Sachin

The trapezium is a quadrilateral that has four sides and one set of parallel sides. The area of trapezium is the region covered by the solid in a two-dimensional plane. Other types of quadrilaterals are parallelogram, rectangle, square, and rhombus. Find the definition, formula, and solved examples on the trapezium area.

Trapezium Definition

The trapezium is a flat closed shape having 4 straight sides with one pair of parallel sides. The parallel sides are called the bases and the non-parallel sides are called the legs. The different types of the trapezium are isosceles trapezium, scalene trapezium, and right trapezium. Some of the properties of the trapezium are along the lines:

• Every trapezium has exactly one pair of parallel and opposite sides.
• The diagonals intersect each other.
• The non-parallel sides in trapezium are unequal except in isosceles trapezium.
• The line that joins the mid-points of the non-parallel sides is always parallel to the bases or parallel sides which is equal to half of the sum of parallel sides.
• In isosceles trapezium, the non parallel sides are congruent
• The sum of interior angles is equal to 360°.
• The sum of two adjacent angles is 180°.

Area of a Trapezium

The trapezium area is the region covered by the trapezium in a two-dimensional plane. It can be measured by taking the average of two bases and multiplied the result by altitude. Therefore, the area of the trapezium formula is given as,

Trapezium Area A = h(a + b)/2 square units

Where,

a, b are the bases

h is the height or altitude

Area of an isosceles Trapezium = 1/2 [√(c² – (a – b)²) (a + b)]

Where,

a = AB, b = CD, c = AD

The perimeter of trapezium P = a + b + c + d units

Perimeter of the Isosceles Trapezium = a + b + 2c units

Prove that:

Area of trapezium ABCD = 1/2[AD + BC] h

Area of trapezium ABCD = area (∆DFA) + area (rectangle DFEC) + area (∆CEB)

= (¹/₂ × AF × DF) + (FE × DF) + (¹/₂ × EB × CE)

= (¹/₂ × AF × h) + (FE × h) + (¹/₂ × EB × h)

= h/2 x (AF + FE + EB)

= h/2 × (AB + FE)

= h/2 x (AB + DC)

= ¹/₂ × (sum of parallel sides) × (distance between them)

Therefore, The area of a trapezium = ¹/₂ × (sum of parallel sides) × (distance between them)

How to find the Trapezium Area?

Follow the steps to calculate the Area of the Trapezium by referring to the below steps. They are as such

• Observe the question carefully and make a note of known parameters.
• Find either height, base sides, or sum of parallel sides, the distance between the parallel sides.
• Substitute these values in the formula to get the area.

Area of a Trapezium Questions and Answers

Example 1.

In a trapezium, the measurement of one parallel side two more than the other parallel side and the height is 6 cm. The area of the trapezium is 216 cm². Find the lengths of the two parallel sides?

Solution:

Let a, b are the parallel sides of the trapezium.

According to the question,

One parallel side is two more than the other parallel side

a = b + 2

Area of the trapezium = 216 cm²

½(a + b)h = 216

Substitute a = b + 2, h = 6 cm in above equation.

½(b + 2 + b)4 = 216

2(2b + 2) = 216

2b + 2 = 216/2

2b + 2 = 108

2b = 108 – 2

2b = 106

b = 106/2

b = 53

Put b = 53 in a = b + 2

a = 53 + 2 = 55

So, the lengths of the two parallel sides are 55 cm, 53 cm.

Example 2.

The area of a trapezium is 148 cm² and its height is 12 cm. If one of the parallel sides is double of the other, find the two parallel sides?

Solution:

Given that,

Area of the trapezium = 148 cm²

Height = 12 cm

Let one side of the trapezium is a, then another parallel side is b.

As per the question,

One of the parallel sides is double of the other.

a = 2b

Area of the trapezium = 148 cm²

½(a + b)h = 148

Substitute a = 2b, h = 12 in above equation.

½(2b + b) x 12 = 148

3b x 6 = 148

18b = 148

b = 148/18

b = 8.2

Put b = 8.2 in a = 2b

a = 8.2 x 2

a = 16.4

Therefore, the two parallel sides of length are 16.4 cm, 8.2 cm.

Example 3.

Find the area of a trapezium whose parallel sides are AB = 42 cm, CD = 16 cm, and non parallel side AD = 15 cm, BC = 15 cm?

Solution:

In trapezium ABCD,

CE = 15 cm

CD = 16 cm, AE = 16 cm

EB = AB – AE = 42 – 16 = 26 cm

Now, in ∆ EBC

S = (15 + 15 + 26)/2

S = 56/2 = 28

Area of triangle = √(s(s – a) (s – b) (s – c))

= √(28 (28 – 15) (28 – 15) (28 – 26))

= √(28 (13) (13) (2))

= √(9464) = 97.2 cm²

Area of ∆EBC = 1/2 × EB × CP

97.2 = 1/2 × 26 × CP

CP x 13 = 97.2

CP = 97.2/13 = 7.48 cm

Therefore, h = 7.48 cm

Area of trapezium = 1/2(p₁ + p₂) × h

= 1/2(58) × 7.48

= 29 x 7.48 = 216.92 cm²

Therefore, area of a trapezium is 216.92 cm².

Example 4.

The length of the parallel sides of a trapezium is in the ratio 5: 8 and the distance between them is 18 cm. If the area of the trapezium is 256 cm², find the length of the parallel sides?

Solution:

Given that,

The ratio of parallel sides of the trapezium is 5: 8.

The distance between parallel sides = 18 cm

Area of the trapezium = 256 cm²

Let the common ratio be x,

Then, parallel sides are 5x, 8x.

Area of trapezium = 1/2 (p₁ + p₂) h

256 = 1/2 (5x + 8x) x 18

256 = 9(13x)

256 = 117x

x = 256/117

x = 2.188

5x = 5 x 2.188 = 10.94, 8x = 8 x 2.188 = 17.5

Therefore, the length of parallel sides area are 10.94 cm and 17.5 cm.

FAQs on Area of a Trapezium

1. What is the formula for the Area of a trapezium?

Area of trapezium = ½ x Sum of parallel sides x Distance between the parallel sides

Trapezium area = ½ x (a+b) x h

2. What are the applications of a trapezium?

It is used in various physics computations and mathematical calculations.

3. What are the properties of Trapezium?
The trapezium has 4 unequal sides, 2 are parallel and 2 are nonparallel. The sum of the interior angles is 360 degrees and diagonals bisect each other.

Students who want to learn the volume of cuboids and cubes can use the Worked-out Problems on Volume of a Cuboid here. Get to see various examples on the cuboid volume in the coming sections. Try to solve the questions and improve your preparation standards. Check out the Cube and Cuboid Word Problems with solutions in the below sections.

Question 1.

Find the volume of a cuboid of length 18 cm, breadth 25 cm, and height 5 cm?

Solution:

Given that,

Length of cuboid = 18 cm

Breadth of cuboid = 25 cm

Height of cuboid = 5 cm

Cuboid volume = length x breadth x height

Volume = 18 x 25 x 5

= 2250

Therefore, the volume of a cuboid is 2250 cm³.

Question 2.

If the area of the base and height of the cuboid is 212 cm², 8 cm, calculate cuboid volume?

Solution:

Given that,

The base of the base = 212 cm²

Height of a cuboid = 8 cm

Cuboid volume = (Area of the base) x height

Volume = 212 x 8 = 1696 cm³.

Question 3.

Find the volume of the cube whose each side is 16 cm?

Solution:

Given that,

Side length of cube a = 16 cm

The volume of the cube V = a³

V = 16³ = 16 x 16 x 16

V = 4096 cm³

Therefore, the cube volume is 4096 cm³.

Question 4.

If the cuboid volume is 512 cm³, its length and height is 8 cm, 7 cm. Find the cuboid breadth?

Solution:

Given that,

Cuboid Volume = 512 cm³

Cuboid length = 8 cm

Cuboid height = 7 cm

Cuboid breadth = Volume / (length) x (height)

= 512 / (8 x 7)

= 512 / 56 = 9.142 cm

Therefore, the breadth of cuboid is 9.142 cm.

Question 5.

The length, breadth, and depth of a lake are 15 m, 20 m, 9 m respectively. Find the capacity of the lake in liters?

Solution:

Given that,

Length of lake = 15 m

Breadth of lake = 20 m

Depth of lake = 9 m

Capacity of lake = (length) x (breadth) x (depth)

= 15 x 20 x 9 = 2700 m³

1000 liter = 1 m³

Capacity of lake in Litres = 2700 x 1000

= 2700000 litres

Therefore, the capacity of lake is 2700000 litres.

Question 6.

The dimensions of the brick are 25 cm x 8 cm x 10 cm. How many such bricks are required to build a wall of 16 m in length, 20 cm breadth, and 8 m in height?

Solution:

Given that,

Length of brick = 25 cm

Breadth of brick = 8 cm

Height of brick = 10 cm

Length of wall = 16 m

Breadth of wall = 20 m

Height of wall = 8 m

Volume of 1 brick = length x breadth x height

= 25 x 8 x 10 = 2000 cm³

Volume of wall = length x breadth x height

= 16 x 20 x 8

= 2560 = 2560 x 100²

Number of bricks required = (2560 x 100²) / 2000

= 1280 x 10 = 12800

So, the required number of bricks are 12800.

Question 7.

External dimensions of a wooden cuboid are 20 cm × 15 cm × 12 cm. If the thickness of the wood is 2 cm all around, find the volume of the wood contained in the cuboid formed.

Solution:

Given that,

External length of cuboid = 20 cm

External breadth of cuboid = 15 cm

External height of the cuboid = 12 cm

External volume of the cuboid = (length x breadth x height)

= (20 x 15 x 12) = 3600 cm³

Internal length of cuboid = 20 – 4 = 16 cm

Internal breadth of cuboid = 15 – 4 = 11 cm

Internal height of the cuboid = 12 – 4 = 8 cm

Internal volume of a cuboid = (length x breadth x height)

= (16 x 11 x 8) = 1408 cm³

Therefore, volume of wood = External volume of the cuboid – Internal volume of a cuboid

= 3600 – 1408 = 2192 cm³

∴ Volume of the wood contained in the cuboid is 2192 cm³.

Question 8.

The volume of a container is 1440 m³. The length and breadth of the container are 15 m and 8 m respectively. Find its height?

Solution:

Given that,

Length of the container = 15 m

The breadth of the container = 8 m

The volume of the container = 1440 m³

(length x breadth x height) = 1440

15 x 8 x height = 1440

120 x height = 1440

height = 1440/120

height = 12

∴ The height of the container is 12 m.

Cube and Cuboids are three-dimensional shapes that have six faces, eight vertices, and twelve edges. The volume of a three-dimensional figure is the number of cubes required to fill the object. It is defined as the side times side. Get the definition, formulas, standard units, and example problems in the following sections.

What is Meant by Volume?

The volume of any three-dimensional solid shape is the measurement of the space occupied by the object. For a hollow 3-dimensional figure, the volume of the object is the difference in space occupied by the body and the amount of space inside a body. The interior capacity of a solid is also called the volume of a substance.

The measuring units of volume are cubic units. Those can be cm³, m³. The volume can also be measured in liters, milliliters, millimeters.

Standard Units of Volume

The volume is always measured in cubic units. The standard unit is cm³, m³, mm³, but various other units of length measurement are m, dm, dam, etc.

Cuboid

The cuboid is a polyhedron made up of six rectangular faces, eight vertices, and twelve edges. The faces of the cuboid are parallel. But all the faces are not equal in dimensions.

The properties of a cuboid are

• Every cuboid has six rectangles, each rectangle is called the face.
• It has top faces, bottom faces. Other than the top, bottom faces are called the adjacent faces.
• It has 12 edges and opposite sides of the rectangle are equal.
• The point of intersection of 3 edges of a cuboid is called the vertex.
• And it has 8 vertices.

Volume of Cuboid

The volume of the cuboid is the product of the area of one surface and height. Let us take l, b, h as the length, breadth, and height of the cuboid.

Area of the rectangular base = (l x b)

The volume of the cuboid = (Area of the base) x height

= (l x b) x h

Volume = lbh

Then,

Length of the Cuboid = (volume) / (breadth x height)

Breadth of the Cuboid = (Volume) / (length x height)

Height of the cuboid = (Volume) / (length x breadth)

Cube

Cube is a special type of cuboid where length, breadth and height are equal.

The properties of Cube are as follows:

• It has six faces and twelve edges of equal length.
• The shape of all faces of the cube is in square shape.
• The angles of the cube in the plane are at a right angle.
• Each face of the cube meets four other faces.
• Opposite edges are parallel to each other.
• The vertex of the cube meets three faces and edges.

Volume of Cube

The volume of a cube is nothing but the number of cubic units occupied by the cube completely.

Volume of cube = a x a x a = a³

Example Questions on Volume of Cubes and Cuboids

Example 1.

If the length, breadth, and height of the cuboid are 10 cm, 14 cm, 9 cm, then find its volume?

Solution:

Given that,

Length of the cuboid l = 10 cm

Breadth of the cuboid b = 14 cm

Height of the cuboid h = 9 cm

The volume of cuboid = lbh

Volume = 10 x 14 x 9 = 1260 cm³.

Example 2.

Find the volume of the cube, having a side length of 6 cm?

Solution:

Given that,

Side length of cube a = 6 cm

The volume of cube = a³

Volume = 6³ = 6 x 6 x 6 = 216 cm³.

Example 3.

Calculate volume, the perimeter of the cube whose side length is 14 cm?

Solution:

Given that,

Side length of cube a = 14 cm

The volume of the cube formula is

Volume = a³

Volume V = 14³

= 14 x 14 x 14 = 2744 cm³

The perimeter of the cube formula is

Perimeter = 12a

= 12 x 14 = 168 cm

Therefore, the perimeter and volume of the cube is 168 cm, 2744 cm³

FAQS on Volume of Cubes and Cuboids

1. What are the formulas of cube and cuboid volume?

The volume of the cuboid is defined as the product of length, height, and breath.

Cuboid volume V = length x breadth x height.

Cube volume = side³

2. What is the difference between cuboid and cube?

The sides of the cube are in equal length but the cuboid sides are different. The faces of the cube are in a square shape, the faces of the cuboid are in a rectangle shape. All diagonals of the cube are equal but the cuboid has equal diagonals for parallel sides.

3. What is the perimeter of the cube and cuboid?

Perimeter of cuboid = 4(length + breadth + height)

Perimeter of cube = 12 x length.

Are you worried about Division on Integers? Check here to know the complete details about the Integers Division. Like integer multiplication, the division of integers is also an important topic to solve high standard questions. Follow various properties, rules, and tricks to solve division problems. Know the several terminologies involved in this concept. Scroll to the below sections to identify important points about the division on integers.

Division of Integers – Introduction

The division is the inverse operation of multiplication. Multiplication is totalling of numbers whereas division is the distribution of numbers. Though it is the inverse operation, the rules for multiplication and division are similar. The division is used in daily life for various purposes like household, funding, trading, living expenses, etc. Division also has various properties that are used in daily life. Before going to solve division problems, you must know the definitions of various terms.

Definitions

Dividend – The number that is to be divided is called a dividend.

Divisor – The number which divides the other number is called the divisor.

Quotient – The result of the division is called the quotient.

Integers Division Rules

There are various rules to be followed while applying division for integers.

1. The quotient of 2 positive integer numbers will always be a positive integer.
2. The quotient of 2 negative integer numbers will always be a negative integer.
3. 1 positive and 1 negative integer number gives negative integer as a quotient.
4. If a positive number is divided by a positive number, then the result will be a positive number.
5. If a negative number is divided by a negative number, then the result will be a positive number.
6. When a positive number is divided by a negative number, the result will be a negative number.
7. When a negative number is divided by a positive number, the result will be a negative number.

Properties of Division of Integers

There are 5 properties which multiplication and addition but the division of integers do not follow. They are:

1. Closure Property of Division
2. Commutative Property of Division
3. Associative Property of Division
4. Identity Property of Division
5. Distributive Property of Division

Property 1: Closure Property of Division

The division of integer does not follow the actual closure property. The quotient of two integers may or may not be an integer.

If x,y, and z are integers

x/y does not belong to z ( x ÷ y ∉ Z)

Property 2: Commutative Property of Division

The division of integer does not allow commutative property also. If the numbers are swapped, the result varies.

x/y is not equal to y/x (x ÷ y ≠ y ÷ x)

Property 3: Associative Property of Division

This division does not allow associative property also.

(x÷y)÷z ≠ x÷(y÷z)

Property 4: Identity Property of Division

x÷1 = x ≠ 1÷x

Property 5: Distributive Property of Division

Division of Integers does not allow distribution property.

As mentioned above, the division of integers does not allow any of the properties whereas multiplication and addition can be possible with all the above properties.

Important Rules for Division of Integers

Rule 1:

The quotient of two positive or negative integers is a positive integer which is equal to the quotient of the corresponding absolute integer values.

1. The quotient of 2 positive numbers is positive and here we divide the dividend numerical value by the divisor numerical value.

Example:

(+ 9) ÷ (+ 3) = + 3
2.  The quotient of 2 negative integers is positive and here, we divide the dividend numerical value by the divisor numerical value and assign a positive (+) sign to the quotient present.

Example:

(-9)÷(- 3) = +3

Therefore, for dividing 2 integers with the same signs, we divide their values and give plus (+) sign to the quotient.

Rule 2:

A positive and a negative integer gives quotient as a negative integer and its value is equal to the quotient of the corresponding values of the integers.

Example:

(+16)÷(-4) = -4

Therefore, for dividing integers with different signs, we divide their values and give the minus(-) sign to the quotient.

Important Properties of Division of Integers

Property 1:

If an integer ‘a’ is divided by another integer ‘b’, then the integer ‘a’ is divided into ‘b’ number of equal parts.

If ‘b’ divides ‘a’ without any intimation, then ‘a’ is evenly divisible by ‘b’.

Property 2:

When an integer ‘a’ is divided by another integer ‘b’, the division algorithm is, the sum of the product of quotient and divisor & the remainder is equal to the dividend.

i.e.,

Dividend = Quotient*Divisor+Remainder

Property 3:

When an integer number is divided 1, the result that is the quotient is the number itself.

Property 4:

When an integer number is divided by itself, the result or quotient is 1.

Property 5:

When any positive or negative integer is divided by zero, the result is undefined. Therefore, division by zero is meaningless.

Property 6:

When zero is divided by a positive or negative integer, the result or quotient is zero.

Property 7:

When an integer number is divided by another integer number which is a multiple of 10 like 10, 100, 1000,10000, etc., the decimal point for the number should be moved to the left.

Solved Problems on Division of Integers

Question 1.

Allen’s score was changed to -120 points in a video game because he missed some targets. He got -15 points for each missed target. How many targets did he miss?

Solution:

As given in the question, Allen’s score was changed to -120 and he got -15 points for each missed target.

To find how many targets he missed = -120/-15 =8

He missed 8 targets and got -15 points for each missed target.

Question 2.

Lousia’s savings change by -$9 each time she goes bowling. In all, it changed by -$99 during the summer. How many times did she go bowling in the summer?

Solution: 

As given in the question, Lousia changed -$99 during the summer and she changes her savings by -$9

To find how many times she went bowling = -$99/9 = 11

The solution is 11 times.

Question 3.

Elisa withdraw $20 at a time from her bank account and withdrew a total of $140. How many times did she withdraw the money?

Solution:

As per the question, Elisa withdrew a total amount of $140 (it will be negative) and she withdrew $20 at a time(it will be negative).

Therefore, to find the number of times she withdraws the money = -140/-20 = 7

The solution is 7 times.

We hope that the provided information about the Division of Integers is sufficient for your preparation. We also provide you with the solved examples and procedures in the next articles. Therefore, stay tuned to our site to get the latest updates on various information. Bookmark our page to get instant information.

Multiplication of Integers Properties, Rules, and Formulae are here. Follow various tips and shortcuts of Integers Multiplication. Explore the various properties of Integers when multiplying. Know the importance and application of each property rule. Go through the below sections to know more about how multiplication affects integer values. Also, find the example questions and procedures to find solutions.

Properties of Multiplication of Integers

The multiplication of integers refers to the product of 2 or 3 integer values. The set of numbers which consists of whole numbers, natural numbers, the additive inverse, and zero are known as integers. Therefore, integers consist of positive or negative and they have sign or magnitude associated with them. Integers are represented with the letter “Z”. In the below sections, you can get the complete information regarding the properties of the multiplication of integers. Moreover, you can also find the solved examples, rules, definitions, formulae, etc.

Before knowing the properties, get perfection in all the fields of Integers. In the previous articles, you can find the important concepts of Integer values and also various operations. Various arithmetic operations can be performed on Integers like addition, subtraction, multiplication, division, etc. Each operation possesses several properties and here we are going to discuss various properties of multiplication.

Integers Multiplication Properties

There are 6 properties of integers multiplication. Each property will be discussed in detail in the further sections. The important properties are

1. Closure Property of Multiplication
2. Commutative Property of Multiplication
3. Associative Property of Multiplication
4. Distributive Property of Multiplication
5. Multiplication by zero
6. Multiplicative Identity

To carry out multiplication, you need to always multiply the exact integer values and there are a few rules to keep in mind to determine the sign of the final result.

1. Closure Property of Multiplication

According to the closure property, if two positive or negative integers are multiplied, then the result is also an integer.

Suppose two integers are x and y, then the product xy is also an integer.

Example:

7*5 = 35

5*(-2)=-10

(-2)*(-5)= 10

In the above example, we can change the operands of 7 and 5, in either way we get the same result. We can multiply 7 with 5 or we can multiply 5 with 7, we get the same result.

2. Commutative Property of Multiplication

According to the commutative property, altering the integers or operands order does not affect the result of the multiplication.

x*y=y*x, for every integer x and y

Example:

7*(-3)=-(7*3)= -21

(-5)*(-6)=(-6)*(-5)=30

In the above example, we can change the operands or sign. We can multiply 7 with -3 or we can multiply 7 and 3 and add a negative sign to it.

3. Associative Property of Multiplication

According to the associative property, the grouping of integers can be changed and it does not affect the final result.

Suppose x, y and z are integers

x*(y*z)=(x*y)*z

Example:

4*(2*5)=(4*2)*5=40

2*(5*3)=(2*5)*3=30

As shown in the above example, we can multiply 2 and 5, then multiply the product with 4 to get the final result or we can multiply 4 and 2, then multiply the product with 5 to get the final answer.

4. Distributive Property of Multiplication

According to the distributive property of multiplication, it explains the ability to distribute an operation over other mathematical operations within the bracket. This property can be of distributive property of multiplication over addition or distributive property of multiplication over subtraction.

Suppose x,y and z are integers

x*(y+z)=(x*y)=(x*z)

Example:

2*(3+5)=(2*3)+(2*5)=16

4*(4+4)=(4*4)+(4*4)=32

In the above example, we can first add 3 and 5 and then divide it by 2 or we can multiply 2 and 3, then multiply 2 and 5, then add both the products to get the final answer.

5. Multiplication by zero

According to multiplication by zero, any integer multiplied by 0 results in zero

Suppose x is an integer

x*0=0

Example:

2*0=0

5*0=0

In the above example, the integer value 2 is multiplied by 0 and the result is 0.

6. Multiplicative Identity

According to the multiplicative identity, any integer multiplied by 1 results in the same integer value.

Suppose x is an integer

x*1=x

Example:

4*1=4

6*1=6

In the above example, the integer 4 is multiplied by 1 and the result is 4.

Key Points to Remember about Integers Multiplication Properties

1. Closure Property of integer multiplication refers to that the product of two or three integer values will be an integer.
2. The commutative property of multiplication refers to that altering two or three integer numbers will not change the value of the final answer
3. The associative property of multiplication refers that the way the grouping of integer values together does not affect the final answer.
4. The distributive property of multiplication refers to the distribution procedure of one operation value over other mathematical integer values within the braces.
5. Multiplication by zero refers that the product of any negative or positive integer by zero
6. Multiplicative Integer refers to the final answer as 1 when any integer is multiplied with 1.

Solved Examples on Properties of Multiplication of Integers

Question 1.

Adam is Scuba diving. He descends 5 feet. He descends the same distance 4 more times. How many feet is Adam’s new distance from the sea level?

Solution:

As given in the question,

Adam descends 5 feet, descends means traveling down. Therefore, it is negative

Also given, he descends 4 more times. Initially, he descended once and again 4 times. Therefore, it is 5 times.

Hence, (-5)*5=-25

The solution is Adam’s new distance is 25 feet from the sea level.

Question 2.

Clark made four of his trucks payments late and was fined four late fees. The total change to his savings account was -$28. What integer represents one late fee?

Solution:

As given in the question, Clark made four truck payments late which means he did 4 payments. It is positive

Also given, his savings account balance is -$28.

Amount of each payment = -28/4 = -$7

Therefore, each payment for one late fee is $7.

The solution is $7.

Question 3.

Jan lost 22 points on her exam. She got 11 questions wrong. What integer represents the point value of each wrong answer?

Solution:

As given in the question, Jan lost 22 points. Therefore it is negative.

To find the points lost for each wrong answer = 22/11 = 2

Therefore, she lost 2 points for each wrong answer.

Question 4.

During the field trip to the zoo, the class was divided into 8 groups of 3. How many students are there in the class?

Solution:

As given in the question,

The class was divided into 8 groups of 3

Therefore, the total number of students in the class are 8*3=24

The solution is 24.

Question 5.

You are making a snack for 4 friends. Each need 4 crackers. How many crackers do you need?

Solution:

As given in the question,

There are 4 friends and each needs 4 crackers to prepare a snack.

Therefore, the total number of crackers you need are 4*4=16

The solution is 16 crackers.

In the above article, we have provided all the information regarding Properties of Multiplication of Integers, its rules, solved examples, procedure, etc. Before solving the questions, know about the various properties and get perfect in them. Hope you liked the article and this will help you in your preparation. If you need any further clarifications, contact us through the comment section or directly message us. Bookmark our page for all the latest updates.

Wanna solve challenging Integer Multiplication Problems? Here is the best guide for you. In the below sections, we have mentioned the important rules and examples on Multiplication of Integers. Get a clear idea of types and rules present in Integer Multiplication and know the easy methods to solve problems. Learn about Multiplication of Integers Definitions, Step by Step Procedure to Multiply Integers, Properties, in the coming modules.

What is Multiplication of Integers?

Integers multiplication is defined as the repeated addition of integers. The rules for multiplication and division are almost similar. Integer multiplication includes 3 cases:

• Multiplication between two positive integers
• The multiplication between two negative integers
• Multiplication between one positive integer and negative integers

Key Points about Multiplication of Integers

Have a glance at the important points to be taken into consideration while multiplying integers.

• A positive integer multiplied by a negative integer is negative.
• A negative integer multiplied by a negative integer is positive.
• Multiplication is commutative whereas division is not.
• When simplifying, calculate the operations of multiplication in order from left to right.

Procedure for Multiplication of Integers

Go through the simple and easy guidelines listed below on How to Multiply Integers. They are as such

1. First of all, read the problem and make sure all the ideas and words are understood.
2. Identify the problem to be solved.
3. Note down the phrase which gives the information for the problem.
4. Translate the given information to an expression.
5. Simplify the given expression.
6. Apply the formulae to the problem and solve it.

Rules for Multiplication of Integers

While Performing the Integer Multiplication keep the below mentioned instructions or rules in mind so that you can solve the problems easily. They are along the lines

• Product of 2 positive integer numbers is an integer number
• Product of 2 negative integer numbers is an integer number
• The addition of 2 positive integer numbers is an integer number
• The addition of 2 negative integer numbers is an integer number
• Sum of an integer number and its inverse equals zero
• Product of an integer number and its reciprocal equals 1

Solved Examples on Product of Integers

Question 1.

Simplify -5*5?

Solution: 

From the given question,

The first number is negative and the second number is positive, so the result must be negative.

5*5 = 25

As the result must be negative.

Therefore, Multiplying -5*5 we get the result as-25.

Question 2.

Simplify -20*-50?

Solution:

From the given question, -20*-50

Hence, both the numbers are negative, the multiplication of two negative numbers will be positive, so the result will be positive.

Therefore, -20*-50 = 1000

Therefore, Multiplying -20*-50 we get the result as 1000.

Question 3.

Simplify 7*0?

Solution:

Any integer multiplied by zero results in zero.

7*0=0

Thus, 7*0 becomes 0.

Question 4.

Simplify the following

a. 4.(5.2)

b. (4.5).2

Solution:

The value of the expression is 40. Changing the group of integers or numbers won’t change the result.

Therefore 4.(5.2) = (4.5).2 = 40.

Question 5.

Simplify 3(-2)(-5)(-1)?

Solution:

From the given question,

3(-2)(-5)(-1)

Multiply 2 numbers as follows:

3(-2)*(-5)(-1)

(-6)(-5)(-1)

30(-1)

-30

As multiplication is commutative, the order of multiplication does not affect the final result. We must work left to right to obtain the correct result.

Therefore, 3(-2)(-5)(-1) is -30.

Properties of Multiplication of Integers

Multiplication of Integers has various properties and each property has a specific rule. All of them are given below.

1. Closure Property for Multiplication
2. Commutative Property for Multiplication
3. Distributive Property for Multiplication
4. Associativity Property for Multiplication
5. Multiplication by zero
6. Multiplicative Identity

Points to Remember Regarding Integer Multiplication Properties

1. Closure Property of multiplication states that the product of two integer values will be an integer.
2. Commutative property of multiplication states that swapping of two integer numbers will not change the value of the final result.
3. Associative property of multiplication states that the way of the grouping of integer values together does not affect the final result.
4. Distributive property of multiplication states the distribution ability of one operation value over other mathematical values within the bracket.
5. Multiplication by zero states that the product of any positive or negative integer by zero is zero.
6. Multiplicative Integer states that when we multiply any integer with one, the final result will be integer itself.

Example Questions on Integer Multiplication

Question 1.

The tank contains a capacity of 30,000 litres and 1,000 litres are taken every day from it. Find the amount of water in the tank 3 days ago?

Solution:

As per the given question, the water in the tank changes by -1000 every day. Therefore, we need to subtract 3 times to calculate the water amount before 3 days, so the change is

-3*-1000=+3000

The complete calculation is :

30,000+(-3*-1000) = 30,000+3,000 = 33,000

Therefore, before 3 days the amount of water in the tank was 33,000 litres.

The solution is 33,000 litres.

Question 2.

Flora made 6 deposits of $7 each from the bank account. What is the overall change in the account?

Solution:

As given in the question, 6 deposits of $7 are made from each bank account.

Therefore 6(7) = 42

The solution is $42.

Question 3.

The temperature dropped by 2 degrees F every hour for 6 hours. What was the total number of degrees the temperature changed in the 6 hours?

Solution:

As given in the question, the temperature is dropped by 2 degrees for 6 hours.

i.e., -2*6 = -12

Therefore, the solution is -12.

Question 4.

The price of one share of Acme company declined $5 per day or 4 days in a row. How much did the price of one share change in total after the 4 days?

Solution:

As given in the question, $5 is declined. Therefore it is negative

Time will always be positive.

Therefore, -5*4 = -20

The solution is $-20.

Question 5.

A mountain climber climbed up a cliff 50 feet at a time. He did this 5 times in one day. What is the overall change in his elevation?

Solution:

As given in the question, The climber climbed up 50 feet. Therefore, it is positive.

He did it 5 times a day.

Hence, 50*5 = 250

The solution is 250 feet.

Question 6.

The price of the jeans was reduced $6 per week for 7 weeks. By how much the price of jeans changed over the 7 weeks?

Solution:

As given in the question, the price is reduced. Therefore, it is negative.

It was reduced for 7 weeks.

Hence, (-6)*7 = -42

The solution is -42$.

Hope the above-given article clears all your doubts regarding Multiplication of Integers. Refer all the mentioned problems to know the various models. Bookmark our site to get all the notifications regarding various competitive exam topics and concepts. In case of any doubts, you can contact us through the below comment section.

Are you one of those candidates looking eagerly to learn about the concept of Integers? If yes, you must check this page to know the complete details about Integers. Integers is a basic and important concept that lays a stronger foundation for your maths. Know definition, rules, numbers, solved questions, symbols, etc.  Go through the below sections to find various methods and formulae.

Integers – Definition

The integer word is derived from the Latin word “Integer” which represents the whole. Integers are the positive, negative numbers, or zero. Integer values cannot be in decimals, fractions, percents, and we can perform various operations(arithmetic operations) like subtraction, addition, multiplication, division, etc. Examples of integers are 1,2,3,-4,-5, etc.  Integers also include various sets like

Integers also include various sets like zero, whole numbers, natural numbers, additive inverses, etc. These are the subset of real numbers.
Example of integer set: -5,-3, -1, 0, 2, 5

Representation of Integers

As integers contain various numbers and sets and are the subset of real numbers, they are represented with the letter “Z”.

Example:
Z= {5,-3, -1, 0, 2, 5}

Types of numbers in Integers

• Natural Numbers
• Whole Numbers
• Real Numbers
• Rational Numbers
• Irrational Numbers
• Odd Numbers
• Even Numbers

Integers Rules

• The Sum of 2 positive integer numbers is an integer number
• The Sum of 2 negative integer numbers is an integer number
• Product of 2 positive integer numbers is an integer number
• Product of 2 negative integer numbers is an integer number
• Sum of an integer number and its inverse equals zero
• Product of an integer number and its reciprocal equals 1

Addition of Integer Numbers

While adding 2 positive or negative integers(with the same sign), add the absolute values and note down the sum of those numbers with the sign provided with numbers.

Example:

(+6)+(+5) = +11
(-5)+(-5)= -10

While adding 2 integers with a different sign, subtract the absolute values and note down the difference of those numbers with the sign provided with numbers.

Example:

(-5)+(+2)= -3
(+6)+(-3)= -3

Subtraction of Integer Numbers

While subtracting we follow the rules of addition but change the 2nd number which is being subtracted.

Example:

(-4)+(-3)= (-4)-(+3) = -11
(+5)-(+4)=(+5)+(-4)= +1

Division and Multiplication of Integer Numbers

The rule is simple while dividing and multiplying 2 integer numbers.

• If both the integers have the same sign, the result is positive.
• If both the integers have a different sign, the result is negative.

Example:

(+3)*(-4) = -12
(+4)*(+3) = 12
(+16)/(+4) = +4
(-6)/(+2) = -3

Integer Properties

There are 7 properties of integers. The major properties are

1. Associative Property
2. Distributive Property
3. Closure Property
4. Commutative Property
5. Identity Property
6. Multiplicative Inverse Property
7. Additive Inverse Property

1. Associative Property

This property refers to grouping and rules can be applied for addition and multiplication.

Associative Property of Addition

Associative property enables the special feature of grouping the numbers in your own way and still, you get the same answer.

(a+b)+c = a+(b+c)

Example:

(-4+2)+3= -2+(3+4)

In the above example, if we consider the first equation you can solve it in either way i.e., First you take the difference of 4 and 2 and then add 3 to it or you can first add 2 and 3 and then subtract 4 from it. In both ways, you get a constant answer.

Associative Property of Multiplication

This property also refers to the same as the addition property. In whatever way you group numbers, you still get the same answer.

(ab)c= a(bc)

Example:

-2(4)*3=-2(4*3)

In the above example, you can solve it 2 ways and still find the same answer. First, you can multiply 2,4 and then multiply that with 3 or you can first multiply 4,3 and then multiply it with 4.

2. Distributive Property

The distributive property is used when the expression involving addition is then multiplied by a number. This property tells us that we can multiply first and then add or add first and multiply then. In both ways, the multiplication is distributed for all the terms in parentheses.

a(b+c) = ab+ac

Example:

-4(2+3)= (-4*2)+(-4*3)

In the above example, we can first add 2 and 3, then multiply it with 4 or we can multiply 4 with 2 and 3 separately and then add it, still you get the same answer.

3. Closure Property

Closure property for addition or subtraction states that the sum or difference of any 2 integers will be an integer value.

a + b = integer
a x b = integer

Example:

6-3= 3
6+(-3)= 3

The closure property for multiplication also states that the product of any two integer numbers is an integer number.

Example:

5*5=25
(-5)*(-5)=25

The closure property for division does not hold true that the division of two integers is an integer value.

Example:

(-3)/(-12)=1/4, which is not an integer

4. Commutative Property

The commutative property for addition states that when two integer numbers undergo swapping, the result remains unchanged.

a+b=b+a
a*b=b*a

Example:

28+5+43=5+43+28=76

The commutative property for multiplication also states the same that if two integers are swapped, the result remains unchanged.

Example:

5*4*2=2*4*5=40

The commutative property doesn’t hold true for subtraction.

5. Identity Property

Identity Property states that any number that is added with zero will give the same number. Zero is called additive identity.

a+0=a
a*1=a

Example:

5+0=5

The identity property for multiplication also states the same that the integer number multiplied by 1 will give the same number. 1 is called the additive identity.

Example:

5*1=5

6. Multiplicative Inverse Property

Consider “a” as an integer, then as per the multiplicative inverse property of integers,

a*(1/a)=1

Here, 1/a is the multiplicative inverse of integer a.

7. Additive Inverse Property

Consider “a” as an integer, then as per the additive inverse property of integers,

a+(-a)= 0

Here, “-a” is the additive inverse of the integer a

Applications of Integers in Real Life

Integers have many real-life applications. We use them in different situations to quantify things. For example, to check the temperature, positive numbers are used to indicate the temperature above zero and negative numbers are used to indicate the temperature below zero. Integers are also mainly used in real-life situations like hockey, football tournaments, rating for a movie, bank credits and debits, etc.

We have mentioned all the important information about Integers. Hope, the above-provided details will help you in your preparation. Stay tuned to our site to get instant updates on various mathematical concepts.

Factoring Terms by Regrouping concept and examples are given in this article. Students who are searching for the best way to solve problems of finding factors can follow this article. All the tricks and tips to learn Factorization problems are given in this article. All the students need to do is solve all the problems and test their knowledge. Score good marks in the exam by solving all the problems given in this article.

Factoring Terms by Regrouping Solved Examples

1. Factorize the expression

(i) p²r + pqr + pc + pqs + q²s + qc

Solution:
Given expression is p²r + pqr + pc + pqs + q²s + qc
Rearrange the terms
p²r + pqr + pqs + q²s + pc + qc
Group the first two terms, middle two terms, and last two terms.
The first two terms are p²r + pqr, middle terms are pqs + q²s, and the last two terms are pc + qc
Take pr common from the first two terms.
pr (p + q)
Take qs common from the second two terms.
qs (p + q)
Take c common from the last two terms.
c (p + q)
pr (p + q) + qs (p + q) + c (p + q)
Then, take (p + q) common from the above expression.
(p + q) (pr + qs + c)

The final answer is (p + q) (pr + qs + c).

(ii) s³k + s²(k – m) – s(m + n) – n

Solution:
Given expression is s³k + s²(k – m) – s(m + n) – n
Rearrange the terms
s³k + s²k – s²m – sm – sn – n
Group the first two terms, middle two terms, and last two terms.
The first two terms are s³k + s²k, the middle terms are – s²m – sm, and the second two terms are – sn – n
Take s²k common from the first two terms.
s²k (s + 1)
Take – sm common from the middle two terms.
– sm (s + 1)
Take -n common from the last two terms.
-n (s + 1)
s²k (s + 1) – sm (s + 1) – n (s + 1)
Then, take (s + 1) common from the above expression.
(s + 1) (s²k – sm – n)

The final answer is (s + 1) (s²k – sm – n).

2. How to factorize by grouping the following expressions?

(i) px – qx + qy + ry – rx – py

Solution:
Given expression is px – qx + qy + ry – rx – py
Rearrange the terms
px – qx – rx + qy + ry – py
Group the first three terms, and last three terms.
The first three terms are px – qx – rx, and the last three terms are qy + ry – py
Take x common from the first three terms.
x (p – q – r)
Take y common from the last three terms.
-y (p – q – r)
x (p – q – r) – y (p – q – r)
Then, take (p – q – r) common from the above expression.
(p – q – r) (x – y)

The final answer is (p – q – r) (x – y).

(ii) a³ – 2a² + ma + a – 2m – 2

Solution:
Given expression is a³ – 2a² + ma + a – 2m – 2
Rearrange the terms
a³ – 2a² + ma – 2m+ a – 2
Group the first two terms, middle two terms, and last two terms.
The first two terms are a³ – 2a², the middle terms are ma – 2m, and the last two terms are a – 2
Take a² common from the first two terms.
a² (a – 2)
Take m common from the middle two terms.
m (a – 2)
Take 1 common from the last two terms.
1 (a – 2)
a² (a – 2) + m (a – 2) + 1 (a – 2)
Then, take (a – 2) common from the above expression.
(a – 2) (a² + m + 1)

The final answer is (a – 2) (a2 + m + 1).

Factorize by Regrouping The Terms to find factors of an algebraic expression. Rewrite the given expression to form different groups and take out the common factor. Finding factors is easy with the regrouping process. Follow all the problems given below and get complete knowledge on Factorization by Regrouping. Find the simplest method to find factors i.e. regrouping method.

Procedure to find Factors by Regrouping

Follow the below process and solve any difficult expression factors in minutes. They are as such

Step 1: Note down the given expression. From the given algebraic expression form the groups of the given expression in such a way that a common factor can be taken out from every group.
Step 2: Factorize each group.
Step 3: At last, take out the common factor of the groups formed.

Solved Examples on Factorization of Algebraic Expressions

1. Factoring the following expressions

(i) mn (a² + b²) – ab (m² + n²)

Solution:
Given expression is mn (a² + b²) – ab (m² + n²)
Rearrange the terms
mna² – abm² + mnb² – abn²
Group the first two terms and last two terms.
The first two terms are mna² – abm²  and the second two terms are mnb² – abn²
Take ma common from the first two terms.
ma (na – bm)
Take -nb common from the second two terms.
-nb (na – bm)
ma (na – bm) -nb (na – bm)
Then, take (na – bm) common from the above expression.
(na – bm) (ma – nb)

The final answer is (na – bm) (ma – nb).

(ii) 2am – 4an – 3bm + 6nb

Solution:
Given expression is 2am – 4an – 3bm + 6nb
Rearrange the terms
2am – 3bm – 4an + 6nb
Group the first two terms and last two terms.
The first two terms are 2am – 3bm and the second two terms are – 4an + 6nb
Take m common from the first two terms.
m (2a – 3b)
Take -2n common from the second two terms.
-2n (2a – 3b)
m (2a – 3b) -2n (2a – 3b)
Then, take (2a – 3b) common from the above expression.
(2a – 3b) (m – 2n)

The final answer is (2a – 3b) (m – 2n).

(iii) – 6 – 12t + 18t²

Solution:
Given expression is – 6 – 12t + 18t²
Rearrange the terms
18t² – 12t – 6
Then, take 6 as common from the above expression.
6 (3t² – 2t – 1)

The final answer is 6 (3t² – 2t – 1).

2. Factorize the expression

(i) mn – m – n + 1

Solution:
Given expression is mn – m – n + 1
Rearrange the terms
mn – n – m + 1
Group the first two terms and last two terms.
The first two terms are mn – n and the second two terms are – m + 1
Take n common from the first two terms.
n (m – 1)
Take -1 common from the second two terms.
-1(m – 1)
n (m – 1) – 1(m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (n – 1)

The final answer is (m – 1) (n – 1).

(ii) pm + pn – qm – qn

Solution:
Given expression is pm + pn – qm – qn
Rearrange the terms
pm – qm + pn – qn
Group the first two terms and last two terms.
The first two terms are pm – qm and the second two terms are pn – qn
Take m common from the first two terms.
m (p – q)
Take n common from the second two terms.
n (p – q)
m (p – q) + n (p – q)
Then, take (p – q) common from the above expression.
(p – q) (m + n)

The final answer is (p – q) (m + n).

Finding factors for an algebraic expression is simple when it consists of only a few terms. But when it comes to an expression that has more than two or three terms, students feel difficult to solve those problems. Students need a better process to solve algebraic expression factorization. Every student who is looking for the best method to solve algebraic expression factorization can follow the grouping method.

Factoring Terms by Grouping is the easy and best method to solve different expressions easily. Also, the process of Factoring by Grouping The Terms is very simple compared to other methods.

Procedure for Factoring Algebraic Expressions by Grouping

Follow the below steps to find the factorization of a given expression using the below steps.

(i) Take out a factor from each group from the groups of the given expression.
(ii) Factorize each group
(iii) Lastly, take out the common factor.

Factoring Terms by Grouping Examples

1. Factoring of algebraic expression

(i) 2ma + mb + 2na + nb

Solution:
Given expression is 2ma + mb + 2na + nb.
Group the first two terms and last two terms.
The first two terms are 2ma + mb and the second two terms are 2na + nb.
Take m common from the first two terms.
m (2a + b)
Take n common from the second two terms.
n (2a + b)
m (2a + b) + n (2a + b)
Then, take (2a + b) common from the above expression.
(2a + b) (m + n)

The final answer is (2a + b) (m + n).

(ii) 3xm – ym – 3xn + yn

Solution:
Given expression is 3xm – ym – 3xn + yn.
Group the first two terms and last two terms.
The first two terms are 3xm – ym and the second two terms are – 3xn + yn.
Take m common from the first two terms.
m (3x – y)
Take -n common from the second two terms.
-n (3x – y)
m (3x – y) – n (3x – y)
Then, take (3x – y) common from the above expression.
(3x – y) (m – n)

The final answer is (3x – y) (m – n).

(iii) 12a² + 6ab – 4ma – 2mb

Solution:
Given expression is 12a² + 6ab – 4ma – 2mb.
Group the first two terms and last two terms.
The first two terms are 12a² + 6ab and the second two terms are – 4ma – 2mb.
Take 6a common from the first two terms.
6a (2a + b)
Take -2m common from the second two terms.
-2m (2a + b)
6a (2a + b) – 2m (2a + b)
Then, take (2a + b) common from the above expression.
(2a + b) (6a – 2m)

The final answer is (2a + b) (6a – 2m).

(iv) am² – bm² + an² – bn² + ar² – br²

Solution:
Given expression is am² – bm² + an² – bn² + ar² – br².
Group the first two terms, middle two terms, and last two terms.
The first two terms are am² – bm², the middle two terms are + an² – bn², and the last two terms are + ar² – br².
Take m² common from the first two terms.
m² (a – b)
Take n² common from the middle two terms.
n² (a – b)
Take r² common from the middle two terms.
r² (a – b)
m² (a – b) + n² (a – b) + r² (a – b)
Then, take (a – b) common from the above expression.
(a – b) (m² + n² + r²)

The final answer is (a – b) (m² + n² + r²).

(v) ax – ay + bx – by

Solution:
Given expression is ax – ay + bx – by.
Group the first two terms and last two terms.
The first two terms are ax – ay and the second two terms are + bx – by.
Take a common from the first two terms.
a (x – y)
Take b common from the second two terms.
b (x – y)
a (x – y) + b (x – y)
Then, take (x – y) common from the above expression.
(x – y) (a + b)

The final answer is (x – y) (a + b).

2. Factoring the following algebraic expression

(i) 4a + 2ab + b + 2

Solution:
Given expression is 4a + 2ab + b + 2.
Group the first two terms and last two terms.
The first two terms are 4a + 2ab and the second two terms are b + 2.
Take 2a common from the first two terms.
2a (2 + b)
Take 1 common from the second two terms.
1 (b + 2)
2a (2 + b) + 1 (2 + b)
Then, take (2 + b) common from the above expression.
(2 + b) (2a + 1)

The final answer is (2 + b) (2a + 1).

(ii) 3m³ + 5m² + 3m + 5

Solution:
Given expression is 3m³ + 5m² + 3m + 5.
Group the first two terms and last two terms.
The first two terms are 3m³ + 5m² and the second two terms are 3m + 5.
Take m² common from the first two terms.
m² (3m + 5)
Take 1 common from the second two terms.
1 (3m + 5)
m² (3m + 5) + 1 (3m + 5)
Then, take (3m + 5) common from the above expression.
(3m + 5) (m² + 1)

The final answer is (3m + 5) (m² + 1).

(iii) b³ + 3b² + b + 3

Solution:
Given expression is b³ + 3b² + b + 3.
Group the first two terms and last two terms.
The first two terms are b³ + 3b² and the second two terms are b + 3.
Take b² common from the first two terms.
b² (b + 3)
Take 1 common from the second two terms.
1 (b + 3)
b² (b + 3) + 1 (b + 3)
Then, take (b + 3) common from the above expression.
(b + 3) (b² + 1)

The final answer is (b + 3) (b² + 1).

(iv) 1 + s + s²t + s³t

Solution:
Given expression is 1 + s + s²t + s³t.
Group the first two terms and last two terms.
The first two terms are 1 + s and the second two terms are s²t + s³t.
Take 1 common from the first two terms.
1 (1 + s)
Take s²t common from the second two terms.
s²t (1 + s)
1 (1 + s) + s²t (1 + s)
Then, take (1 + s) common from the above expression.
(1 + s) (1 + s²t)

The final answer is (1 + s) (1 + s²t).

(v) m – 1 – (m – 1)² + bm – b

Solution:
Given expression is m – 1 – (m – 1)² + bm – b.
Group the first two terms, middle and last two terms.
The first two terms are m – 1, the middle term is – (m – 1)², and the last two terms is bm – b.
Take 1 common from the first two terms.
1 (m – 1)
Take 1 common from the middle term.
– (m – 1)²
Take b common from the last two terms.
b (m – 1)
1 (m – 1) – (m – 1)²+ b (m – 1)
Then, take (m – 1) common from the above expression.
(m – 1) (1 – m + 1 + b)
(m – 1) (2 + b – m)

The final answer is (m – 1) (2 + b – m).

Factorize by grouping the terms consists of two or more product terms as resultant value. To divide the expression into product terms, we need to identify the greatest common factor which divides all the remaining terms in the expression. Step by step process on how to do Factorization by Grouping the Terms are clearly given in this article.

Procedure to Factorize by Grouping the Terms

Refer to the below-mentioned step by step process and learn the Factoring by Grouping Terms. They are along the lines

1. Note down the given expression.
2. Group the first two terms and last two terms.
3. Factor out the greatest common factor from each group.
4. Finally, you will get two or more product terms as result.

Solved Examples on Factorization by Grouping Terms

1. Factorize grouping the following expressions?

(i) 18x³y³ – 27x²y³ + 36x³y²

Solution:
The given expression is 18x³y³ – 27x²y³ + 36x³y²
Factor out the greatest common factor from the above expression.
That is, 9x²y²(2xy – 3y + 4x).

Therefore, the final solution for the expression 18x³y³ – 27x²y³ + 36x³y² is 9x²y²(2xy – 3y + 4x).

(ii) 12a²b³ – 21a³b²

Solution:
The given expression is 12a²b³ – 21a³b²
Factor out the greatest common factor from the above expression.
3a²b²(4b – 7a).

Finally, the solution for the expression 12a²b³ – 21a³b² is 3a²b²(4b – 7a).

(iii) a³ – a² + a – 1.

Solution:
The given expression is a³ – a² + a – 1
Group the first two terms and last two terms.
Here, first two terms are a³ – a² and the last two terms are a – 1.
So, (a³ – a²) + (a – 1).
Now, factor out the greatest common factor from each group.
That is, a²(a – 1) + (a – 1).
(a – 1) (a² + 1).

Therefore, solution for the expression a³ – a² + a – 1 is (a – 1) (a² + 1).

(iv) prs + qurs – pt – qut

Solution:
The given expression is prs + qurs – pt – qut.
Group the first two terms and last two terms.
Here, the first two terms are prs + qurs and last two terms are – pt – qut.
Then, (prs + qurs) – (pt + qut).
Now, factor out the greatest common factor from the above two groups.
That is, rs(p + qu) – t(p + qu).
(p + qu) (rs – t).

Therefore, the solution for the expression prs + qurs – pt – qut is (p + qu) (rs – t).

(v) a² – 3a – ab + 3b
Solution:

The given expression is
a² – 3a – ab + 3b.
Group the first two terms and last two terms.
Here, the first two terms are a²-3a and the last two terms are ab + 3b.
Then, (a² – 3a) – (ab – 3b).
Now, factor out the greatest common factor from the above two groups.
That is a(a – 3) – b(a – 3).
(a – 3) (a – b).

Therefore, solution for the expression is a² – 3a – ab + 3b is (a – 3) (a – b).

2. How to factorize by grouping the following expressions?
(i) 2q^4 – q³ + 4q – 2
Solution:
The given expression is
2q^4 – q³ + 4q – 2.
Group the first two terms and last two terms.
Here, the first two terms are 2q^4 – q³ and the last two terms are 4q – 2.
Then, (2q^4 – q^3) + (4q-2).
Now, factor out the greatest common factor from the above two groups.
That is, q^3(2q – 1) +2(2q – 1).
(2q – 1) (q^3 + 2).

Therefore, solution for the expression is 2q4 – q3 + 4q – 2 is (2q-1) (q^3+2).

(ii) ac + bc – ad – bd.

Solution:
The given expression is ac + bc – ad – bd.
Group the first two terms and last two terms.
Here, the first two terms are ac+bc and the last two terms are – ad-bd.
Then, (ac + bc) – (ad + bd).
Now, factor out the greatest common factor from the above two groups.
That is, c(a+b) -d(a+b).
(a+b) (c-d).

Therefore, solution for the expression is ac + bc – ad – bd is (a+b) (c-d).

(iii) pa – pb – qa –qb.

Solution:
The given expression is pa – pb – qa–qb.
Group the first two terms and last two terms.
Here, the first two terms are pa – pb, and the last two terms are qa – qb.
Then, (pa – pb) – (qa -qb).
Now, factor out the greatest common factor from the above two groups.
That is, p(a – b) – q(a – b).
(p – q) (a – b).

Therefore, the solution for the expression is pa – pb – qa – qb is (p – q) (a – b).

3. How to factorize by grouping the algebraic expressions?
(i)x^2z^2 + xzw + xyz + yw

Solution:
The given expression is
x^2z^2 + xzw + xyz + yw.
Group the first two terms and last two terms.
Here, the first two terms are x^2z^2 + xzw and the last two terms are xyz + yw.
Then, (x^2z^2 + xzw) + (xyz + yw).
Now, factor out the greatest common factor from the above two groups.
That is, xz(xz + w) + y(xz + w).
(xz + y) (xz + w).

Therefore, solution for the expression is x^2z^2 + xzw + xyz + yw is (xz + y) (xz + w).

(ii) 5x + xy + 5y + y^2

Solution:
The given expression is 5x + xy + 5y + y^2
Group the first two terms and last two terms.
Here, the first two terms are 5x + xy, and the last two terms are 5y + y^2.
Then, (5x + xy) + (5y + y^2).
Now, factor out the greatest common factor from the above two groups.
That is, x(5 + y) + y(5 + y).
(5 + y) (x + y).

Therefore, solution for the expression is 5x + xy + 5y + y^2 is (5 + y) (x + y).

(iii) xy – yz – xz + z^2

Solution:
The given expression is xy – yz – xz + z^2
Group the first two terms and last two terms.
Here, the first two terms are xy – yz and the last two terms are – xz + z^2.
Then, (xy – yz) – (xz – z^2).
Now, factor out the greatest common factor from the above two groups.
That is, y(x – z) – z(x – z).
(x – z) (y – z).

Therefore, solution for the expression xy – yz – xz + z^2 is (x – z) (y – z).

4. Factorize the expressions
(i) a^4 + a^3 + 2a + 2

Solution:
The given expression is a^4 + a^3 + 2a + 2.
Group the first two terms and last two terms.
Here, the first two terms are a^4 + a^3 and the last two terms are 2a + 2.
Then, (a^4 + a^3) + (2a + 2).
Now, factor out the greatest common factor from the above two groups.
That is, a^3(a + 1) + 2(a + 1).
(a + 1) (a^3 + 2).

Therefore, solution for the expression a^4 + a^3 + 2a + 2 is (a + 1) (a^3 + 2).

(ii) a²b² + d²b² – cd² – ca²

Solution:
The given expression is a²b² + d²b² – cd² – ca²
Group the first two terms and last two terms.
Here, the first two terms are a²b² + d²b² and the last two terms are – cd² – ca².
Then, (a^2b^2 + d^2b^2) – (cd^2 + ca^2).
Now, factor out the greatest common factor from the above two groups.
That is, b^2(a^2 + d^2) – c(d^2 + a^2).
(a^2 + d^2) (b^2 – c).

Therefore, solution for the expression a2b2 + d2b2 – cd2 – ca2 is(a^2 + d^2) (b^2 – c).

5. Factorize by grouping the terms
(x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y

Solution:
Thegiven expression is (x^2 + 3x)^2 – 2(x^2 + 3x) – y(x^2 + 3x) + 2y.
Group the first two terms and last two terms.
here, first two terms are (x^2 + 3x)^2 – 2(x^2 + 3x) and the last two terms are –y(x^2 + 3x) + 2y.
Then, [(x^2 + 3x)^2 – 2(x^2 + 3x)] –[y(x^2 + 3x) – 2y].
That is, (x^2 + 3x)(x^2 + 3x – 2) – y(x^2 + 3x – 2).
(x^2 + 3x – 2)(x^2 + 3x – y).

Therefore, solution for the expression (x2 + 3x)2 – 2(x2 + 3x) – y(x2 + 3x) + 2y is (x^2 + 3x – 2)(x^2 + 3x – y).

Are you looking for the different problems on Factorization by Grouping? Then, you are in the right place. We have given all types of factorization problems on our website. Students can learn to Factorize by Grouping the Terms in these articles. While solving Factorization by grouping problems, students need to group the terms with common factors before factoring.

How to Factor by Grouping?

Have a look at the Factoring by Grouping Steps and learn how to solve related problems easily. Follow the guidelines provided and perform factorization by grouping method. They are as under

• Note down the given expression, group the first two terms and last two terms.
• Find out the greatest common factor(GCF) from the first term and second term.
• Now, find the common factor from the above two groups.
• Finally factor out the terms in terms of product.

Factorization by Grouping Examples

1. Factor grouping the expressions?
1 + x + xy + x²y.

Solution: Given Expression is 1 + x + xy + x²y.
Group the first two terms and last two terms.
First two terms are 1 + x and the last two terms are xy + x²y.
(1+ x) + (xy + x²y).
Find out the common factor from the above two groups.
(1+x) + xy(1+x).
Factor out the terms in terms of product.
(1+x) (1+xy).

By factor grouping the expression 1 + x + xy + x²y., we will get (1+x) (1+xy).

2. How to factor by grouping the following algebraic expressions?

(i) x² – xy + xz – zy.

Solution:
Given Expression is x² – xy + xz – zy
Group the first two terms and last two terms.
First two terms are x² – xy and the last two terms are xz – zy.
(x² – xy) + (xz – zy)
Find out the common factor from the above two groups.
x(x – y) + z(x – y)
Factor out the terms in terms of product.
(x + z) (x – y)

Factor by grouping the expression x² – xy + xz – zy, we will get the result as (x+z) (x-y).

(ii) x² + 3x + xy + 3y.

Solution:
Given Expression is x² + 3x + xy + 3y.
Group the first two terms and last two terms.
First two terms are x² + 3x and the last two terms are xy + 3y.
(x² + 3x) + (xy + 3y)
Find out the common factor from the above two groups.
x(x + 3) + y(x + 3)
Factor out the terms in terms of product.
(x+y) (x+3).

Factor by grouping the expression x² + 3x + xy + 3y, we will get solution as (x+y) (x+3).

3. Factorize the algebraic expressions.

(i) 2a + ba + 2b + b²

Solution:
Given Expression is 2a + ba + 2b + b²
Group the first two terms and last two terms.
First two terms are 2a + ba and the last two terms are 2b + b².
(2a + ba)+ (2b + b²).
Find out the common factor from the above two groups.
a(2 + b) + b(2 + b).
Factor out the terms in terms of product.
(a + b) (2 + b).

By factorizing the expression 2a + ba + 2b + b², we will get (a + b) (2 + b).

(ii) b² – yb + 5b– 5y.

Solution:
Given Expression is b² – yb + 5b– 5y.
Group the first two terms and last two terms.
First, two terms are b² – yb and the last two terms are 5b – 5y.
(b² – yb)+ (5b – 5y).
Find out the common factor from the above two groups.
b(b – y) + 5(b – y).
Factor out the terms in terms of product.
(5 + b) (b – y).

By factorizing the expression b² – yb + 5b– 5y, we will get (5 + b) (b – y).

(iii) pq – rq – ps + rs.

Solution:
Given Expression is pq – rq – ps + rs.
Group the first two terms and last two terms.
First two terms are pq – rq and the last two terms are – ps + rs.
(pq – rq) – (ps – rs).
Find out the common factor from the above two groups.
q(p – r) -s(p – r).
Factor out the terms in terms of product.
(p – r) (q – s).

By factorizing the expression pq – rq – ps + rs, we will get (p – r) (q – s).

(iv) ab – 2ac – db + 2dc.

Solution:
The given expression is ab – 2ac – db + 2dc.
Group the first two terms and last two terms.
Fist two terms are ab – 2ac and the last two terms are – db + 2dc.
(ab – 2ac) – (db – 2dc).
Find out the common factor from the above two groups.
a(b – 2c) -d(b – 2c).
Factor out the terms in terms of product.
(b – 2c) (a – d).

By factorizing the expression ab – 2ac – db + 2dc., we will get (b – 2c) (a – d).

(v) pq² – 3rqs – pqs + 3rs²

Solution:
The given expression is pq² – 3rqs – pqs + 3rs²
Group the first two terms and last two terms.
First two terms are pq^2 – 3rqs and the last two terms are –pqs + 3rs².
(pq² – 3rqs) – (pqs – 3rs²).
Find out the common factor from the above two groups.
q(pq – 3rs) -s(pq – 3rs).
Factor out the terms in terms of product.
(pq – 3rs) (q – s).

By factorizing the expression pq² – 3rqs – pqs + 3rs², we will get (pq – 3rs) (q – s).

4. Factor each of the following expressions by grouping

(i) a² – 3a – ab + 3b.

Solution:
Given expression is a² – 3a – ab + 3b
Group the first two terms and last two terms.
First two terms are a² – 3a and the last two terms are – ab + 3b.
(a² – 3a) – (ab – 3b).
Find out the common factor from the above two groups.
a(a – 3) – b(a – 3).
Factor out the terms in terms of product.
(a – 3) (a – b).

By factorizing the expression a² – 3a – ab + 3b, we will get (a – 3) (a – b).

(ii) pq² + rq² + 2p + 2r.

Solution:
Given expression is pq² + rq² + 2p + 2r
Group the first two terms and last two terms.
First two terms are pq² + rq² and the last two terms are 2p + 2r.
(pq² + rq²) + (2p + 2r).
Find out the common factor from the above two groups.
q²(p +r) + 2(p+r).
Factor out the terms in terms of product.
(q²+ 2) (p + r).

By factorizing the expression pq² + rq² + 2p + 2r, we will get (q²+ 2) (p + r).

(iii) 2pq² + 3pqr – 2sqr – 3sr²

Solution:
Given expression is 2pq² + 3pqr – 2sqr – 3sr²
Group the first two terms and last two terms.
First two terms are 2pq² + 3pqr and the last two terms are – 2sqr – 3sr².
(2pq² + 3pqr) – ( 2sqr + 3sr²).
Find out the common factor from the above two groups.
pq(2q + 3r) – sr(2q + 3r).
Factor out the terms in terms of product.
(2q + 3r) (pq – sr).

By factorizing the expression 2pq² + 3pqr – 2sqr – 3sr², we will get (2q + 3r) (pq – sr).

(iv) par² + qars – pnrs – qns²

Solution:
The given expression is par² + qars – pnrs – qns²
Group the first two terms and last two terms.
First two terms are par² + qars and the last two terms are – pnrs – qns².
(par² + qars) – (pnrs – qns²).
Find out the common factor from the above two groups.
ar(pr + qs) – ns(pr + qs).
Factor out the terms in terms of product.
(pr + qs) (ar – ns).

By factorizing the expression par² + qars – pnrs – qns², we will get (pr + qs) (ar – ns).

5. Factorize
(i) (a + b) (2a + 5) – (a + b) (a + 3).

Solution:
The given expression is (a + b) (2a + 5) – (a + b) (a + 3).
Find out the common factor from the above expression.
(a + b) [(2a + 5) – (a + 3)].
Expand the terms.
(a + b)(2a + 5 – a – 3).
Simplify the second term.
(a + b) (a + 2).

By factorizing the expression (a + b) (2a + 5) – (a + b) (a + 3), we will get (a + b)(a + 2).

(ii) 6xy – y² + 12xz – 2yz.

Solution:
The given expression is 6xy – y² + 12xz – 2yz.
Group the first two terms and last two terms.
First two terms are 6xy – y² and the last two terms are 12xz – 2yz.
(6xy – y²) + (12xz – 2yz).
Find out the common factor from the above two groups.
y(6x – y)+ 2z(6x – y).
Factor out the terms in terms of product.
(6x – y) (y + 2z).

By factorizing the expression 6xy – y² + 12xz – 2yz, we will get (6x – y) (y + 2z).

Students can find the factors for an algebraic expression when one of its factors is binomial. Find given algebraic expression Factorization when Binomial is Common Factor. All related problems are included in this article along with a clear explanation. Therefore, students can practice every problem available in this article and improve their knowledge. The process of solving the factorization problem is very simple if you follow the procedure we explained below. Go through the complete article and learn different methods to solve factorization problems.

Factorization of Algebraic Expressions when Binomial is Common

Follow the simple and easy guidelines on Factorization of Algebraic Expressions When Binomial is Common. They are as follows

Step 1: In the first step, find the common binomial factor.
Step 2: Note down the given expression as the product of this binomial and the quotient obtained on dividing the given expression by this binomial.

Solved Examples on Factorization When Binomial is a Common

1. Factorize the algebraic expressions.

(i) 7b(3x – 4y) + 3a(3x – 4y)

Solution:
Given expression is 7b(3x – 4y) + 3a(3x – 4y)
In the given expression, the binomial factor is (3x – 4y) as it is common in both terms.
Take the (3x – 4y) common and multiply it with the remained terms.
(3x – 4y) (7b + 3a)

The final answer is (3x – 4y) (7b + 3a)

(ii) 12(9a + 6b)² – 4(9a + 6b)

Solution:
Given expression is 12(9a + 6b)² – 4(9a + 6b)
12 (9a + 6b) (9a + 6b) -4(9a + 6b)
In the given expression, the binomial factor is (9a + 6b) as it is common in both terms.
Take the (9a + 6b) common and multiply it with the remained terms.
(9a + 6b)(12(9a + 6b) – 4)
(9a + 6b)(108a + 72b – 48)
The final answer is (9a + 6b)(108a + 72b – 48)

2. Factorize the expression 10r(m – 2n) – 8m + 16n

Solution:
Given expression is 10r(m – 2n) – 8m + 16n
Lets take -8m  + 16n from the above equation.
Take -8 common from -8m + 16n
-8(m – 2n)
Place -8(m – 2n) in 10r(2m – 4n) – 8m + 16n equation.
10r(m – 2n) -8(m – 2n)
In the above expression, the binomial factor is (m – 2n) as it is common in both terms.
Take the (m – 2n) common and multiply it with the remained terms.
(m – 2n)(10r – 8)

The final answer is (m – 2n)(10r – 8)

3. Factorize (a – 4b)^2 – 7a + 28b

Solution:
(a – 4b)(a – 4b) – 7a + 28b
Given expression is (a – 4b)(a – 4b) – 7a + 28b
Lets take – 7a + 28b from the above equation.
Take -7 common from – 7a + 28b
-7(a – 4b)
Place -7(a – 4b) in (a – 4b)(a – 4b) – 7a + 28b equation.
(a – 4b)(a – 4b) – 7(a – 4b)
In the above expression, the binomial factor is (a – 4b) as it is common in both terms.
Take the (a – 4b) common and multiply it with the remained terms.
(a – 4b)(a – 4b – 7)

The final answer is (a – 4b)(a – 4b – 7)

All solved examples of Factorization problems when Binomial is a common factor are included in this article. Check how to find the binomial factor and practice all problems given here. Compare the terms in the given expression and factor out the greatest common factor term from the expression. Based on the greatest common factor, we will get the solution for the expression that is in terms of the product of two or more terms.

How to Factorize taking out the Binomial Common Factor?

Go through the below step by step procedure and do Factorization when a Binomial is a Common Factor. They are along the lines

1. Note down the given expression
2. Note down the first term and second term from the expression
3. Compare the two terms and observe the greatest common factor
4. Now, factor out the greatest common factor from the expression
5. Finally, we will get the result in the form of the product of two or more terms.

Solved Binomial Factor Examples

1. Factorize the expression (6x + 1)² – 5(6x + 1)?

Solution:
The given expression is (6x + 1)² – 5(6x + 1)
Here, the first term is (6x + 1)² and second term is – 5(6x + 1)
By comparing the above two terms, we can observe the greatest common factor and that is (6x + 1)
Now, factor out the greatest common factor from the expression
That is, (6x + 1) [(6x + 1) – 5]
(6x + 1)(6x – 4)

Therefore, the resultant value for the expression (6x + 1)² – 5(6x + 1) is (6x + 1)(6x – 4)

2. Factorize the algebraic expression 4x(y – z) + 3(y – z)?

Solution:
The given expression is 4x(y – z) + 3(y – z)
Here, the first term is 4x(y – z) and the second term is 3(y – z)
By comparing the above two terms, we can observe the greatest common factor and that is (y – z)
Now, factor out the greatest common factor from the expression
That is, (y – z)(4x + 3)

Therefore, the resultant value for the expression 4x(y – z) + 3(y – z) is (y – z)(4x + 3)

3. Factorize the expression (3x – 4y) (p – q) + (3x – 2y) (p – q)?

Solution:
The given expression is (3x – 4y) (p – q) + (3x – 2y) (p – q)
Here, the first term is (3x – 4y) (p – q) and the second term is (3x – 2y) (p – q)
By comparing the above two terms, we can observe the greatest common factor and that is (p – q)
Now, factor out the greatest common factor from the expression
That is, (p – q)[(3x – 4y) + (3x – 2y)]
= (p – q)(6x – 6y)

Therefore, the resultant value for the expression (3x – 4y) (p – q) + (3x – 2y) (p – q) is (p – q)(6x – 6y)

4. Factorize the expression 6(2x + y)² +8(2x + y)?

Solution:
The given expression is 6(2x + y)² +8(2x + y)
Here, the first term is 6(2x + y)² and the second term is 8(2x + y)
By comparing the above two terms, we can observe the greatest common factor and that is (2x + y)
Now, factor out the greatest common factor from the expression
That is, (2x + y)[6(2x + y) + 8]
(2x + y)(12x + 6y + 8)

Therefore, the resultant value for the expression 6(2x + y)² +8(2x + y) is (2x + y)(12x + 6y + 8)

While finding the factorization when monomial is a common factor, an algebraic expression can be expressed as the sum or the difference of monomials. Look at the different problems that have an algebraic expression with Monomial as a common factor. Also, find all Factorization concepts on our website with a detailed explanation.

How to Find Factorization of an Expression?

Below is the step by step guidelines on how to factorize an expression. They are along the lines

Step 1: Note Down the algebraic expression.
Step 2: Find the different terms of given algebraic expressions.
Step 3: Find the HCF of all the separate terms of the given algebraic expression.
Step 4: Express each term of the algebraic expression as the product of H.C.F and divide each term of the given expression by the HCF.
Step 5: Now use the distributive property of multiplication over addition or subtraction to express the algebraic expression as the product of H.C.F and the quotient of the expression divided by the H.C.F.
Step 6: In the final step, keep the H.C.F. outside the bracket and the quotients obtained within the bracket.

Solved Examples of Factorization when Monomial is Common

1. Factorize each of the following

(i) 4b + 20

Solution:
The given expression is 4b + 20
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 4 and 20 is 4.
HCF of literal coefficients:
The lowest power of b is 0.
Therefore, the HCF of literal coefficients is nothing.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 4b + 20 is 4.
Multiply and divide each term of the given expression 4b + 20 with 4.
4(4b/4 + 20/4) = 4(b + 5)

The final answer is 4(b + 5)

(ii) 3m² + 2m

Solution:
The given expression is 3m² + 2m
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 3 and 2 is nothing.
HCF of literal coefficients:
The lowest power of m is 1.
Therefore, the HCF of literal coefficients is m.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 3m² + 2m is m.
Multiply and divide each term of the given expression 3m² + 2m with m.
m(3m²/m + 2m/m) = m(3m + 2)

The final answer is m(3m + 2)

(iii) 6a²b – 3ab²

Solution:
The given expression is 6a²b – 3ab²
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 6 and 3 is 3.
HCF of literal coefficients:
The lowest power of a is 1.
The lowest power of b is 1.
Therefore, the HCF of literal coefficients is ab.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 6a²b – 3ab² is 3ab.
Multiply and divide each term of the given expression 3m² + 2m with m.
3ab(6a²b/3ab – 3ab²/3ab) = 3ab(2a – b)

The final answer is 3ab(2a – b)

(iv) 9cd – 6bc

Solution:
The given expression is 9cd – 6bc
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 9 and 6 is 3.
HCF of literal coefficients:
The lowest power of b is 0.
The lowest power of c is 1.
The lowest power of d is 0.
Therefore, the HCF of literal coefficients is c.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 9cd – 6bc is 3c.
Multiply and divide each term of the given expression 9cd – 6bc with 3c.
3c(9cd/3c – 6bc/3c) = 3c(3d – 2b)

The final answer is 3c(3d – 2b)

2. Factorize 18x²y²z + 12xyz.

Solution:
The given expression is 18x²y²z + 12xyz
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 18 and 12 is 3.
HCF of literal coefficients:
The lowest power of x is 1.
The lowest power of y is 1.
The lowest power of z is 1.
Therefore, the HCF of literal coefficients is xyz.
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 18x²y²z + 12xyz is 3xyz.
Multiply and divide each term of the given expression 18x²y²z + 12xyz with 3xyz.
3xyz(18x²y²z/3xyz + 12xyz/3xyz) = 3xyz(6xy + 4)

The final answer is 3xyz(6xy + 4)

3. Factorize the expression: 8m³ – 32m²n

Solution:
The given expression is 8m³ – 32m²n
Firstly, find the HCF of given terms.
HCF of their numerical coefficients 8 and 32 is 8.
HCF of literal coefficients:
The lowest power of m is 2.
The lowest power of n is 0.
Therefore, the HCF of literal coefficients is m².
HCF of two or more monomials = (HCF of their numerical coefficients) × (HCF of their literal coefficients)
HCF of 8m³ – 32m²n is 8m².
Multiply and divide each term of the given expression 8m³ – 32m²n with 8m².
8m²(8m³/8m² – 32m²n/8m²) = 8m²(m – 4n)

The final answer is 8m²(m – 4n)