𝒏 = 𝒂𝟏 +(𝒏−𝟏)𝒅
where an = nth term of the arithmetic sequence a1
= first term of the arithmetic sequence
n = the term position d
= common difference
Example 1: Find the 10th term of the arithmetic sequence 5, 8, 11, …
Solution: a1 = 5, n = 10, and d = 3 an = a1 + (n-1)d
a10 = 5 + (10-1)(3)
a10 = 5 + (9)(3)
a10 = 5 + 27
a10 = 32
Answer: The 10th term of the arithmetic sequence is 32.
Example 2: Solve for the 57th term of the Arithmetic Sequence 15, 11, 7, 3, …
Solution: a1 = 15, n = 57, and d = -4 an = a1 + (n-1)d
a57 = 15 + (57-1)(-4)
a57 = 15 + (56)(-4)
a57 = 15 + (-224) a57
= -209
Answer: The 57th term of the arithmetic sequence is -209.
Activity 1: Find the first 5 terms of the sequence given the general rule or the
nth term.
1. an = n + 7
2. an = 5n – 4
3. an = n3
4. an = n2 – 3
5. an = (−3)n
Activity 2: Find the missing term and common difference of the arithmetic
sequence.
1. 3, 10, 17, 24, __, 38,...
2. __, 3, 8, 13, 18,...
3. 6, 0, __, -12, -18, -24,...
4. 7, 19, 31, __, 55,...
5. -9, -12, __, -18, -21,…
Activity 3: Answer the following questions about arithmetic sequence.
1. What is the 15th term of the arithmetic sequence 10, 5, 0, -5…?
2. Solve for the 100th term of the arithmetic sequence 2, 9, 16, 23,…
3. Find a1 if a7 = 79 and a8 = 86.
4. How many terms are there in an arithmetic sequence with a common
difference of 6, first term of 14, and last term of 68?
5. The first three terms of an arithmetic sequence are 2x + 6, 4x – 2, and 3x + 8,
respectively. Find the value of x.
Real-life Arithmetic Sequence: Complete the concept map by creating five
situations that show arithmetic sequence in real life.
Arithmetic
Sequence in Real
Life Situations
Module 2
This module was designed and written with you in mind. It is here to help you
master on how to determine arithmetic means and sum of the terms of a given
arithmetic sequence. 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 two lessons, namely:
Lesson 2.1 – Arithmetic Means
Lesson 2.2 – Arithmetic Series
After going through this module, you are expected to:
1. Determine arithmetic means
2. Determine the sum of the terms of a given arithmetic sequence
Lesson
2.1
Arithmetic Means
An arithmetic sequence is a sequence where every term after the first is
obtained by adding a constant called the common difference (d). The sequence 3,
9, 15, 21, 27, 33, 39 is an arithmetic sequence with seven terms and a common
difference of 6. And since it is a finite arithmetic sequence, it also has a first term of
3 and a last term of 39. The terms between 3 and 39 which are 9, 15, 21, 27, and 33
are called arithmetic means.
The term/s between any two nonconsecutive terms of an arithmetic sequence is/are
known as arithmetic mean/s.
Examples of Arithmetic Mean/s:
Ex 1: In the arithmetic sequence 4, 11, 18
between 4 and 18.
11 is the arithmetic mean
Ex 2: In the arithmetic sequence -7, -2, 3, 8, 13
arithmetic means between -7 and 13.
-2, 3, and 8 are the
Ex 3: In the arithmetic sequence 20, 16, 12, 8
arithmetic means between 20 and 8.
16 and 12 are the
The rule in arithmetic sequence an = a1 + (n-1)d can also be used to find the
arithmetic mean/s.
Example: Insert 3
arithmetic means
between 15 and
31.
Solution:
a1 = 15, n = 5, and an = 31
an = a1 + (n-1)d
31 = 15 + (5-1)d
31 = 15 + 4d
4d = 31-15
d=4
Answer:
The arithmetic
sequence is 15,
19, 23, 27, 31
with d=4. 19, 23,
and 27 are the
arithmetic means.
Activity 1: Find the missing terms of the arithmetic sequence.
1. 3, _, _, _, _, _, 33
2.
6, _, _, _, _, -9
3.
2, _, _, _, 34
4.
5, _, _, _, _, 25
5.
3, _, _, _, _, 43
6.
-4, _, _, _, _, -19
7.
1.5, _, _, _, 5.5
8.
6, _, _, _, -18
9.
-5, _, _, _, _, 30
10. 10, _, _, _, 40
Activity 2: Answer the following questions about arithmetic means.
1. Insert 3 arithmetic means between 9 and 41.
2. What is the second arithmetic mean: 16, _, _, _, _, _, 46?
3. The arithmetic mean between two terms in arithmetic sequence is 62. If one
term is 47, find the other term.
4. Find the common difference and insert 3 arithmetic means between 7 and 27.
5. What is the common difference of the arithmetic sequence given that the third
term is 36 and the eight term is -14?
Pyramid Kingdom: Applying the concept of arithmetic sequence, construct a
pyramid or make a pyramid kingdom using cards and determine the number of cards
that will be used in each row. (Examples are shown below. No limit for the number
of cards.)
Example 1: Pyramid
Lesson
2.2
Example 2: Pyramid Kingdom
Arithmetic Series
Arithmetic series is the sum of the arithmetic sequence. To determine the
arithmetic series given the first term and last term, the formula can be used:
𝒏
𝑺𝒏 = (𝒂𝟏 +𝒂𝒏)
𝟐
where Sn = sum of the arithmetic sequence an
= nth term of the arithmetic sequence
a1 = first term of the arithmetic sequence
n = the term position
Example: Find the sum of the arithmetic sequence 12, 17, 22, 27, 32.
Solution:
a1 = 12, n = 5, and an = 32
𝒏
𝑺𝒏 = (𝒂𝟏 + 𝒂𝒏)
𝟐
S5 =
(12 + 32)
S5 = (44)
S5 = 110
Answer: The sum of the arithmetic sequence is 110.
However, if the last term of an arithmetic sequence is not given, use the formula:
𝒏
𝑺𝒏 = [𝟐𝒂𝟏 +(𝒏−𝟏)𝒅]
𝟐
where Sn = sum of the arithmetic
sequence
a1 = first term of the arithmetic sequence
n = the term position d
= common difference
Example: Find the sum of the first 80th term of the arithmetic sequence 5, 8,
11, 14,…
Solution:
a1 = 5, n = 80, and d = 3
𝒏
𝑺𝒏 = [𝟐𝒂𝟏 + (𝒏 − 𝟏)𝒅]
𝟐
S80 =
[2(5) + (80 − 1)(3)]
S80 = 40[10 + (79)(3)]
S80 = 40[10 + (79)(3)]
S80 = 40(10 + 237)
S80 = 40(247)
S80 = 9880
Answer: The sum of the arithmetic sequence is 9880.
Activity 1: Find the sum of the terms of each finite sequence.
1. 2, 6, 10, 14
2. 5, 8, 11, 14, 17
3. 7, 4, 1, -2, -5, -8
4. 37, 27, 17, 7, -3, -13
5. -4, -6, -8, -10, -12
Activity 2: Answer the following questions about arithmetic series.
1. Find the sum of the first 100th term of the arithmetic sequence 3, 5, 7, 9,
11,… 2. Solve for S50: 4, 9, 14, 19,…
3. The sum of the first 20 terms of an arithmetic sequence is -300. What is the
first term if the common difference is -4?
4. What is the sum of the first 42 terms of an Arithmetic Sequence where the
first term is 3 and the last term is 167?
5. The second term of an arithmetic sequence is -2 and the sixth term is 14.
What is the sum of the first 8 terms?
Pyramid Kingdom V2.0: Refer to Week 2, Lesson 1 “What I Can Do”. Using the same
pyramid or pyramid kingdom, determine the total number of cards that were used.
(Examples are shown below. No limit for the number of cards.)
Example 1: Pyramid (10 Cards)
Example 2: Pyramid Kingdom (30 Cards)
Module 3
This module was designed and written with you in mind. It is here to help you master
on how to illustrate a geometric sequence, to differentiate a geometric sequence from
an arithmetic sequence, and to determine the nth term of a geometric sequence. 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 one lesson, namely:
Lesson 3 – Geometric Sequence
After going through this module, you are expected to:
1. Illustrate a geometric sequence (M10AL-ld-1)
2. Differentiate a geometric sequence from an arithmetic sequence (M10AL-
ld-2)
3. Determine the nth term of a geometric sequence
Lesson
3
Geometric Sequence
Geometric Sequence is a sequence where each term after the first is obtained by
multiplying the preceding term by a nonzero constant called the common ratio (r).
The common ratio can be obtained by dividing any term in the sequence by the
preceding term.
Ex 2: 15, -30, 60, -120,…
r= 2 -2 -2
Ex 1: 15, 30, 60, 120,…
r= 2
2
2
Ex 3: 120, 60, 30, 15
r=
Ex 4: 120, -60, 30, -15
r=
−
−
−
If the terms in geometric sequence are increasing, then the common ratio is positive.
If the terms are decreasing, then the common ratio is a fraction. If the geometric
sequence has an alternate sign for the terms, then it is negative.
In finding any term in the geometric sequence, use the rule:
𝒂𝒏 = 𝒂𝟏𝒓(𝒏−𝟏)
where an = nth term of the geometric sequence
first term of the geometric sequence
n = the term position
r = common ratio
a1
=
Example: What is the 10th term of the geometric sequence 4,8,16,32…?
Solution: a1 = 4, n = 10, and r = 2 an = a1r(n-1)
a10 = (4)(210-1)
a10 = (4)(29)
a10 = (4)(512)
a10 = 2048
Answer: The 10th term of the geometric sequence is 2048.
Comparison of Arithmetic Sequence and Geometric Sequence:
Basis for Comparison
Arithmetic Sequence
Geometric Sequence
Definition
Sequence where every
term after the first is
obtained by adding a
constant called the
common difference (d).
Sequence where each
term after the first is
obtained by multiplying
the preceding term by a
nonzero constant called
the common ratio (r).
Process in Finding the
Next Term/s
Addition or Subtraction
Multiplication or Division
Variation of Terms
Linear
Exponential
Rule or Formula
an = a1 + (n-1)d
an = a1r(n-1)
Activity 1: Find the ratio of the second number to the first number.
1. 8
12
2. -3
-18
3. 15
-25
4. 7
13
5.
Activity 2: State whether the given sequence is geometric or not. If geometric,
give the common ratio and next term.
1. 5, 15, 45, 135,...
2. 0, 1, 1, 1, 1, 1,...
3. 1, -1, 1, -1, 1,...
4. -80, 40, -20, 10, -5,...
5.
, , , ,
,...
Activity 3: Answer the following questions about geometric sequence.
1. What is the 9th term of the geometric sequence 2, 8, 32…?
2. Find the 10th term of the Geometric Sequence 4, 2, 1,…
3. In the geometric sequence 4, 8, 16…, which term is 2048?
4. Find the value of k so that the terms 3k-2, 6k, and 17k+2 form a geometric
sequence.
5. The common ratio of a geometric sequence 14, m, 56, n… is -2. Find the values
of m and n.
LET’S GO, MATH CLUB: A new school year has begun, and the Math Club president
needs to enlist two new members. Each member needs to enlist another two
members, and so on. Suppose the process of enlisting continues, how many
members will be recruit in the 3rd enlistment? 5th enlistment? 7th enlistment?
Module 4
This module was designed and written with you in mind. It is here to help you
master on how to determine geometric means and sum of the terms of a given finite
or infinite geometric sequence. 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 two lessons, namely:
Lesson 4.1 – Geometric Means
Lesson 4.2 – Geometric Series
After going through this module, you are expected to:
1. Determine geometric means
2. Determine the sum of the terms of a given finite or infinite geometric
sequence
Lesson
4.1
Geometric Means
Geometric Sequence is a sequence where each term after the first is obtained by
multiplying the preceding term by a nonzero constant called the common ratio (r).
The sequence 1, 3, 9, 27, 81 is a geometric sequence with five terms and a common
ratio of 3. And since it is a finite geometric sequence, it also has a first term of 1 and
last term of 81. The terms between 1 and 81 which are 3, 9, and 27 are called
geometric means.
The term/s between any two nonconsecutive terms of a geometric sequence is/are
known as geometric mean/s.
Examples of Geometric Mean/s:
Ex 1: In the geometric sequence 3, -6, 12, -24
geometric means between 3 and -24.
Ex 2: In the geometric sequence 100, 10, 1,
are
the geometric means between 100 and
,
-6 and 12 are the
10, 1, and
.
Ex 3: In the geometric sequence 1, 0.5, 0.25, 0.125
0.5 and 0.25
are the geometric means between 1 and 0.125.
The rule in geometric sequence an = a1r(n-1) can also be used to find the geometric
mean/s.
Example: Insert 2
geometric means
between 4 and
256.
Solution:
a1 = 4, n = 4, and an = 256
an = a1r(n-1)
256 = (4)(r4-1)
256 = (4)(r3)
Answer:
The geometric
sequence is 4, 16,
64, 256 with r=4.
16 and 64 are the
geometric means.
r=4
Activity 1: Find the missing term and common ratio.
1.
10, __, 40, 80, 160
2.
3.
90, 30, __,
a2, a4, a6, __, a10
4.
1000x, 100x, 10x, __
5.
3y, __, 27y, 81y
6.
7.
, __, 4, 16, 64
64, -48, 36, __
8.
2, 6, 18, 54, __, 486
9.
1, -5, __, -125, 625
10. 80, 20, __,
Activity 2: Answer the following questions about geometric means.
1. Insert 5 geometric means between 4 and 256.
2. What is the third geometric mean if 3 geometric means were inserted between
and ?
3. What are the two geometric means between 4 and 500?
4. The geometric mean between the first and third term of a geometric sequence
is 18. If the first term is 3, find the third term.
5. Insert 2 geometric means between 5x and 625x4.
Compare and Contrast: Complete the Venn diagram to show the similarities and
differences of arithmetic means and geometric means.
Arithmetic Means
Lesson
4.2
Geometric Means
Geometric Series
Geometric series is the sum of the geometric sequence. There are two types of
geometric series: the finite geometric series (with last term) and the infinite
geometric series (without last term).
To determine the finite geometric series, the formula can be used:
𝒂𝟏 (𝟏−𝒓𝒏)
𝑺𝒏 =
𝟏−𝒓
where Sn = sum of the finite geometric sequence
a1 = first term of the geometric sequence n =
the term position r = common ratio
Example 1: Find the sum of the first 8th terms of the geometric sequence 2,
4, 8, 16,…
Solution:
a1 = 2, n = 8, and r = 2
𝒂𝟏 (𝟏 − 𝒓𝒏)
𝑺𝒏 =
𝟏−𝒓
S8 =
S8 =
−1
2(−255)
S8 =
−1
−510
S8 =
−1
S8 = 510
Answer: The sum of the finite geometric sequence is 510.
Example 2: What is the sum of the first 6 terms of Geometric Sequence 1, -5,
25,...?
Solution:
a1 = 1, n = 6, and r = -5
𝒂𝟏 (𝟏 − 𝒓𝒏)
𝑺𝒏 =
𝟏−𝒓
S6 =
S6 =
6
−
15624
S6 =
S6 =
S6 = −2604
Answer: The sum of the finite geometric sequence is -2604.
To determine the infinite geometric series, the formula can be used:
𝒂𝟏
𝑺=
𝟏−𝒓
where S = sum of the infinite geometric sequence a1
= first term of the geometric sequence
r = common ratio
Example: What is the sum to infinity of
, ,
,…?
Solution:
a1 =
and r =
𝒂𝟏
𝑺=
𝟏−𝒓
S=
1
3
1−
S=
1
3
1
2
2
S=
𝟐
Answer: The sum of the infinite geometric sequence is .
𝟑
Activity 1: Identify if the geometric sequence is finite or infinite by grouping
them in the table below.
Finite Geometric Sequence
Infinite Geometric Sequence
5, 10, 20, 40
27, 9, 3, 1,…
1, -1, 1, -1, 1,…
𝑥 𝑥 𝑥
3x, x, 3 , 9 , 27 ,…
The first 10 terms
of 1,5, 25
1, 10, 100,…
a2, a5, a8, a11
2x, 6x, 18x, 54x
1 1 1
2, 1, 2, 4, 8
10, 5, 2.5,…
Activity 2: Solve the following problems. Show your solution in the box.
Given 5, -10, 20... Solve for S7.
1
Solve: 8+4+2+1+2,...
LET’S GO, MATH CLUB V2.0: A new school year has begun, and the Math Club
president needs to enlist two new members. Each member needs to enlist another
two members, and so on. Suppose the process of enlisting continues, what is the
total number of members will the Math Club has in the 3rd enlistment? 5th
enlistment? 7th enlistment?
Module 5
This module was designed and written with you in mind. It is here to help you
master on how to solve problems involving sequences. 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 one lesson, namely:
Lesson 5 – Problems Involving Sequences
After going through this module, you are expected to:
1. Solve problems involving sequences (M10AL-lf-2)
Lesson
Problems Involving Sequences
5
Arithmetic sequence is a sequence where the difference between any two
consecutive terms is a constant called the common difference while geometric
sequence is a sequence in which the ratio of consecutive terms is a nonzero constant
called the common ratio.
In solving problems involving sequences, it is important to know if the question
involves arithmetic sequence or geometric sequence. Then after knowing, use the
appropriate formula.
Rule or Formula for:
Arithmetic Sequence
Geometric Sequence
Finding the nth term of
the sequence
an = a1 + (n-1)d
an = a1r(n-1)
Finding the terms
between two
nonconsecutive terms
of the sequence (Means)
an = a1 + (n-1)d
an = a1r(n-1)
Arithmetic series given
the first term and last
n term: Sn
= 2 (a1 + an)
Finding the sum of the
terms of the sequence
(Series)
Arithmetic Series if the
last term is not given: n
Sn = [2a1 + (n − 1)d] 2
Finite
Geometric
Series: a
Sn =
r
Infinite Geometric Series:
a
S=
r
Activity 1: Find the common difference of the following arithmetic sequence.
1. 10, 7, 4, 1, -2,...
2. 18, 29, 40, 51,...
3. -4, -8, -12, -16, -20, -24
4. , 1, , 2, ,…
5. 3x+2, 3x, 3x-2, 3x-4
Activity 2: Find the common ratio of the following geometric sequence.
1. 3, -9, 27, -81
2. 16, 8, 4, 2, 1,…
3. 2x, 2x2, 2x3, 2x4, 2x5
4. 100, 10, 1, ,
5. -8, 2, -0.5, 0.125,…
Activity 3: Describe the pattern in the sequence by identifying if it is arithmetic
or geometric, then find the next three terms.
Sequences
Arithmetic or
Geometric
Next Three Terms
13, 24, 35, 46, 57
-128, -64, -32, -16
-6, -15, -24, -33
1.2, 2.2, 3.2, 4.2
2, -1, 0.5, -0.25
WHAT’S YOUR PROBLEM: Identify if the real-life situation is arithmetic or geometric
sequence, then solve for what is asked by using the appropriate formula.
The student has 300 sheets of colored paper to use for school requirements. After
the first week of school year, there were 283 sheets of colored paper left. After the
second week, there were 266 sheets of colored paper left. After the third week, there
were 249 sheets of colored paper left. If this pattern continues, how many sheets of
colored paper will be left after 16 weeks?
Module 6
This module was designed and written with you in mind. It is here to help you
master on how to perform division of polynomials using long division and synthetic
division and to prove Remainder Theorem and Factor Theorem. 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 two lessons, namely:
Lesson 6.1 – Division of Polynomials
Lesson 6.2 – Remainder Theorem and Factor Theorem
After going through this module, you are expected to:
1. Perform division of polynomials using long division and synthetic division
(M10AL-lg-1)
2. Prove Remainder Theorem and Factor Theorem
Lesson
6.1
Division of Polynomials
A polynomial expression P(x) is an expression of the form anxn + an-1xn-1 +
an2xn-2 + … + a1x + a0, an 0, where the nonnegative integer n is called the degree of
the polynomial and coefficients a0, a1,…, an are real numbers. It is a special type of
algebraic expression that consists of variables, constants, nonnegative exponents,
and is combined using mathematical operations like addition, subtraction,
multiplication, and division. Polynomial expressions can be classified according to
the number of terms: as monomials, binomials, trinomials, and multinomials. It can
also be classified according to degree: constant, linear, quadratic, cubic, quartic,
quintic, 6th degree, 7th degree, and so on.
Examples of Polynomial
Expression
Classification
According to the
Number of Terms
5x2 + 7x -3
3x
7x4 – 4x3 + 3x2 – 9x + 2
Trinomial
Monomial
Multinomial
–x
Classification
According to Degree
Quadratic
Linear Quartic
Quadratic
Binomial
Cubic
Binomial
The terms of polynomial may be written in any order. However, when the terms of
expression are ordered from the highest degree to the lowest degree, then the
polynomial expression is said to be in standard form.
Example of Polynomial Expression in Standard Form:
3x3 – 6x2 – x + 8
The terms of the polynomial expression are in decreasing powers of the variable
x. And since the highest degree is 3, then it is classified as cubic. 3x3 is the
leading term (term of the polynomial expression with the highest degree), 3 is
the leading coefficient (numerical coefficient of the leading term), and 8 is the
constant term (term that contains only a number).
Dividing Polynomials: Two methods can be used in dividing polynomials, long
division method and synthetic division method. For long division method, the
procedure for dividing a polynomial by another polynomial is the same as the
procedure when dividing whole numbers.
Here are the steps in long division method:
Step 1: Identify the divisor and dividend then write the polynomials in
descending order of their degrees. If any terms are missing, use zero
as coefficient to fill in the missing term.
Step 2: Divide the leading term of the dividend by the leading term of the divisor
to obtain the leading term of the quotient.
Step 3: Multiply the answer obtained in the previous step by the divisor.
Step 4: Subtract the product in the previous step to the dividend and bring down
the next term.
Step 5: Repeat steps 2, 3, and 4 until reaching the last term and there are no
more terms to bring down.
Step 6: Write the final answer. If there is a remaining term after the last
subtract step, then this will be the remainder and must be written as
a fraction (over the divisor) in the final answer.
If the dividend P(x) and the divisor D(x) are polynomial expressions with D(x) 0, then
𝐏(𝐱)
𝐑(𝐱)
= Q(x) +
𝐃(𝐱)
or P(x) = Q(x)•D(x) + R(x), where Q(x) is the quotient and R(x) is the
𝐃(𝐱)
remainder. R(x) is either 0 or its degree is less than the degree of D(x).
The shorter method of dividing polynomial by a binomial of the form (x – r) is
synthetic division. In this procedure, only coefficients of the terms are used in the
division process. Here are the steps in synthetic division method with example:
Divide (x3 + 10x2 + 16x – 14) by (x + 3).
Step 1: Identify the divisor and dividend then write the numerical coefficients
of the dividend in descending order of their degrees. If any terms are
missing, use zero as coefficient to fill in the missing term.
(x3 + 10x2 + 16x – 14) is the dividend and (x + 3) is the divisor.
1
10
16
-14
Step 2: Place the value of r (from x – r of the divisor) in the upper left corner and
bring down the numerical coefficient of the leading term.
-3
1
10
16
-14
1
Step 3: Multiply the numerical coefficient of the third row to the value of r,
then write the answer on the second row. Get the sum of the product
and the term in the dividend. Repeat the procedure until the last term.
-3
1
1
10
-3
7
16
-21
-5
-14
15
1
Quotient
Step 4: Write the quotient Q(x) using the numerical coefficients in the third
row. Take note that the degree of Q(x) is one less than the degree of
the dividend. The last numerical coefficient is the remainder R(x).
𝟏
Answer: x2 + 7x – 5 +
𝐱+𝟑
Notice that the quotient obtained in long division method is the same as the
quotient in synthetic division method.
Activity 1: Use long division method to find the quotient and remainder of the
following polynomials.
1. (x3 + 3x2 – 9x + 5) ÷ (x + 5)
2. (x3 – x2 – 8x + 1) ÷ (x – 2)
3. (x3 + x2 – 7x – 7) ÷ (x + 1)
4. (2x4 + 13x3 + 22x2 – 9) ÷ (x + 3)
5. (x3 – 20) ÷ (x – 3)
Activity 2: Complete the method of synthetic division by filling the missing
parts of the solution. Then identify the divisor, dividend, quotient, and
remainder.
1.
2
2
-9
11
2
-4
4
-10
-5
-2
-4
0
Dividend:
Quotient:
Divisor:
Remainder:
2.
3
1
0
-9
-16
-3
-3
0
0
-5
15
-1
Dividend:
Quotient:
Divisor:
Remainder:
Booth Everywhere: Use long division method or synthetic division method to find
what is asked in this real-life situation. Then do the task that follows.
The area of rectangular school field is given by the polynomial (x 3 + 6x2 – 19x + 6)
square meters. If the width is (x – 2) meters, find the length of the field.
TASK: School foundation day is coming. The school head tasked the ten school clubs
to set a booth. Using the measurement in the situation, assign any value of x that
will satisfy the polynomial expressions and design a possible model of the school field
partition for each booth. Include the mathematical concepts of the partition.
Lesson
6.2
Remainder Theorem and
Factor Theorem
Remainder Theorem: If the polynomial P(x) is divided by (x – r), the remainder R is
a constant and is equal to P(r).
R = P(r)
There are three ways to find the remainder when P(x) is divided by (x – r), that is: a.
long division,
b. synthetic division, or
c. substituting r in the polynomial expression P(x)
Example: Find the remainder when (3x2 + 13x – 12) is divided by (x + 5).
Solution using Remainder Theorem:
P(x) = 3x2 + 13x – 12
P(-5) = 3(-5)2 + 13(-5) – 12
P(-5) = 3(25) + (-65) – 12
Therefore, the remainder when P(-5) = 75
2
+ (-65) – 12
P(x) = 3x + 13x – 12 is divided by P(-5) = -2
(x + 5) is -2.
If remainder when P(x) is divided by (x – r) is 0, then (x – r) is a factor of P(x). This
means that P(r) = 0. This concept is called Factor Theorem.
Factor Theorem: The polynomial P(x) has x – r as factor if and only if P(r)=0.
Example: Show that (x3 + 7x2 + 10x – 8) is a factor of (x + 4).
Solution using Factor Theorem:
P(x) = x3 + 7x2 + 10x – 8
P(-4) = (-4)3 + 7(-4)2 + 10(-4) – 8
P(-4) = (-64) + 7(16) + (-40) – 8
P(-4) = (-64) + 112 + (-40) – 8
Since P(-4) = 0, then (x + 4)
is a factor of (x3 + 7x2 + 10x – 8).
P(-4) = 0
Activity 1: Use remainder theorem to find the remainder when the given
polynomial is divided by each binomial. Then determine the letter that matches
your answer to decode the word.
x3 – 4x2 + 2x + 5
1. x – 2 (N)
2. x + 1 (S)
HIDDEN WORD:
3. x + 2 (I)
4. x – 5 (D)
5. x – 3 (V)
6. x – 1 (O)
40
-23
2
-23
-2
-23
4
1
Activity 2: Use the Factor Theorem to determine which expression is a factor
of the given polynomial. Write yes if it is a factor and give the remainder if not.
x3 + 9x2 + 14x – 24
1. x + 1
2. x + 6
3. x – 1
4. x – 2
5. x + 4
Think Outside the Box: Create a real-life situation or problem in which Remainder
Theorem or Factor Theorem can be applied. Then solve it by using the appropriate
solution for the theorem.
Real-Life Problem
Solution
Module 7
This module was designed and written with you in mind. It is here to help you
master on how to factor polynomials. 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 one lesson, namely:
Lesson 7 – Factoring polynomials
After going through this module, you are expected to:
1. Factor polynomials (M10AL-lh-1)
Lesson
7
Factoring Polynomials
Factoring polynomials is the reverse process of multiplying the factors of
polynomials.
Example:
3x2 + 6x = 3x (x + 2)
product
factors
Here are some methods in factoring polynomials:
1. Greatest Common Factor (GCF) method
2. Grouping method
3. Sum or difference of two cubes method
4. Difference of two squares method
5. Perfect square trinomial method
6. Trinomial method
Activity 1: Determine which method is appropriate to use to factor the
following polynomials.
1. 2x – 10
6. 25x2 – 49
2. 3x2 + 9x
7. 8x3 + 1
3. 2x3 + 4x2 + 3x + 6
8. 9x2 – 12x + 4
2
4. x – 81
9. 2x(x + 4) + 5(x + 4)
2
5. x + 22x + 121
10. 3x2 + 14x + 8
Activity 2: Factor the following polynomials.
1. 3x + 21
2. 4x2 – 6x
3. x2 – 8x – 20
4. x3 + 8
5. 6x2 + 13x – 5
6. x(x + 2) – 4(x + 2)
7. x4 – 36
8. 6x3 + 3x2 + 10x + 5
9. x2 – 8x + 16
10. 3x2 + 7x – 6
Your Turn: Complete the tree diagram by giving examples of factoring polynomials
using each method.
Examples of Factoring
Polynomials Using Each
Method
Module 8
This module was designed and written with you in mind. It is here to help you master
on how to illustrate polynomial equations and to prove Rational Root Theorem. 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 one lesson, namely:
Lesson 8 – Polynomial Equations and Rational Root Theorem
After going through this module, you are expected to:
1. Illustrate polynomial equations (M10AL-li-1)
2. Prove Rational Root Theorem
Lesson
8
Polynomial Equations and
Rational Root Theorem
Polynomial Equation is an equation of the form anxn + an-1xn-1 + an-2xn-2 + … + a1x
+ a0 = 0, an 0, where n is a nonnegative integer called degree and a0, a1,…, an are
real numbers called coefficients. anxn is the leading term, an is the leading coefficient,
and a0 is the constant term. It is a mathematical statement formed with variables,
exponent, and coefficients together with operations and an equal sign.
Examples of Polynomial Equation
5x2 + 2 = 0
𝑥2
3x +
2x3 + x2 + 4x4
4=0
– 3x + 1 = 0
–
x+2=
0 x(x – 5) =0
Zero Product Property states that the product is zero if and only if at least one of
the factors is equal to zero. If the polynomial can be factored, then this property can
be used to find the root/s of a polynomial equation. Root/s of a polynomial equation
refer/s to the value/s of a variable for which the given polynomial is equal to zero.
Solving Polynomial Equation using Zero
Product Property
x2 + 3x – 18 = 0
(x – 3)(x + 6) = 0
x–3=0
x+6=0
Fundamental
Theorem of Algebra
x=3
x = -6
Roots
A polynomial of degree n can have at most n distinct real roots.
Examples:
x–5=0
at most 1 distinct real root
x4 – 5x2 + 4 = 0
at most 4 distinct real roots
x3 – x2 – 10x – 8 =0
at most 3 distinct real roots
Root of Multiplicity of n is the number of times a root occurs.
Example: x3 + 5x2 + 3x – 9 =0
Standard Form (x
– 1)(x + 3)(x + 3) = 0
Factored
Form x = 1, x = -3, x = -3
Roots
The root 1 has a multiplicity of 1. The root -3 has a multiplicity of 2.
More examples of polynomial equation where the leading term and constant term
were given. It is important in finding the roots of polynomial equation using Rational
Root Theorem.
Polynomial Equation
Leading
Coefficient
Constant Term
Roots
x–7=0
1
-7
7
x3 – 2x2 – 4x + 8 = 0
1
8
-2, 2, 2
2x3 + x2 – 6x + 2 = 0
2
2
, 1, -2
Rational Root Theorem
Let anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 = 0, an 0 and ai is an integer for all
𝒑
I, 0
i
n, be a polynomial equation of degree n. If , in lowest terms, is a rational
𝒒
root of the equation, then p is a factor of ao and q is a factor of an.
Example: Find the roots of x3 – 4x2 + x + 6 = 0.
Solution: The equation has at most 3 distinct real roots. The leading coefficient
is 1, and its factors are 1 (q). The constant term is 6, and its factors are 1,
𝒑
2, 3, 6 (p). The possible roots of the equation are 1, 2, 3, 6 ( ).
𝒒
Note: To find the roots of polynomial equation, test the possible roots of the
equation using synthetic division.
2
1
-4
2
1
-4
6
-6
1
-2
-3
0
x=2
Since the remainder is 0, therefore 2 is one of the roots of the polynomial
equation. The quotient in the synthetic division is a quadratic, hence, the
other roots can obtain by using quadratic formula or factoring.
x2 – 2x – 3 = 0
(x – 3)(x + 1) = 0 x
= 3, x = -1
Answer: The roots of x3 – 4x2 + x + 6 = 0 are 2, 3, and -1.
Forming Polynomial Equations
Steps in forming polynomial equation given the roots:
1. Write the roots as factors by changing the signs and putting each factor
inside parentheses.
2. Equate the factors to zero.
3. Multiply the factors formed.
4. Combine similar terms.
5. Arrange the polynomial equation in standard form.
Example:
Given the roots: 2, -3, 1
(x – 2)(x + 3)(x + 1)(x – 1) = 0 x4
+ x3 – 7x2 – x + 6 = 0
Activity 1: Identify the root/s of the following factored form equations.
1. (x – 11) = 0
2. (x + 4) = 0
3. (x – 6)(x + 3) = 0
4. (2x + 1)(x – 1) = 0
5. (x + 2)2(x – 2) = 0
6. (x + 3)3(x – 2)3 = 0
7. x(x + 7) = 0
8. (4x – 3)(x + 5)(x – 5) = 0
9. x2(x + 6) = 0
10. (2x – 1)(3x + 2) = 0
Activity 2: Complete the table below.
Polynomial
Equation
x+9=0
Leading
Coefficient
Constant
Term
Number of
Possible
Roots
Roots
x2 + 2x – 15 = 0
3x2 – 7x + 2 = 0
x3 – 5x2 + 3x + 9 = 0
2x3 + x2 – 5x + 2 = 0
Activity 3: Create a polynomial equation given the roots.
1. 5, -3
2. 1, 4
3. 3, 0, -7
4.
, 1, -6
5. 1, − ,
2
Check It Out: The volume of the musical box is 24 cubic centimeters. If the length
is 5 cm greater than its width and the height is 2 cm less than the width.
a. Write an equation to model the volume of the musical box.
b. What are the measurements of the musical box?
Module 9
This module was designed and written with you in mind. It is here to help you
master on how to solve problems involving polynomials and polynomial equations.
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 one lesson, namely:
Lesson 9 – Problems Involving Polynomials and Polynomial Equations
After going through this module, you are expected to:
1. Solve problems involving polynomials and polynomial equations
(M10AL-lj-2)
Lesson
9
Problems Involving Polynomials
and Polynomial Equations
A polynomial expression P(x) is an expression of the form anxn + an-1xn-1 +
an2xn-2 + … + a1x + a0, an 0, where the nonnegative integer n is called the degree of
the polynomial and coefficients a0, a1,…, an are real numbers. A polynomial
equation is a mathematical equation that contains a polynomial expression.
In solving problems involving polynomials and polynomial equations, it is
important to determine what is asked and to analyze the questions. Some common
problems that involve polynomial equations are problems about number, age, money,
measurement, area, and volume.
It is important to use some strategies to solve problems involving polynomials and
polynomial equations:
1. Read the problem carefully.
2. Find what is asked.
3. Identify the given in the problem. Represent it using variable/s.
4. Create an equation based on the given and what is asked.
5. Solve the equation using appropriate algebra methods.
6. Verify the answer in the problem.
7. Answer the question with a complete sentence.
Activity 1: Use Zero Product Property to identify the root/s of the following
factored form equations.
Polynomial Equation
(x + 1)2(x – 2)(x + 7) = 0
(3x – 2 )(x + 5)3 = 0
x(x + 2)3(x – 2)2 = 0
x3(5x + 3) = 0
Roots
(x + 4)(x – 4)2(2x + 1) = 0
Activity 2: Solve the following problems about polynomials and polynomial
equations. Show your solution.
1. The product of two consecutive odd integers is 483. Find the odd integers.
Solution:
2. The area of the table is 112 square feet. The length of the table is six feet more
than the width. Find the length and width of the table.
Solution:
3. Ryza is 3 years older than Arianne. If the product of their ages is eight less
than six times the sum of their ages, how old are Ryza and Arianne?
Solution:
4. The height of the triangle is 4 cm less than the base. If the area is 48 cm 2,
what is the height and base of the triangle?
Solution:
5. The area of a rectangle is two times the area of a square. If the width of the
rectangle is the same as the length of a side of the square and its length is 6
decimeters, what are the dimensions rectangle and the square?
Solution:
STEP-BY-STEP: Create a question about polynomials and polynomial equations and
solve it using the strategies in problem solving.
Read the problem carefully: (Write the question here)
Find what is asked:
Identify the given in the problem. Represent it using variable/s:
Create an equation based on the given and what is asked:
Solve the equation using appropriate algebra methods:
Verify the answer in the problem:
Answer the question with a complete sentence:
DIRECTIONS: Read the questions carefully. Choose the letter of the correct answer.
1. What is the common difference of the arithmetic sequence , 1 , 1 , 1 ,...
a.
b. −
c.
d. −
2. What is the next term of the geometric sequence: 16, -8, 4, -2, 1, __?
a. 0
b. -1
c.
d. −
3. Which is an example of geometric sequence?
a. 5, 0, 0, 0, 0,...
c. 1, 4, 9, 16, 25,...
b. 2, 4, 6, 8, 10,...
d. 1, 1, 1, 1, 1,...
4. If three geometric means are inserted between 8 and 128, what is the common
ratio?
a. 2
b. 3
c. 4
d. 5
5. What is the sum of this infinite geometric sequence: , , , ,…?
a. -1
b. 0
c. 1
d. 2
6. Find the value of r so that r + 15, 5r + 4, 8r – 3… form an arithmetic sequence.
a. 3
b. 4
c. 5
d. 6
7. A snail crawls 3 cm after 1 minute, 6 cm after 2 minutes, 9 cm after 3 minutes,
and 12 cm after 4 minutes. If the snail continues to crawl at this rate, how
many centimeters will it crawl after 14 minutes?
a. 41
b. 42
c. 43
d. 44
8. The remainder when 4x123 + 5 is divided by (x – 1) is
a. 1
b. 4
c. 5
.
d. 9
9. What is the constant term of the polynomial equation x(x – 3)(x + 3)=0?
a. 0
b. 1
c. -3
d. 3
10. Which of the following cubic polynomial equations has roots -1, -5, and 5?
a. x3 – x2 + 25x – 25 = 0
c. x3 + x2 – 25x + 25 = 0
b. x3 + x2 – 25x – 25 = 0
d. x3 + x2 + 25x + 25 = 0