Mathematics 7 Quarter 2 Module 5: Operations Involving Polynomials

Mathematics

Quarter 2 – Module 5: Operations Involving Polynomials

7

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CO_Q2_Mathematics 7_Module 5

Mathematics– Grade 7 Alternative Delivery Mode Quarter 2 – Module 5: Operations Involving Polynomials First Edition, 2020

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Development Team of the Module

Writers: Alfredo T. Ondap Jr., Venus Jane E. Rosete Editor: Arnel C. Lapitan Reviewers: Reynaldo C. Tagala Illustrator: Layout Artist: Maylene F. Grigana, Darven G. Cinchez Management Team: Allan G. Farnazo, CESO IV – Regional Director Gilbert B. Barrera – Chief, CLMD Arturo D. Tingson Jr. – REPS, LRMS Peter Van C. Ang-ug – REPS, ADM Donna S. Panes – CID Chief Elizabeth G. Torres– EPS – LRMS Judith B. Alba – Division ADM Coordinator

Introductory Message

This Self-Learning Module (SLM) is prepared so that you, our dear learners, can continue your studies and learn while at home. Activities, questions, directions, exercises, and discussions are carefully stated for you to understand each lesson.

Each SLM is composed of different parts. Each part shall guide you step-by-step as you discover and understand the lesson prepared for you.

Pre-tests are provided to measure your prior knowledge on lessons in each SLM. This will tell you if you need to proceed on completing this module or if you need to ask your facilitator or your teacher’s assistance for better understanding of the lesson. At the end of each module, you need to answer the post-test to self-check your learning. Answer keys are provided for each activity and test. We trust that you will be honest in using these.

In addition to the material in the main text, Notes to the Teacher are also provided to our facilitators and parents for strategies and reminders on how they can best help you on your home-based learning.

Please use this module with care. Do not put unnecessary marks on any part of this SLM. Use a separate sheet of paper in answering the exercises and tests. And read the instructions carefully before performing each task.

If you have any questions in using this SLM or any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator.

Thank you.

What I Need to Know

[illustration: hand pressing a circular button]

This module was designed and written with you in mind. It is here to help you master the Operations Involving 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.

This module is comprised of only one lesson:

• Lesson – Operations Involving Polynomials

After going through this module, you are expected to:

1. add and subtract polynomials;
2. solve problems involving addition and subtraction of polynomials;
3. derive the laws of exponents;
4. apply the laws of exponents in simplifying expressions;
5. multiply polynomials such as:
   a. monomial by monomial;
   b. monomial by polynomial with more than one term;
   c. binomial by binomial;
   d. polynomial with more than one term to polynomial with three or more terms;
6. divide polynomials such as:
   a. polynomial by a monomial;
   b. polynomial by a polynomial with more than one term; and
7. solve problems involving multiplication and division of polynomials.

What I Know

[illustration: hand holding a pencil]

Choose the letter of the best answer. Write the chosen letter on a separate sheet of paper.

1. Which of the following below are like terms?

   • A. 2𝑥, 3𝑥, 4𝑥, 5𝑥
   • B. −4𝑎, −3𝑎, −2𝑎, 𝑎𝑏
   • C. 3𝑥, 𝑥², 𝑥³, 𝑥⁴
   • D. 𝑥𝑦, 3𝑥𝑦, 5𝑦, 4𝑧

2. Which of the following would result to a sum of 5𝑎𝑏?

   • A. 2𝑎 + 3𝑏
   • B. 4𝑎 + 𝑏
   • C. 2𝑎𝑏 + 3𝑎𝑏
   • D. 3𝑎 + 2𝑏

3. What is the simplified term of 3𝑥 + 𝑥 – 2𝑥 + 5𝑥?

   • A. 6𝑥
   • B. 7𝑥
   • C. 8𝑥
   • D. 9𝑥

4. What must be subtracted from 8𝑎 + 3 to make a difference of 5𝑎 + 1?

   • A. 3𝑎 + 2
   • B. 3𝑎 – 2
   • C. 3𝑎 + 4
   • D. 3𝑎 – 4

5. Mark saved 15𝑥 + 8 from his allowance. How much did he have after buying a gift worth 10𝑥 + 3 for his mother’s birthday?

   • A. 5𝑥 − 5
   • B. 5𝑥 + 5
   • C. 5𝑥 − 11
   • D. 5𝑥 + 11

6. Which of the following illustrates the law 𝑎ⁿ = 𝑎 ∙ 𝑎 ∙ 𝑎 ∙ 𝑎 … ∙ 𝑎 (𝑛 𝑡𝑖𝑚𝑒𝑠)?

   • A. 𝑥⁵ = 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥
   • B. 2³ = 3 ∙ 3
   • C. 3⁴ = 3 ∙ 4
   • D. 𝑥⁵ = 𝑥 ∙ 5

7. Which of the following statements illustrates the law: 𝑎ⁿ ∙ 𝑎ᵐ = 𝑎ⁿ⁺ᵐ?

   • A. 2³ ∙ 3⁴ = 6⁷
   • B. 3³ ∙ 2⁵ = 6¹⁵
   • C. 2³ ∙ 2⁴ = 2¹²
   • D. 3² ∙ 3⁴ = 3⁶

8. [transcriber’s note: the first letter of this question is missing in the printed text; it reads “hich of the following is true?”]

   • A. 5² + 5³ = 5⁵
   • B. 2²/2³ = 1/2
   • C. (3²)⁹ = 9¹⁸
   • D. (4³)² = 4⁵

9. Divide (4𝑥² − 24𝑥 + 35) ÷ (2𝑥 − 5).

   • A. 2x – 7
   • B. 2x + 12
   • C. 2x – 12
   • D. 2x + 7

10. The length of the rectangle is 2x – 3 and its width is equal to x – 1. Find the area of the rectangle.

    
    

• A. x – 2
• B. 3x – 4
• C. 2x² – x – 3
• D. 2x² – 5x + 3

11. Which of the following is the result when (3𝑎²𝑏³) is multiplied to (4𝑎³𝑏²)?

• A. 7𝑎⁵𝑏⁵
• B. 7𝑎⁶𝑏⁶
• C. 12𝑎⁶𝑏⁶
• D. 12𝑎⁵𝑏⁵

12. Which of the following describes the product of (𝑏 + 5)(𝑏 + 5)?

• A. a monomial
• B. a binomial
• C. a trinomial
• D. a multinomial

13. Which of the following is the result when (𝑥 + 7) is multiplied to (𝑥 + 4)?

• A. 𝑥² + 11𝑥 + 28
• B. 𝑥² + 28𝑥 + 11
• C. 𝑥² + 7𝑥 + 28
• D. 𝑥² + 8𝑥 + 28

14. Divide 12𝑥³𝑦⁴ + 9𝑥²𝑦³ − 3𝑥⁵𝑦² by 3𝑥²𝑦.

• A. 6𝑥²𝑦⁴
• B. 6𝑥𝑦³ + 𝑦² − 𝑥³𝑦
• C. 3𝑥⁴ + 2𝑦⁴
• D. 4𝑥𝑦³ + 3𝑦² − 𝑥³𝑦

15. The product of 4𝑥²𝑦⁵ and a certain monomial is 28𝑥⁷𝑦⁹𝑧². What is the missing factor?

• A. 7𝑥⁵𝑦⁴
• B. 7𝑥⁹𝑦¹⁴𝑧²
• C. 7𝑥⁵𝑦²𝑧²
• D. 7𝑥⁵𝑦⁴𝑧²

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

Lesson 1: Operations Involving Polynomials

[illustration: lesson number badge “Lesson 1”]

A great day to start with another module! Do you know that there are a lot of real-life experiences that you can apply using operations of polynomials? Engineers use it in designing roads, bridges, building and other structures. It is also applied in predicting traffic patterns to design appropriate traffic control measures, and even in determining the best combination of grocery items and its quantity that will suit one’s budget.

Simplifying expressions by making use of the laws of exponents is also an exciting thing to do!

What’s In

[illustration: hand holding a puzzle piece]

In order to add, subtract, multiply and divide polynomials, you need to review the operations on integers.

Rules for Adding Integers

1. If the integers have the same sign, copy the sign then add the numbers.
   Examples: 5 + 2 = 7; -3 + (-5) = -8; 6 + 5 = 11; -8 + (-4) = -12
2. If the integers have different signs, subtract the numbers then copy the sign of the larger number.
   Examples: -7 + 4 = -3; 8 + (-8) = 0; 6 + (-5) = 1; -2 + 2 = 0

Rules for Subtracting Integers

1. Change the sign of the subtrahend then proceed to addition.
2. Follow the rules in adding integers.
   Examples: 5 – 2 = 3; -5 – (-2) = -5 + 2 = -3; 6 – 8 = 6 + (-8) = -2; 6 – (-2) = 6 + 2 = 8

Rules for Multiplying Integers

1. When multiplying two integers with the same sign, the product is positive.
   Examples: (7)(8) = 56; (-8)(-5) = 40; (6)(2) = 12; (-6)(-2) = 12
2. When you multiply two integers with different signs, the product is negative.
   Examples: (7)(-2) = -14; (-2)(3) = -6
3. Any number multiplied by 0 gives a product of 0.
   Examples: (0)(5) = 0; (-2)(0) = 0

For a product with no zero factors:

4. If the number of negative factors is odd, the product is negative.
   Examples: (-2)(3)(1) = -6; (4)(-3)(-2)(-1) = -24

   
   

5. If the number of negative factors is even, the product is positive.
   Examples: (-2)(-3)(1) = 6; (-1)(-2)(3)(2)(-3)(-2) = 72

   
   

Rules for Dividing Integers

1. When two integers with the same sign are divided, the quotient is always positive.
   Examples: 42 ÷ 7 = 6; -45 ÷ (-9) = 5
2. When two integers with different signs are divided, the quotient is always negative.
   Examples: 63 ÷ (-9) = -7; -56 ÷ 7 = -8
3. The rules for dividing zero by a non-zero number and for division by zero still hold.
   (0) / (non−zero number) = 0; (any number) / (0) = undefined
   Examples: 0 / -2 = 0; 5 / 0 = undefined

For a quotient with no zero divisor and dividend:

4. If the number of negative integers to be divided is odd, the product is negative.
   Examples: -4 ÷ 2 ÷ 1 = -2; -10 ÷ (-2) ÷ 1 ÷ (-1) = -5
5. If the number of negative integers to be divided is even, the product is negative.
   Examples: -2 ÷ 1 ÷ (-1) ÷ 1 = 2; -4 ÷ 4 ÷ (-1) ÷ (-1) ÷ (-1) = 1

[transcriber’s note: in rule 5 under “Rules for Dividing Integers”, the printed text says the product is negative for an even number of negative integers; the examples given show positive results]

Can you still remember these rules? Well, let’s see! Let’s apply these rules in doing the next activity. Enjoy!

Activity 1: Who Am I?

Reveal who is being referred to in the statements above by performing the indicated operation in Column A and matching it to the answers in Column B.

“I am the father of Archimedes. Do you know my name?” Find it out by decoding the hidden message below.

No.	Column A	Column B
1.	8 + (−5)	S −2
2.	(−12) + (−8)	H −20
3.	(−2) + 1	I −1
4.	(4)(−7)	P 3
5.	(−5) ÷ (5)	A 6
6.	(−2)(−3)	E −13
7.	(−6) ÷ (3)	D −28
		V 28

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1 2 3 4 5 6 7

Great! The activity revealed who is the father of Archimedes. If you were able to name him correctly, then you really have remembered the operations on integers which is very useful in this module.

Activity 2: Agree or Disagree!

In this activity, you will guess whether the following statements is correct or not. Write A if you agree with the statements; otherwise, write D. Write your answer on a separate sheet of paper.

Statement	Response
1. Any number raised to zero is equal to one (1).
2. An expression with exponents CANNOT be simplified even if we follow the laws of exponents.
3. $2^3$ is equal to $\frac{1}{8}$.
4. Laws of exponents may be used to simplify expressions with exponents.
5. $\frac{1}{3^2} = 9$
6. $3^0 4^2 = 16$
7. $\frac{1}{(32x^3y^5)^2}$ may be written as $(32x^3y^5)^2$ where $x \neq 0$ and $y \neq 0$.
8. $3^2 \cdot 4^0 + 1^2 \cdot 5^0 = 11$

Nice one! You are now ready to discover more about operations of polynomials.

What’s New

Familiarize yourself with the tiles below:

[illustration: small white square tile] Stands for (+1)

[illustration: small dark square tile] Stands for (-1)

[illustration: white rectangular tile] Stands for (+x)

[illustration: dark rectangular tile] Stands for (-x)

[illustration: large white square tile] Stands for ($x^2$)

[illustration: large dark square tile] Stands for ($-x^2$)

Can you represent the following quantities using the above tiles?

a. $x – 2$ b. $4x + 1$

The tiles can make operations on polynomials easy to understand and follow. Look and observe the following examples.

a. To represent $x – 2$, we get one (+$x$) and two (-1) tiles.

[illustration: one white rectangular tile and two dark small square tiles]

b. To represent $4x + 1$, we get four (+$x$) and one (+1) tiles.

Try this!

Use the tiles to perform the operations on the following polynomials.

1. $5x + 3x$
2. $(3x – 4) – 6x$
3. $(2x^2 – 5x + 2) + (3x^2 + 2x)$
4. $(3x)(x)$
5. $(3 – x)(x + 2)$
6. $(x^2 + 7x + 6) \div (x + 1)$

Solutions:

1. $5x + 3x =$

[illustration: five white (+$x$) tiles, a plus sign, three white (+$x$) tiles, an equal sign, and eight white (+$x$) tiles]

Since there are eight (+$x$) altogether, therefore, $5x + 3x = 8x$.

2. $(3x – 4) – 6x$

[illustration: tiles for (3x − 4) shown as two white (+$x$) tiles, one dark small tile and two white tiles, and tiles for 6x shown as six dark (−$x$) tiles, with a minus sign between the two groups]

Recall that combining (+$x$) and (−$x$) will result to zero.

[illustration: one (+$x$) tile + one (−$x$) tile = 0]

[illustration: one (+$x$) tile + one (−$x$) tile = 0]

[illustration: one (+$x$) tile + one (−$x$) tile = 0]

Note: There are 3 pairs of (+$x$) and (−$x$) on the illustration above. After combining the (+$x$) and (−$x$), notice that there are no other tiles to be paired. The tiles that are left are:

[illustration: three dark (−$x$) tiles and four dark small (−1) tiles]

There are three(3) (−$x$) tiles and four(4) (−1) tiles

−3$x$

−4

Therefore, $(3x – 4) – 6x = -3x – 4$

3. $(2x^2 – 5x + 2) + (3x^2 + 2x)$

[illustration: tiles for $2x^2$, −5$x$ and +2 added to tiles for $3x^2$ and +2$x$, shown as two large white squares, two dark (−$x$) rectangles, two small white squares and one dark (−$x$) rectangle, with a plus sign, then three large white squares and two white (+$x$) rectangles]

Note: Combining (+$x$) and (−$x$) will result to zero.

[illustration: one (+$x$) tile + one (−$x$) tile = 0]

[illustration: one (+$x$) tile + one (−$x$) tile = 0]

From the illustration above, it can be noticed that there are only two pairs of (+$x$) and (−$x$). There are no other tiles that can be paired that will be equal to zero. The tiles that are left are:

[illustration: five large white ($x^2$) tiles]

five(5) (+$x^2$) tiles = $5x^2$

[illustration: three dark (−$x$) tiles and two small white (+1) tiles]

three(3) (−$x$) tiles = −3$x$ and two(2) (+1) tiles = +2

Therefore, $(2x^2 – 5x + 2) + (3x^2 + 2x) = 5x^2 – 3x + 2$

4. $(3x)(x)$

[chart: rectangular tile grid for (3x)(x); labels as printed: top row “Construct 3x by forming three(3) x- tiles” with three $x$ tiles shown above the grid; left column “Construct x by forming one(1) x-tiles” with one $x$ tile shown; the inner area is labeled “You are going to put your answer here. This area forms a rectangle. Therefore, you need to construct a rectangle with length 3x and x”]

Notice that you need to put one(1) −$x^2$-tile, two(2) −$x$-tiles, three(3) $x$-tiles, and six(6) 1-tiles that will fit together and will form a rectangle.

[chart: tile grid for (3 − x)(x + 2); labels as printed: top “Construct (x + 2) by forming one(1) x tile and two(2) 1-tiles”; left “Construct (3 − x) by forming one(1) −x-tile and three(3) 1-tiles”; inner area “You are going to put your answer here. This area forms a rectangle. Therefore, you need to create a rectangle using the tiles.”]

Multiplying white tiles to black tiles will result to black tiles.

Multiplying white tiles to white tiles will result to white tiles.

Notice that you need to put one(1) −$x^2$-tile, two(2) −$x$-tiles, three(3) $x$-tiles, and six(6) 1-tiles that will fit together and will form a rectangle.

Therefore, $(3x)(x) = 3x^2$

5. $(3 – x)(x + 2)$

Construct $(x + 2)$ by forming one(1) $x$ tile and two(2) 1-tiles

Construct $(3 – x)$ by forming one(1) −$x$-tile and three(3) 1-tiles

You are going to put your answer here. This area forms a rectangle. Therefore, you need to create a rectangle using the tiles.

Take note that combining (+$x$) tiles and (−$x$) will result to zero.

[illustration: two dark (−$x$) tiles + two white (+$x$) tiles = 0]

Therefore, $(3 – x)(x + 2) = -x^2 + x + 6.$

(Take note: multiplying white tile to a white tile will also result to a white tile.)

Notice that you need to put three(3) $x^2$ tile that will fit together and will form a rectangle.

6. $(x^2 + 7x + 6) \div (x + 1)$

You are going to put your answer here. Use the tiles that will perfectly fit the rectangle.

You are going to form the divisor $(x + 1)$ here by putting one(1) $x$-tile and one(1) 1-tile.

The dividend $(x^2 + 7x + 6)$ is placed in this rectangular area formed by putting one(1) $x^2$-tile, seven(7) $x$-tiles and six(6) 1-tiles.

[chart: division tile layout for $(x^2 + 7x + 6) \div (x + 1)$; labels as printed: divisor “$x$” and “1”; dividend tiles “$x^2$”, seven “$x$” tiles and six “1” tiles]

(Take note: Dividing white tile by a white tile will also result to a white tile.)

Notice that the tile that will fit on the length of the dividend ($+x^2$)- tile and length of the divisor $x$-tile is the $x$-tile. So it is placed on this area of the quotient. Take note that you should be able to create a rectangle.

You need to put some more tiles in order to create a rectangle whose length is the same as the length of the whole dividend. Now, it is clear that you are going to put (+1)-tile.

Continuing the process, you need to put five(5) more (+1)-tile in order to create a rectangle that will fit together.

Quotient

After performing the process, it can be noticed that rectangle on the quotient portion is composed of these tiles:

[illustration: one $x$ tile and six 1 tiles]

Therefore, $(x^2 + 7x + 6) \div (x + 1) = x + 6$

What is It

ADDITION AND SUBTRACTION OF POLYNOMIALS

It is important that we know what are like or similar terms before we proceed with the addition and subtraction of polynomials. Terms with the same literal coefficients are called like terms or similar terms.

2ab, 6ab and -ab are similar terms because they have the same literal coefficients which is $ab$.

2a, 3ab and -5b are not similar terms because their literal coefficients are not the same (the literal coefficients are $a$, $ab$, and $b$, respectively)

Try this!

Write inside the box the terms that are similar. The first one is done for you.

Given	Similar terms
$3xy, -6x, xy, \frac{1}{2}xy$	$3xy$, $xy$ and $\frac{1}{2}xy$
1. $-ab, 0.5ab, 8by, \frac{3ab}{10}$
2. $4(x + y), -7xy, (x + y), 13(x + y)$
3. $a^2b^3, -a^2b^3, -a^3b^2$

In this procedure, align all the terms with the same literal coefficients and then combine them.

From the exercise, the similar terms in Item No.1 are $-ab, 0.5ab$ and $\frac{3ab}{10}$ because they have the same literal coefficient which is $ab$. Meanwhile in Item No. 2, $4(x + y)$, $(x + y)$, and $13(x + y)$ are similar terms because they have the same literal coefficient which is $(x + y)$. Furthermore, in Item No. 3 only $a^2b^3$ and $-a^2b^3$ are similar terms because they have common literal coefficient which is $a^2b^3$. The term $-a^3b^2$ is not similar to $a^2b^3$ and $-a^2b^3$ for the reason that their exponents are not the same.

Thus, for them to be called similar terms, they should have the same literal coefficients including its exponents.

Rules for Adding Polynomials

To add polynomials, simply combine like terms or similar terms. To combine similar terms, get the sum of the numerical coefficients and copy the same literal coefficients. For convenience, write similar terms in the same column.

Example 1:

Find the sum.

a. $5a + (2a + 6)$ b. $(2x + 4) + (4x – 1)$

Solution:

a. $5a + (2a + 6) = (5a + 2a) + 6$ Group like terms. $= (5 + 2)a + 6$ Add the numerical coefficients of the like terms, then simplify. $= 7a + 6$

b. $(2x + 4) + (4x – 1) = (2x + 4x) + (4 – 1)$ Group like terms. $= (2 + 4)x + (4 – 1)$ Add the numerical coefficients of the like terms, then simplify. $= 6x + 3$

Example 2:

Simplify each.

a. $(5a + 8b – 2c) + (8a + 7b – 8c)$ b. $(4x + 10y – 6z) + (7x – 4y + 7z) + (3x + 9z)$

Solution:

a. By removing the parentheses and grouping and combining like terms, we get:

$(5a + 8b – 2c) + (8a + 7b – 8c)$ $= (5a + 8a) + (8b + 7b) + (-2c – 8c)$ Group like terms. $= (5 + 8)a + (8 + 7)b + (-2 – 8)c$ Combine the numerical coefficients of each term, then simplify. $= 13a + 15b – 10c$

By aligning the terms vertically, we get:

   5a + 8b - 2c
+  8a + 7b - 8c
  13a + 15b - 10c

In this procedure, align all the terms with the same literal coefficients and then combine them.

Using any of the two procedures will result to $13a + 15b – 10c$.

Notice that the middle part is left blank because the expression has no term which is similar to the term aligned above it.

Therefore, $(5a + 8b – 2c) + (8a + 7b – 8c) = 13a + 15b – 10c$.

b. By removing the parentheses and grouping and combining like terms, we get:

$(4x + 10y – 6z) + (7x – 4y + 7z) + (3x + 9z)$ $= (4x + 7x + 3x) + (10y – 4y) + (-6z + 7z + 9z)$ Group like terms. $= (4 + 7 + 3)x + (10 – 4)y + (-6 + 7 + 9)z$ Combine the numerical coefficients of each term, then simplify. $= 14x + 6y + 10z$

By aligning the terms vertically, we get:

 4x + 10y - 6z
 7x -  4y + 7z
+ 3x         + 9z
 14x +  6y + 10z

Using any of the two procedures will result to $14x + 6y + 10z$.

Therefore, $(4x + 10y – 6z) + (7x – 4y + 7z) + (3x + 9z) = 14x + 6y + 10z$.

Example 3:

The lengths of the sides of quadrilateral MATH are shown in the diagram. Find the perimeter of the quadrilateral MATH.

[chart: quadrilateral MATH with vertices M (top left), A (top right), T (bottom right), H (bottom left); side labels as printed: top “4a − b”, right “3a”, left “4b”, bottom “5a + 2b”]

Solution:

To find the perimeter, add the lengths of the sides.

$P = (4a – b) + (3a) + (5a + 2b) + (4b)$ $= (4a + 3a + 5a) + (-b + 2b + 4b)$ $= (4 + 3 + 5)a + (-1 + 2 + 4)b$ $= 12a + 5b$

Therefore, the perimeter is $12a + 5b$.

Example 4:

Vincent Vann saved (2x + 3y)pesos from his allowance on Monday and (6x – y)pesos on Tuesday. What is his total savings for two days?

Solution:

To find his savings for two days, add how much he has saved on Monday and Tuesday.

Total Savings = (2x + 3y) + (6x – y) = 2x + 6x + 3y + (-y) = 8x + 2y

Therefore, he has saved a total of (8x + 2y)pesos in two days.

Rule for Subtracting Polynomials

To subtract a polynomial from another polynomial, change the sign of the subtrahend and proceed to addition. Keep in mind that in adding polynomials, you can only add those that are similar or like terms.

Example 1.

$a – b = a + (-b)$

minuend subtrahend

Subtract.

a. (5𝑥 + 4) – (7𝑥 + 3) b. (9𝑥 + 3) – (−4𝑥 + 5) c. (6𝑥² − 4𝑥 + 8) − (3𝑥² − 𝑥 − 5) d. (8𝑥³ − 2𝑥² + 2𝑥 − 2) − (3𝑥² + 4𝑥 + 7)

Solution:

a. (5𝑥 + 4) – (7𝑥 + 3) Given = (5𝑥 + 4) + (−7𝑥 − 3) Definition of Subtraction = (5𝑥 − 7𝑥) + (4 − 3) Group like terms = −2𝒙 + 1 Simplify

b. (9𝑥 + 3) – (−4𝑥 + 5) Given = (9𝑥 + 3) + (4𝑥 − 5) Definition of Subtraction = (9𝑥 + 4𝑥) + (3 − 5) Group like terms = 13𝒙 − 2 Simplify

c. Applying horizontal subtraction, we get

(6𝑥² − 4𝑥 + 8) − (3𝑥² − 𝑥 − 5) Given minuend subtrahend = 6𝑥² − 4𝑥 + 8 − 3𝑥² + 𝑥 + 5 Remove the grouping symbols and change the sign of the subtrahend = 3𝒙² − 3𝒙 + 13 Simplify

Applying vertical subtraction, we get

(align like terms) (change the sign of the subtrahend and proceed to addition) 6𝑥² − 4𝑥 + 8 6𝑥² − 4𝑥 + 8 − (3𝑥² − 𝑥 − 5) + − 3𝑥² + 𝑥 + 5 3𝒙² − 3𝒙 + 13

d. (8𝑥³ − 2𝑥² + 2𝑥 − 2) − (3𝑥² + 4𝑥 + 7) Given = (8𝑥³ − 2𝑥² + 2𝑥 − 2) + (−3𝑥² − 4𝑥 − 7) Definition of Subtraction = 8𝑥³ + (−2𝑥² − 3𝑥²) + (2𝑥 − 4𝑥) + (−2 − 7) Group like terms = 8𝒙³ − 5𝒙² − 2𝒙 − 9 Simplify

Example 2.

Subtract 6𝑥² − 5𝑥 − 2 from 4𝑥² − 2𝑥.

Solution:

“Subtract 6𝑥² − 5𝑥 − 2 from 4𝑥² − 2𝑥” means (6𝑥² − 5𝑥 − 2) − (4𝑥² − 2𝑥). Thus,

(4𝑥² − 2𝑥) – (6𝑥² − 5𝑥 − 2) = (4𝑥² − 2𝑥) + (−6𝑥² + 5𝑥 + 2) = (4𝑥² − 6𝑥²) + (−2𝑥 + 5𝑥) + (2) = −2𝒙² + 3𝒙 + 2

To do it vertically, we have

(align like terms) (change the sign of the subtrahend and proceed to addition) 4𝑥² − 2𝑥 4𝑥² − 2𝑥 − (6𝑥² − 5𝑥 − 2) + − 6𝑥² + 5𝑥 + 2 −2𝒙² + 3𝒙 + 2

Example 3:

There are 14𝑎² + 2𝑎 + 16 children in a classroom. If the number of boys is 5𝑎² − 9𝑎 + 7, how many girls are in the classroom?

Solution:

In order to determine the number of girls in the classroom, we should subtract the number of boys from the total number of children in the classroom. Hence,

Number of girls = total number of children – number of boys = (14𝑎² + 2𝑎 + 16) − (5𝑎² − 9𝑎 + 7) = 14𝑎² + 2𝑎 + 16 − 5𝑎² + 9𝑎 − 7 = 14𝑎² − 5𝑎² + 2𝑎 + 9𝑎 + 16 − 7 = 9𝒂² + 11𝒂 + 9

Therefore, there are 9𝑎² + 11𝑎 + 9 number of girls in the classroom. Addition and subtraction of polynomials is just easy, right? Try this exercise to check your understanding before proceeding to the next topic.

Exercise:

Simplify each of the given expressions.

1. (3𝑥 − 7) + (−4𝑥 − 2)
2. (9𝑥² − 2𝑥) − (8𝑥² + 4𝑥)
3. (−3𝑥² − 2𝑥 + 5) + (5𝑥² + 2𝑥 − 8)
4. (8𝑥³ − 2𝑥² + 2𝑥 − 2) − (3𝑥² + 4𝑥 + 7)
5. (3𝑥² − 7𝑥𝑦 + 2𝑦²) + (5𝑥² + 6𝑥𝑦 − 4𝑦²)

Now, let us derive the laws of exponents. These will be used in simplifying expressions and in multiplying and dividing polynomials.

LAWS OF EXPONENTS

Activity 3.1: Answer Me. Faster!

Give the product of each of the following as fast as you can. Write it on the space provided. Do not forget to record the time you spent in answering.

1. 3 𝑥 3 = ________ 4. 2 𝑥 2 𝑥 2 = _______
2. 4 𝑥 4 𝑥 4 = ________ 5. 2 𝑥 2 𝑥 2 𝑥 2 = _______
3. 5 𝑥 5 𝑥 5 = ________ 6. 2 𝑥 2 𝑥 2 𝑥 2 𝑥 2 = _______

Time Spent in Answering	____sec/mins

This leads us to the definition of exponents.

The exponent will tell how many times we are going to multiply the base by itself.

Examples:

1. Which of the following is/are correct?
   Answers:
   a. 4² = 4 𝑥 4 = 16 CORRECT
   b. 2⁴ = 2 𝑥 2 𝑥 2 𝑥 2 = 8 INCORRECT
   c. 2⁵ = 2 𝑥 5 = 10 INCORRECT

   
   

2. Give the value of each of the following as fast as you can.
   Answers
   a. 2³ = _____ 8
   b. 2⁵ = _____ 32
   c. 3⁴ = _____ 81

   
   

Note: Do not multiply the exponent by its base. INCORRECT: 4² = 4 ∙ 2 CORRECT: 4² = 4 ∙ 4

Activity 3.2: Let’s Investigate!

Evaluate the following by applying the law that we have discussed. Investigate the result. Make a simple guess on it after completing the table. The first two were done for you.

Given	Factored Form Using the Previous Law	Long Process	Result
1. (2³)²	2³ ∙ 2³	2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2	64
2. (𝑥⁴)³	𝑥⁴ ∙ 𝑥⁴ ∙ 𝑥⁴	𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥 ∙ 𝑥	𝑥¹²
3. (3²)²	3² ∙ 3²
4. (2²)³		2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2

The activity leads us to the next law on exponents:

LAW OF EXPONENTS (Power of Powers)
B. (𝒂ⁿ)ᵐ = 𝒂ⁿᵐ

When a power is raised to another power or when an exponential expression is raised to another exponent, multiply the exponents.

Examples:

1. (𝑥¹⁰⁰)³ = 𝑥⁽¹⁰⁰⁾⁽³⁾ Using the Power of Powers Law, multiply the exponents.
   = 𝑥³⁰⁰ Simplify.

   
   

2. (𝑦¹²)⁵ = 𝑦⁽¹²⁾⁽⁵⁾ Using the Power of Powers Law, multiply the exponents.
   = 𝑦⁶⁰ Simplify.

   
   

LAW OF EXPONENTS (n times)
A. 𝒂ⁿ = 𝒂 ∙ 𝒂 ∙ 𝒂 ∙ 𝒂 … ∙ 𝒂 where 𝑎 ≠ 0 and n is an integer. In 𝑎ⁿ, 𝑎 is called the base and 𝑛 is called the exponent.

Note: The table above is the Law of Exponents, part A, as printed at the top of this page.

Activity 3.3: Can You Notice Something?

Evaluate the following by completing the table. Notice that the bases are the same. The first example is done for you.

Given	Long Form	Exponential Form	Result
1. (2³)(2²)	(2 ∙ 2 ∙ 2) ∙ (2 ∙ 2)	2⁵	32
2. (𝑥⁵)(𝑥⁴)		𝑥⁹
3. (3²)(3³)	(3 ∙ 3) ∙ (3 ∙ 3 ∙ 3)
4. (2⁴)(2⁵)			512

Did you notice something? What can you conclude about 𝒂ⁿ ∙ 𝒂ᵐ? What will you do with 𝒂, 𝒏 and 𝒎?

LAW OF EXPONENTS (Product of a Power)
C. 𝒂ⁿ ∙ 𝒂ᵐ = 𝒂ⁿ⁺ᵐ

In multiplying exponential expressions whose bases are the same, just copy the common base and then add its exponents.

Examples:

1. (𝑥³²)(𝑥²⁵) = 𝑥³²⁺²⁵ Since they have common base which is “x”, copy the common base and add the exponents
   = 𝑥⁵⁷ Simplify the result

   
   

2. (𝑦⁵⁹)(𝑦⁵¹) = 𝑦⁵⁹⁺⁵¹ Copy the common base “y” and add the exponents.
   = 𝑦¹¹⁰ Simplify the result.

   
   

Activity 3.4: Evaluate us!

Evaluate each of the following. Notice that the bases are the same. The first example is done for you.

Given	Solution	Result	Equivalent Exponential Form
1. 2⁷/2³	= (2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2 ∙ 2) / (2 ∙ 2 ∙ 2)	16	2⁴
2. 3⁵/3³		9
3. 4³/4²	= (4 ∙ 4 ∙ 4) / (4 ∙ 4)		4¹

Transcriber’s note: in the Solution column of row 1 and row 3, the cancelled factors are struck through in the original; the expression above shows the full factors before cancellation.

Did you notice something? What can you conclude about 𝑎ⁿ/𝑎ᵐ? What will you do with 𝒂, 𝒏 and 𝒎?

In dividing exponential expressions with the same bases, copy the common base and subtract its exponents.

Examples:

1. 𝑥²⁰/𝑥¹³ = 𝑥²⁰⁻¹³ Copy the common base “x” and subtract the exponents.
   = 𝑥⁷ Simplify the result.

   
   

2. 𝑦¹⁰⁵/𝑦⁸⁷ = 𝑦¹⁰⁵⁻⁸⁷ Copy the common base “y” and subtract the exponents.
   = 𝑦¹⁸ Simplify the result.

   
   

Note: The law of exponent that we have just discussed applies only to expressions in which the exponent on the numerator is greater than the exponent on the denominator. There is another law that will be followed if the exponent on the numerator is lesser than the exponent on the denominator. However, this law will be discussed when you will be in Grade 9.

Notice what happens on the expressions inside the table.

No	Given (Start Here)	Applying a law of exponent	Result	Answer	Reason
1.	5/5	5¹⁻¹	5⁰	1	Any number divided by itself is equal to 1.
2.	100/100	100¹⁻¹	100⁰	1
3.	𝑥/𝑥	𝑥¹⁻¹	𝑥⁰	1
4.	𝑎⁵/𝑎⁵	𝑎⁵⁻⁵	𝑎⁰	1

Transcriber’s note: the Reason text is printed once for the four rows, spanning them in the original; it is placed in the first row only.

LAW OF EXPONENTS (Law For Zero Exponent)
E. 𝒂⁰ = 1 where 𝑎 ≠ 0

Always take note that any nonzero number raised to a zero exponent is always equal to 1.

Examples:

1. (7,654,321)⁰ = 1
2. 3⁰ + 𝑥⁰ + (3𝑦)⁰ = 1 + 1 + 1

Let us summarize the laws of exponents discussed in this lesson.

LAW OF EXPONENTS (Quotient of a Power)
D. 𝒂ⁿ/𝒂ᵐ = 𝒂ⁿ⁻ᵐ
Note: This is only applicable if the exponent on the numerator (denoted by n) is greater than the exponent on the denominator (denoted by m).

LAWS ON EXPONENTS
For any real number 𝑎 ≠ 0 and positive integers 𝑚 and 𝑛:
1. (𝒂ⁿ)ᵐ = 𝒂ⁿᵐ Power of Powers
2. 𝒂ⁿ ∙ 𝒂ᵐ = 𝒂ⁿ⁺ᵐ Product of a Power
3. 𝒂ⁿ/𝒂ᵐ = 𝒂ⁿ⁻ᵐ, where 𝒏 > 𝒎 Quotient of a Power
4. 𝒂⁰ = 1 where 𝑎 ≠ 0 Law for Zero Exponent

There are still other Laws of Exponents which you will learn in Grade 9. For now, you have to learn first the basic laws discussed above so that you will be able to understand other laws to be discussed in higher grade level.

MULTIPLYING POLYNOMIALS

Unlike the processes of addition and subtraction, multiplication of polynomials considers not only the numerical coefficients but also the literal coefficients.

Rules in Multiplying Polynomials

A. Monomial by Monomial

To multiply a monomial with another monomial, simply multiply the numerical coefficients then multiply the literal coefficients by applying the basic laws of exponents.

Examples:

1. (𝑥³)(𝑥⁵) = 𝑥³⁺⁵ Applying the law of exponent, copy the common base “x” and add the exponents “3” and “5”
   = 𝒙⁸ Simplify the result.

   
   

2. (3𝑥²)(−5𝑥¹⁰) = (3)(−5)(𝑥²⁺¹⁰) Multiply the numerical coefficients “3” and “-5” and copy the common base “x” then add the exponents “2” and “10”
   = −15𝒙¹² Simplify the result.

   
   

3. (−8𝑥²𝑦³)(−9𝑥𝑦⁸)
   = (−8)(−9)(𝑥²⁺¹)(𝑦³⁺⁸) Multiply the numerical coefficients “-8” and “-9” and copy the common base “x” then add the exponents “2” and “1” and the common base “y” and add the exponents “3” and “8”
   = 72𝒙³𝒚¹¹ Simplify the result.

   
   

B. Monomial by a Polynomial

To multiply a monomial with a polynomial, simply apply the distributive property and follow the rule in multiplying monomial by a polynomial. Multiplication of monomial by a polynomial could also be done vertically. In the examples below, the solutions are presented into two ways:

Examples:

1. (3𝑥)(𝑥² − 5𝑥 + 7)

Solutions:

Using the distributive property, (3𝑥)(𝑥² − 5𝑥 + 7) = (3𝒙)(𝑥²) + (3𝒙)(−5𝑥) + (3𝒙)(7) = (3𝑥¹⁺²) + (−15𝑥¹⁺¹) + (21𝑥) = 3𝒙³ − 15𝒙² + 21𝒙

Using the Vertical Solving, 𝑥² − 5𝑥 + 7 x 3𝑥 3𝒙³ − 15𝒙² + 21𝒙

2. (−5𝑥²𝑦³)(2𝑥²𝑦 − 3𝑥 + 4𝑦⁵)

Solutions:

Using the distributive property, (−5𝑥²𝑦³)(2𝑥²𝑦 − 3𝑥 + 4𝑦⁵) = (−5𝒙²𝒚³)(2𝑥²𝑦) + (−5𝒙²𝒚³)(−3𝑥) + (−5𝒙²𝒚³)(4𝑦⁵) = (−5)(2)(𝑥²⁺²)(𝑦³⁺¹) + (−5)(−3)(𝑥²⁺¹)(𝑦³) + (−5)(4)(𝑥²)(𝑦³⁺⁵) = −10𝒙⁴𝒚⁴ + 15𝒙³𝒚³ − 20𝒙²𝒚⁸

Solving vertically, 2𝑥²𝑦 − 3𝑥 + 4𝑦⁵ x −5𝑥²𝑦³ −10𝒙⁴𝒚⁴ + 15𝒙³𝒚³ − 20𝒙²𝒚⁸

C. Binomial by a Binomial

To multiply a binomial with another binomial, you can use F-O-I-L Method or Smile Method. F-O-I-L stands for F-First, O-Outer, I-Inner, and L-Last. This acronym represents the order in which you should multiply the binomials’ terms to get the product. You can draw arcs from the first to first terms and outer to outer terms over the top of the expression and draw arcs from the inner to inner terms and last to last terms on the bottom. When you do, you create a smiley face!

Another way is the vertical way of multiplying which is the conventional one.

Examples:

1. (𝑥 + 3)(𝑥 + 5)

Solutions:

Using F-O-I-L Method, [illustration: arcs drawn over and under the expression (𝑥 + 3)(𝑥 + 5) labeled First terms, Last terms, Inner terms, and Last terms]

F-irst terms → (𝑥)(𝑥) = 𝑥² O-outer terms → (𝑥)(5) = 5𝑥 I-nner terms → (3)(𝑥) = 3𝑥 L-ast terms → (3)(5) = 15

Since 5𝑥 and 3𝑥 are similar terms, we can combine them. → 5x + 3x = 8x. The final answer is 𝒙² + 𝟖𝒙 + 𝟏𝟓.

Using the vertical way, 𝑥 + 3 x 𝑥 + 5 𝑥² + 3𝑥 → Multiply 𝑥 to (𝑥 + 3)

• 5𝑥 + 15 → Multiply 5 to (𝑥 + 3). Align similar terms.
  𝒙² + 𝟖𝒙 + 𝟏𝟓

Notice that any of the two solutions will give you the same answer which is 𝒙² + 𝟖𝒙 + 𝟏𝟓.

2. (𝑥 − 5)(𝑥 + 5)

Solution:

[illustration: F O I L arrows drawn over and under the expression (𝑥 − 5)(𝑥 + 5)]

(𝑥 − 5)(𝑥 + 5) = 𝑥² + 5𝑥 − 5𝑥 − 25 Since “5x” and “-5x” are similar terms, they will be combined and will result to 0. = 𝒙² − 𝟐𝟓

Therefore, (𝑥 − 5)(𝑥 + 5) = 𝒙² − 𝟐𝟓

3. (𝑥 + 6)²

Solution:

(𝑥 + 6)² = (𝑥 + 6)(𝑥 + 6) = 𝑥² + 6𝑥 + 6𝑥 + 36 Combine like terms: “6x” and “6x” = 𝒙² + 𝟏𝟐𝒙 + 𝟑𝟔

Therefore, (𝑥 + 6)² = 𝒙² + 𝟏𝟐𝒙 + 𝟑𝟔

4. (3𝑎 − 5𝑏)(4𝑎 + 7)

Solutions:

Using the FOIL Method,

[illustration: arrows labeled F, O, I, L joining the pairs of terms in (3𝑎 − 5𝑏)(4𝑎 + 7)]

(3𝑎 − 5𝑏)(4𝑎 + 7) = (3𝑎)(4𝑎) + (3𝑎)(7) + (−5𝑏)(4𝑎) + (−5𝑏)(7) = 12𝑎² + 21𝑎 − 20𝑎𝑏 − 35𝑏

Using the vertical way,

      3𝑎 − 5𝑏
    x  4𝑎 + 7
    ----------
      12𝑎² − 20𝑎𝑏
   +  + 21𝑎 − 35𝑏
    ----------
  12𝑎² − 20𝑎𝑏 + 21𝑎 − 35𝑏

Multiply 4𝑎 to (3𝑎 − 5𝑏) and 7 to (3𝑎 − 5𝑏), respectively.

Notice that no terms are aligned because there are no terms which are similar.

D. Polynomial with more than one term to Polynomial with three or more terms

To multiply a polynomial with more than one term to a polynomial with three or more terms, simply apply the Distributive Property of Multiplication over Addition. Multiply the first term of the first polynomial to each term of the other polynomial. Repeat the procedure up to the last term and simplify the results by combining similar terms.

Another way of multiplying this is using the vertical way which is the conventional one.

Examples:

1. (𝑥 + 3)(𝑥² − 2𝑥 + 3)

Solutions:

Using the Distributive Property,

(𝑥 + 3)(𝑥² − 2𝑥 + 3) = 𝒙(𝑥² − 2𝑥 + 3) + 𝟑(𝑥² − 2𝑥 + 3) = (𝑥³ − 2𝑥² + 3𝑥) + (3𝑥² − 6𝑥 + 9) = 𝒙³ + 𝒙² − 𝟑𝒙 + 𝟗

Using the vertical way,

      𝑥² − 2𝑥 + 3
    x     𝑥 + 3
    ------------
      𝑥³ − 2𝑥² + 3𝑥
   +   3𝑥² − 6𝑥 + 9
    ------------
      𝒙³ + 𝒙² − 𝟑𝒙 + 𝟗

2. (𝑥² + 3𝑥 − 4)(4𝑥³ + 5𝑥 − 1)

Solutions:

Using the distributive property,

(𝑥² + 3𝑥 − 4)(4𝑥³ + 5𝑥 − 1) = 𝒙²(4𝑥³ + 5𝑥 − 1) + 𝟑𝒙(4𝑥³ + 5𝑥 − 1) − 𝟒(4𝑥³ + 5𝑥 − 1) = 4𝑥⁵ + 5𝑥³ − 𝑥² + 12𝑥⁴ + 15𝑥² − 3𝑥 − 16𝑥³ − 20�x + 4 = 4𝑥⁵ + 12𝑥⁴ + 5𝑥³ − 16𝑥³ − 𝑥² + 15𝑥² − 3𝑥 − 20𝑥 + 4 = 𝟒𝒙⁵ + 𝟏𝟐𝒙⁴ − 𝟏𝟏𝒙³ + 𝟏𝟒𝒙² − 𝟐𝟑𝒙 + 𝟒

Notice that no terms are aligned because there are no terms which are similar.

Using the vertical way,

          𝑥² + 3𝑥 − 4
    x    4𝑥³ + 5𝑥 − 1
    -------------------
    4𝑥⁵ + 12𝑥⁴ − 16𝑥³
  +        5𝑥³ + 15𝑥² − 20𝑥
  +             − 𝑥² − 3𝑥 + 4
    -------------------
    𝟒𝒙⁵ + 𝟏𝟐𝒙⁴ − 𝟏𝟏𝒙³ + 𝟏𝟒𝒙² − 𝟐𝟑𝒙 + 𝟒

3. (2𝑥 − 3)(3𝑥 + 2)(𝑥² − 2𝑥 − 1)

Solution:

Multiply (2𝑥 − 3) and (3𝑥 + 2) first. By FOIL Method,

(2𝑥 − 3)(3𝑥 + 2) = 6𝑥² + 4𝑥 − 9𝑥 − 6 = 𝟔𝒙² − 𝟓𝒙 − 𝟔

The result will then be multiplied to (𝑥² − 2𝑥 − 1).

(6𝑥² − 5𝑥 − 6)(𝑥² − 2𝑥 − 1) = 𝟔𝒙²(𝑥² − 2𝑥 − 1) − 𝟓𝒙(𝑥² − 2𝑥 − 1) − 𝟔(𝑥² − 2𝑥 − 1) = 6𝑥⁴ − 12𝑥³ − 6𝑥² − 5𝑥³ + 10𝑥² + 5𝑥 − 6𝑥² + 12𝑥 + 6 = 6𝑥⁴ − 12𝑥³ − 5𝑥³ − 6𝑥² + 10𝑥² − 6𝑥² + 5𝑥 + 12𝑥 + 6 = 𝟔𝒙⁴ − 𝟏𝟕𝒙³ − 𝟐𝒙² + 𝟏𝟕𝒙 + 𝟔

Alright! Let us apply the skills we have learned from multiplying polynomials into solving real-life problems. Let’s study this!

Tina has a square garden with a side of length s feet. If she increased the length by 5 feet and then decreased the width by 2 feet, what trinomial represents the area of the new rectangular garden?

Solution:

The length of the new garden is 𝑠 + 5 and the width is 𝑠 − 2.

[illustration: rectangle with the left side labeled 𝑠 − 2 and the bottom side labeled 𝑠 + 5]

Multiply the length and width of the new garden by following the procedure in multiplying binomials.

(𝑠 + 5)(𝑠 − 2) = 𝒔(𝑠 − 2) + 𝟓(𝑠 − 2) = 𝑠² − 2𝑠 + 5𝑠 − 10 = 𝑠² + 3𝑠 − 10

Therefore, the trinomial that represents the area of the new rectangular garden is 𝒔² + 𝟑𝒔 − 𝟏𝟎.

Subtract the product 𝒙² + 𝟐𝒙 (subtrahend) from 𝒙² − 𝟑𝒙 (minuend). Remember to change the sign of the subtrahend then proceed to addition.

Bring down the remaining term −𝟏𝟎. Then, divide −𝟓𝒙 by 𝒙.

Divide 𝒙² by 𝒙 and write the result on top.

Multiply the result 𝒙 to 𝒙 + 𝟐.

Multiply the result −𝟓 to 𝒙 + 𝟐.

Subtract the product −𝟓𝒙 − 𝟏𝟎 from −5𝒙 − 𝟏𝟎. Thus, the remainder is 0.

DIVIDING POLYNOMIALS

Like in multiplication, the division of polynomials applies the laws of exponents.

Rules in Dividing Polynomials

A. Polynomial by Monomial

To divide a polynomial by a monomial, simply divide each term of the polynomial by the given divisor.

Examples:

1. Divide 12𝑥⁴ − 16𝑥³ + 8𝑥² by 4𝑥².

Solution:

12𝑥⁴ − 16𝑥³ + 8𝑥² ÷ 4𝑥² = 12𝑥⁴/4𝑥² − 16𝑥³/4𝑥² + 8𝑥²/4𝑥² = 3𝑥⁴⁻² − 4𝑥³⁻² + 2𝑥²⁻² = 3𝑥² − 4𝑥 + 2𝑥⁰ = 3𝑥² − 4𝑥 + 2(1) = 𝟑𝒙² − 𝟒𝒙 + 𝟐

B. Polynomial by a Polynomial with more than one term

To divide a polynomial by a polynomial with more than one term by long division, simply follow the procedure in dividing numbers by long division:

1. Check the dividend and the divisor to see if they are in standard form.
2. Set-up the long division by writing the division symbol where the divisor is outside the division symbol and the dividend inside it.
3. You may now start the division process.
4. You can stop the cycle when:
   a. The quotient (answer) has reached the constant term.
   b. The exponent of the divisor is greater than the exponent of the dividend.

Examples:

1. Divide 𝑥² − 3𝑥 − 10 by 𝑥 + 2

Solution:

𝑥² − 3𝑥 − 10 and 𝑥 + 2 are already in standard form.

              𝒙 − 5
    𝒙 + 2 ) 𝒙² − 3𝑥 − 10
            𝑥² + 2𝑥
            ---------
              −5𝑥 − 10
              −5𝑥 − 10
              ---------
                    0

Annotations beside the division:

• Divide 𝑥² by 𝑥 and write the result on top.
• Multiply the result 𝑥 to 𝑥 + 2.
• Subtract the product 𝑥² + 2𝑥 (subtrahend) from 𝑥² − 3𝑥 (minuend). Remember to change the sign of the subtrahend then proceed to addition.
• Bring down the remaining term −10. Then, divide −5𝑥 by 𝑥.
• Multiply the result −5 to 𝑥 + 2.
• Subtract the product −5𝑥 − 10 from −5𝑥 − 10. Thus, the remainder is 0.

Therefore, the quotient is 𝒙 − 𝟓.

2. Divide 𝑥³ − 6𝑥² + 11𝑥 + 6 by 𝑥 − 3

Solution:

Since the terms in the dividend and divisor are in standard form, then,

                𝑥² − 3𝑥 + 2
    𝑥 − 3 ) 𝑥³ − 6𝑥² + 11𝑥 − 6
            𝑥³ − 3𝑥²
            ---------
              − 3𝑥² + 11𝑥
              − 3𝑥² + 9𝑥
              -----------
                    2𝑥 − 6
                    2𝑥 − 6
                    ------
                         0

Annotations beside the division:

• Divide 𝑥³ by 𝑥 and write the result on top.
• Multiply the result 𝑥² to 𝑥 − 3.
• Subtract the product 𝑥³ − 3𝑥² (subtrahend) from 𝑥³ − 6𝑥² (minuend). Remember to change the sign of the subtrahend then proceed to addition. Bring down 11𝑥. Then, divide −3𝑥² by 𝑥.
• Multiply the result −3𝑥 to 𝑥 − 3.
• Subtract the product −3𝑥² + 9𝑥 from −3𝑥² + 11𝑥 and bring down −6. Then, divide 2𝑥 by 𝑥.
• Multiply the result 2 to 𝑥 − 3.
• Subtract the product 2𝑥 − 6 from 2𝑥 − 6. Thus, the remainder is 0.

Therefore, the quotient is 𝒙² − 𝟑𝒙 + 𝟐.

Doing great! Now, let us solve related problems on dividing polynomials.

1. There are 28𝑥² + 36𝑥 − 16 oranges in a basket. Mother divides them equally among 4 children. How many oranges does each child get?

Solution:

In order to determine how many oranges will each child get, we need to divide the total number of oranges in a basket by 4 since it will divided equally among 4 children. In doing so, we have

(28𝑥² + 36𝑥 − 16)/4 = 28𝑥²/4 + 36𝑥/4 − 16/4 = (28/4)𝑥² + (36/4)𝑥 − 16/4 = 7𝑥² + 9𝑥 − 4

Therefore, each child will get 7𝑥² + 9𝑥 − 4 of oranges equally.

2. The area of a parallelogram is 8𝑥² + 10𝑥 − 3. If its base is 2𝑥 + 3, what is its height?

Solution:

𝐴 = 𝑏ℎ Area formula of a parallelogram ℎ = 𝐴/𝑏 Divide both sides by b. = (8𝑥² + 10𝑥 − 3)/(2𝑥 + 3) Substitute the values.

[illustration: parallelogram with a dashed height line labeled h, and the base labeled 2𝑥 + 3]

Use long division to solve for h.

                4𝑥 − 1
    2𝑥 + 3 ) 8𝑥² + 10𝑥 − 3
             8𝑥² + 12𝑥
             ----------
               −2𝑥 − 3
               −2𝑥 − 3
               ----------
                      0

Annotations beside the division:

• Divide 8𝑥² by 2𝑥.
• Multiply the result 4𝑥 to the divisor 2𝑥 + 3.
• Subtract 8𝑥² + 12𝑥 from 8𝑥² + 10𝑥. Remember to change the sign of the subtrahend and proceed to addition. Then divide −2𝑥 by 2𝑥.
• Multiply the result −1 to 2𝑥 + 3.
• Subtract −2𝑥 − 3 from −2𝑥 − 3. The remainder is 0.

Therefore, the height of the parallelogram is 𝟒𝒙 − 𝟏.

What’s More

Let us try to answer more challenging set of problem and activities about performing operations involving polynomials.

A. Find the term that makes each statement true. An example is done for you.

Example:

Given : −8𝑎 + ______ = 4𝑎

Solution : We need to determine what should be written on the blank in order to make the statement correct. So, we have

−8𝑎 + ______ = 4𝑎 −8𝑎 + ______ + 8𝑎 = 4𝑎 + 8𝑎 Add 8a to both sides of the equation to eliminate -8a at the right side of the equation. ______ = 12𝑎 Combining like terms.

This means that we need to write 12𝑎 on the blank in order to make the statement correct.

Answer : −8𝑎 + 𝟏𝟐𝒂 = 4𝑎

1. (−6𝑎) + ____ = 2𝑎
2. 5𝑚 + (−9𝑚) + ____ = 2𝑚
3. 4𝑎𝑏 – ____ = 2𝑎𝑏
4. 3𝑥 + ___ + (−5𝑥) = 2𝑥
5. (−6𝑥𝑦²) – 10𝑥𝑦² = _____
6. 3𝑦² + ____ + (−2𝑦²) = 5𝑦²
7. (−2𝑥𝑦) + (−8𝑥𝑦) + ____ = 3𝑥𝑦
8. (−3𝑐) – (5𝑐) = ________
9. 10𝑥² – ____ = 5𝑥²
10. 8𝑏 + (−4𝑏) + ____ = 5𝑏

B. sssTell whether the given statement is True or False. Write True if the statement is correct. Otherwise, write False. Write your answer on a separate sheet.

1. 3² · 3⁴ = 9⁶
2. 4² · 4⁴ = 4⁸
3. (𝑥²)³ = 𝑥⁸
4. (2³ · 3²)² = 6¹²
5. (𝑚²)² = (𝑚⁶)³
6. −7² = −49
7. (𝑡³)² = 𝑡⁹
8. (2² · 4)³ = 2¹²
9. (4 + 5) = 3⁴
10. (3 + 4)² = 3² + 4²

C. Perform the indicated operations. Choose your answer inside the box.

Choices: −24𝑦⁶ | 2𝑥 + 1 | 𝑥 − 3 | 12𝑥⁵𝑦⁵ | 𝑥² − 6𝑥 + 27 | 2𝑥² − 𝑥 − 6 | 𝑥 − 4 | 𝑥² + 5𝑥 + 6 | 36𝑥³ − 61𝑥² − 72𝑥 − 15 | 8𝑥²𝑦 + 4𝑥𝑦² | 𝑥² + 6𝑥 − 27 | 6𝑥⁵ − 21𝑥⁴ − 24𝑥³

1. (3𝑥²𝑦³)(4𝑥³𝑦²)
2. (3𝑦²)(−2𝑦)(4𝑦³)
3. (3𝑥³)(2𝑥² − 7𝑥 − 8)
4. (𝑥 + 3)(𝑥 + 2)
5. (𝑥 − 3)(𝑥 + 9)
6. (4𝑥² − 9𝑥 − 3)(9𝑥 + 5)
7. (2𝑥² − 3𝑥 − 2) ÷ (𝑥 − 2)
8. (8𝑥³ − 4𝑥² − 24𝑥) ÷ 4𝑥
9. (𝑥² − 8𝑥 + 16) ÷ (𝑥 − 4)
10. (𝑥² − 9) ÷ (𝑥 + 3)

Good job! Get ready for another learning battle ahead!

What I Have Learned

Write the correct word or phrase make the statement true. Choose your answer inside the box.

Choices: 1 | numerical coefficient | general | standard form | add | Distributive Property | literal coefficient | subtract | Associative Property | divisor | positive | subtrahend

1. To add like polynomials, add their ____________________ and copy the ____________________.
2. To subtract a polynomial from another polynomial, change the sign of the ____________________ and proceed to addition.
3. To multiply powers having the same base, keep the base and ____________________ the exponents.
4. To divide powers having the same base, keep the base and ____________________ the exponents.
5. Any number, excluding zero, raised to zero is always equal to ____________________.
6. To multiply a monomial by a polynomial with more than one terms, we will make use of ____________________.
7. In dividing polynomials by a monomial, divide each term in the dividend by the ____________________.
8. In dividing polynomial with more than one term by another polynomial with more than one term, the first thing that we are going to do is to see to it that both dividend and the divisor are in ____________________.

Nice work! Now you’re up for the next challenge of this lesson.

What I Can Do

Here is another activity that will let you apply what you have learned about performing operations on polynomials by simplifying it to real-life situations.

Solve the following problems.

1. The three sides of a triangle measures (3𝑥² + 𝑥 − 2), (−5𝑥² + 4) and (15𝑥² − 4𝑥 − 9) respectively. What is the perimeter of the triangle?
2. There are (12𝑎² + 4𝑎 + 15) children in a classroom. If the number of boys is 4𝑎² − 9𝑎 + 8, how many girls are in the classroom?
3. John Ike made a layout of the garden where he planned to put his Aglonema and Sanseveria Plants for business. In order to protect his plants from direct sunlight, he needs to cover the garden with nets on top of it. To determine the dimensions of the net he needs to buy to cover the top of a garden, he must be able to compute the area of the garden first. The length of the garden is 𝑥⁵ and its width is 𝑥³. What expression represents the area of the garden? (Hint: The area of a rectangle is A = lw)
4. Yvanna has a square garden with a side of length s feet. If she increases the length by 5 feet and then decreased the width by 2 feet, what trinomial represents the area of the new rectangular garden?

Assessment

For you to determine how much you’ve learned, please answer the questions by choosing the letter of the best answer.

1. What should be added to 5𝑎 + 2 to get a sum of 10𝑎² + 4?
   A. 5𝑎 + 2
   B. 5𝑎² + 2
   C. 10𝑎² + 5𝑎 + 2
   D. 10𝑎² − 5𝑎 + 2

   
   

2. What is the sum of 3𝑎² − 5𝑎 + 2 and 𝑎² − 2𝑎 + 4?
   A. 3𝑎² + 7𝑎 + 6
   B. 3𝑎² − 7𝑎 + 6
   C. 4𝑎² + 7𝑎 + 6
   D. 4𝑎² − 7𝑎 + 6

   
   

3. What must be added to 5𝑏 − 3 to obtain 0?
   A. −5𝑏 − 3
   B. −5𝑏 + 3
   C. 5𝑏 − 3
   D. 5𝑏 + 3

   
   

4. If the length of the rectangle in terms of x is 8𝑥² + 4𝑥 + 7 and its width is 7𝑥 + 8, what is the perimeter of the rectangle?
   A. 16𝑥² + 22𝑥 + 30
   B. 16𝑥⁴ + 22𝑥² + 30
   C. 16𝑥² + 30𝑥 + 15
   D. 16𝑥⁴ + 22𝑥² + 15

   
   

5. What must be subtracted from 9𝑥² − 2𝑥 to make a difference of 5𝑥² + 3?
   A. 4𝑥² − 2𝑥 − 3
   B. 4𝑥² + 2𝑥 + 3
   C. 4𝑥² − 𝑥 − 3
   D. 4𝑥² + 𝑥 − 3

   
   

6. Which of the following illustrates the law (𝑎ⁿ)ᵐ = 𝑎ⁿᵐ?
   A. (2³)⁸ = 2²⁴
   B. (3²)⁴ = 3²⁺⁴
   C. (𝑥⁴)² = 𝑥⁴⁻²
   D. (2²)⁴ = 4⁽²⁾⁽⁴⁾

   
   

7. Which of the following expressions when simplified is equal to 1 ?
   A. (4𝑎⁵)⁰
   B. (5𝑎⁰)²
   C. 3²/3⁰
   D. 0³

   
   

8. Which of the following statements follows the law: 𝑎ⁿ/𝑎ᵐ = 𝑎ⁿ⁻ᵐ?
   A. 2⁶/3⁴ = 6²
   B. 2³/2⁴ = 2
   C. 3⁴/3² = 9
   D. 3²/3⁴ = 9

   
   

9. Which of the following is the result when (3𝑎²𝑏³) is multiplied to (4𝑎³𝑏²)?
   A. 12𝑎⁶𝑏⁶
   B. 12𝑎⁵𝑏⁵
   C. 7𝑎⁵𝑏⁵
   D. 7𝑎⁶𝑏⁶

   
   

10. Which of the following laws of exponents will be useful in simplifying the expression: (𝑏⁵/𝑏³)³?
    I. (𝑎ⁿ)ᵐ = 𝑎ⁿᵐ III. 𝑎ⁿ/𝑎ᵐ = 𝑎ⁿ⁻ᵐ
    II. 𝑎ⁿ · 𝑎ᵐ = 𝑎ⁿ⁺ᵐ IV. 𝑎⁻ⁿ = 1/𝑎ⁿ
    A. I and II
    B. II and IV
    C. III and IV
    D. I and III

    
    

11. What is the simplified form of (4⁰)(5²)(6¹)(100⁰)?
    A. 1
    B. 10
    C. 25
    D. 150

    
    

12. What is the result when (𝑥 − 5) is multiplied to (𝑥 + 5)?
    A. a multinomial
    B. a trinomial
    C. a binomial
    D. a monomial

    
    

13. Which of the following is the quotient when $(x^2 + 8x + 15)$ is divided by $(x + 5)$?

    
    

a. $x + 3$ b. $x + 5$ c. $x + 5 + \frac{3}{x+5}$ d. $x + 3 + \frac{5}{x+3}$

14. The product of $4x^2y^5$ and a certain monomial is $28x^7y^9z^2$. What is the missing factor?

a. $7x^9y^{14}z^2$ b. $7x^5y^2z^2$ c. $7x^5y^4z^2$ d. $7x^5y^4$

15. The length and width of the top of the rectangular study table is represented by the expression $(x + 5)$ meter and $(x + 3)$ meter, respectively. What is the area of the top of the table?

a. $x^2 + 15$ square meter b. $9x + 15$ square meter c. $x^2 + 8x + 15$ square meter d. $x^2 + 15x + 8$ square meter

Good Job! You did well on this module! Keep going!

Additional Activities

This section includes supplementary activities related to operations involving polynomials.

Do the following activities.

1. Use tiles to illustrate the sum of the following expressions.

a. $5x + 2x$ c. $(x – 1) + (x + 2)$ b. $3x^2 + 4x – x$ d. $(x^2 – 2x + 3) + (x^2 + 2x – 5)$

2. Write an expression for the area of the shaded region below. Choose the answer inside the box opposite to the figure.

[illustration: rectangle with a shaded region on the left and an unshaded strip on the right; labels “1” (width of the unshaded strip), “2m” (height), “4m” (length)]

a. $8m^2 + 2m$ b. $8m^2 – 2m$ c. $8m^2 + 4m$ d. $8m^2 – 4m$

Answer: _____________________

3. Supply the missing term in B that will make the division procedure in A correct.

A

        (1) − (3) − (6)
x − 3 ) 6x³ − 19x² + x + 6
        6x³ − (2)
        −x² + x
        (4) + 3x
        (5) + 6
        −2x + (7)
              (8)

[transcriber’s note: the numbers in circles are blanks, shown here as (1) through (8); the layout of the division procedure is kept as printed]

B

a. $-2x$ b. $6$ c. $6x^2$ d. $0$ e. $-x$ f. $-2$ g. $18x^2$ h. $-x^2$

[transcriber’s note: this page is printed upside down; it is transcribed here in reading order after rotation. Some entries are hard to read and are marked with [?]]

Answer Key

What I Know

1. A
2. C
3. B
4. A
5. B
6. A
7. D
8. B
9. A
10. D
11. D
12. C
13. A
14. D
15. D

What’s In

Activity 1: PHIDIAS

Activity 2:

1. A
2. D
3. D
4. A
5. D
6. A
7. D
8. B

What Is It

Activity 3.1

1. 9
2. 64
3. 125
4. 8
5. 16
6. 32

Activity 3.2

1. $3 \cdot 3 \cdot 3 \cdot 3 = 81$
2. [?]
3. $3 \cdot 3 \cdot 3 \cdot 3 = 81$ [?]
4. $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 64$ [?]

Activity 3.3

1. [illegible]
2. $x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x \cdot x = x^9$
3. $3^5 = 243$
4. $2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^9$ [?]

Activity 3.4

1. [illegible]
2. $\frac{243}{27}$, $3^2$ [?]
3. $4$ [?]

What’s More

A.

1. $8a$
2. $6m$
3. $2ab$
4. $4x$
5. $-16xy^2$
6. $4y^2$
7. $13xy$
8. $-8c$
9. $5x^2$
10. $b$

B.

1. False
2. False
3. False
4. False
5. False
6. True
7. False
8. True
9. False
10. False

C.

1. $12x^5y^5$
2. $-24y^6$
3. $6x^5 – 21x^4 – 24x^3$
4. $x^2 + 5x + 6$
5. $x^2 + 6x – 27$
6. $36x^3 – 61x^2 – 72x$
7. $2x + 1$ [?]
8. $2x^2 – x – 6$
9. $x + 2 + \frac{x-4}{4}$ [?]
10. $x – 3$ [?]

What I Have Learned

1. numerical coefficient, literal coefficient
2. subtrahend
3. add
4. subtract
5. 1
6. Distributive Property
7. divisor
8. standard form

What I Can Do

1. $13x^2 – 3x – 7$
2. $8a^2 + 13a + 7$
3. $x^8$
4. $s^2 + 3s – 10$

Assessment

1. D
2. D
3. B
4. A
5. A
6. A
7. A
8. C
9. B
10. D
11. D
12. C
13. A
14. C
15. C

Additional Activities

a. $7x$ b. $3x^2 + 3x$ c. $2x + 1$ d. $2x^2 – 2$

2. B

   
   

4. C

   
   

5. G

   
   

6. E

   
   

7. H

   
   

8. A

   
   

9. F

   
   

10. B

    
    

11. D

    
    

References

De Leon, Cecile M., et.al, Integrated Mathematics Textbook for First Year, 2000, ISBN: 971-07-1389-2

Jose-Dilao, Soledad, et.al, Intermediate Algebra Textbook for Second Year Revised Edition,2009, ISBN: 978-971-0315-55-0

Department of Education-Bureau of Learning Resources (DepEd-BLR) (2016) Grade 7 Mathematics Teacher’s Guide. Lexicon Press Inc., Philippines

Oronce, Orlando A., et.al, E-Math Elementary Algebra Revised Edition 2010, ISBN: 978-971-23-5479-3, pp.192-195, pp.208-211

Oronce, Orlando A., et.al, E-Math Intermediate Algebra Revised Edition 2010, ISBN:978-971-23-5480-9, pp.70-72, pp.76-83, pp.86-95

For inquiries or feedback, please write or call:

Department of Education – Bureau of Learning Resources (DepEd-BLR)

Ground Floor, Bonifacio Bldg., DepEd Complex Meralco Avenue, Pasig City, Philippines 1600

Telefax: (632) 8634-1072; 8634-1054; 8631-4985

Email Address: blr.lrqad@deped.gov.ph * blr.lrpd@deped.gov.ph