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RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management are part of RBSE Solutions for Class 10 Science. Here we have given Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Science
Chapter	Chapter 13
Chapter Name	Waste and its Management
Number of Questions Solved	36
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management

Textbook Questions Solved

I. Multiple Choice Questions

Question 1:
Which method is ideal for the management of bio-medical waste?
(a) Landfill
(b) Incineration
(c) Recycling
(d) Disposal in water bodies
Answer:
(b) Incineration

Question 2:
Recycling is ideal for which type of waste?
(a) Metallic waste
(b) Medical waste
(c) Household waste
(d) Agricultural waste
Answer:
(a) Metallic waste

Question 3:
Which of the following is a greenhouse gas?
(a) Hydrogen
(b) Carbon monoxide
(c) Carbon dioxide
(d) Sulphur dioxide
Answer:
(c) Carbon dioxide

Question 4:
What is the average amount of waste per person in big cities of India?
(a) 0.1 kg
(b) 0.1 – 0.2 kg
(c) 0.2 – 0.4 kg
(d) 0.4 – 0.6 kg
Answer:
(d) 0.4 – 0.6 kg

Question 5:
Which of the following can be turned into organic manure?
(a) Household waste
(b) Agricultural waste
(c) Both ‘a’ and ‘b’
(d) None of these
Answer:
(c) Both ‘a’ and ‘b’

Waste and its Management Very Short Answer Type Questions

Question 6:
How is biogas made?
Answer:
Biogas is made bv decomposition of farm waste and some domestic waste.

Question 7:
What is waste?
Answer:
Useless product at the end of any process is called waste.

Question 8:
Write the names of greenhouse gases.
Answer:
Carbon dioxide and methane

Question 9:
What is vermicompost?
Answer:
Compost which is prepared with help from earthworms is called vermicompost.

Question 10:
Which diseases are spread by stagnant water in drainage?
Answer:
Malaria, dengue

Waste and its Management Short Answer Type Questions

Question 11:
Explain waste management.
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management. Different types of waste need different kinds of management practices. Waste management is highly important for a comfortable and disease-free life in cities and villages.

Question 12:
What do you understand by solid waste?
Answer:
Waste which is in solid form is called solid waste. Management of solid waste requires a different approach. One needs to physically transport solid waste to the dumping site. A huge amount of manpower and capital is required for management of solid waste.

Question 13:
Differentiate between biodegradable and non-biodegradable waste.
Answer:

Biodegradable Waste	Non-biodegradable Waste
Can be decomposed by biological agents.	Cannot be decomposed by biological agents.
They are generally environment- friendly.	They generally harm the environment.
Examples: fruit peels, stale food, paper, etc.	Examples: cans, bottles, plastic, etc.

Question 14:
What do you understand by landfill?
Answer:
Waste is dumped at landfill sites. When amount of waste attains a significant figure, it is covered with soil. After that, waste is allowed to decompose. Pipeline system is installed in landfill to channelize and utilize the methane gas which is produced during decomposition. This gas is utilized for power generation at many places. A landfill site is usually developed outside the city limits.

Question 15:
What is the meaning of recycling?
Answer:
Recyling is the process of treatment of discarded material to make it available fo resue. Certain items from waste are segregated for recycling. Items like plastic, aluminium cans, metals, etc. are sent to recycling unit so that new items can be made from them. This not only helps in reducing the amount of waste but also in conservation of resources.

Question 16:
What is the use of incineration?
Answer:
In this method, waste is burnt under controlled conditions so that production of polluting gases can be minimized. This method is ideal for places where land is in short supply. This method is also ideal for disposing off the medical waste.

Waste and its Management Long Answer Type Questions

Question 17:
Explain different types of waste.
Answer:
Waste can be solid, liquid or gas but on the basis of ability to undergo decomposition, waste is classified into two types.

• Biodegradable Waste: Waste which can be decomposed by biological factors is called biodegradable waste, e.g. fruit peels, stale food, paper, etc. Biodegradable is generally environment-friendly. Biogas is produced on decomposition of biodegradable waste. This biogas can be utilized for many purposes like as domestic fuel, for power generation, etc. Non-
• biodegradable Waste: Waste which cannot be decomposed by biological factors is called non-biodegradable waste, e.g. aluminium can, polythene bag, etc. Non-biodegradable waste is generally harmful for environment. However, many non-biodegradable waste can be recycled to make useful products.

Question 18:
Write an essay on waste management.
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management. Different types of waste need different kinds of management practices. While gaseous waste goes up in air on its own, we need to make proper drainage system for handling liquid waste. For handling solid waste, we need a proper system of transportation and people. Even a small village produces a large amount of waste per day. Daily production of garbage could be in thousands of tons in mega cities.

Methods of Waste Management:

1. Landfill: Waste is dumped at landfill site. When amount of waste attains a significant figure, it is covered with soil. After that, waste is allowed to decompose. Pipeline system is installed in landfill to channelize and utilize the methane gas which is produced during decomposition. This gas is utilized for power generation at many places. A landfill site is usually developed outside the city limits.
2. Incineration: In this method, waste is burnt under controlled conditions so that production of polluting gases can be minimized. This method is ideal for places where land is in short supply. This method is also ideal for disposing off the medical waste.
3. Recycling: Certain items from waste are segregated for recycling. Items like plastic, aluminium cans, metals, etc. are sent to recycling unit so that new items can be made from them. This not only helps in reducing the amount of waste but also in conservation of resources.
4. Chemical Methods: Many wastes can be managed by using chemical methods.

Question 19:
Write an essay on sources of waste.
Answer:
Following are different sources of waste:

• Household Source: Paper, cardboard, fruit peels, vegetable peels, soapy water, rags, etc.
• Municipality: Municipality looks after management of solid and liquid waste from households, offices and factories. Municipal waste also includes human excreta.
• Industry and Mining: Many harmful wastes are produced by industries, e.g. harmful chemicals, grease, etc. Mining activity also produces wastes like dust.
• Agriculture: Hay, dry leaves, husk, cowdung, etc. are examples of farm waste.
• Medical Waste: Hospitals produce many wastes like syringe, cotton, bandage, catheters, strips, etc. Some of the hospital waste may contain harmful germs.

Question 20:
Make a list of various wastes from your surroundings and classify them.
Answer:

Biodegradable Waste	Non-biodegradable Waste
Fruit peels, vegetable peels, dry leaves, stale food, paper, empty cartons, old clothes, jute bags, etc.	Empty cans, plastic bottles, bubble wrap, discarded toys, glass items, etc.

Question 21:
What will you do for waste management in your village or neighbourhood?
Answer:
I will take following steps for waste management in my neighbourhood:

• I will urge people to use garbage bins for throwing garbage. I will organize awareness campaign for this.
• I will request the municipal authorities for regular maintenance of drainage system.
• I will take a team of volunteers for ‘Swachhata Abhiyan’ on a weekly basis.
• I will educate people on benefits of segregating waste so that some items can be recycles.

Waste and its Management Additional Questions Solved

I. Multiple Choice Questions

Question 1:
Which of the following poses the risk of more number of diseases than any other?
(a) Domestic waste
(b) Hospital waste
(c) Farm waste
(d) Industrial waste
Answer:
(b) Hospital waste

Question 2:
Which of the following is biodegradable waste?
(a) Polythene bag
(b) Aluminium can
(c) Fruit peel
(d) Plastic furniture
Answer:
(c) Fruit peel

Question 3:
Which gas is produced on decomposition of farm waste?
(a) Methane
(b) Carbon dioxide
(c) Hydrogen
(d) Nitrogen oxide
Answer:
(a) Methane

Question 4:
Which epidemic can spread due to stagnant drainage?
(a) Jaundice
(b) Hepatitis
(c) AIDS
(d) Malaria
Answer:
(d) Malaria

Question 5:
Drinking contaminated water can cause which disease?
(a) Jaundice
(b) Cholera
(c) Diarrohea
(d) All of these
Answer:
(d) All of these

Waste and its Management Very Short Answer Type Questions

Question 1:
List some wastes which come from household.
Answer:
Paper, cardboard, fruit peels, vegetable peels, soapy water, rags, etc.

Question 2:
List some wastes from farming.
Answer:
Hay, dry leaves, husk, cowdung, etc.

Question 3:
List some pastes from hospital.
Answer:
Syringe, cotton, bandage, catheters, strips, etc.

Question 4:
What is waste management?
Answer:
The practice of transportation and dumping of waste, recycling of useful material from waste, etc. are integral part of waste management.

Question 5:
Which civic body looks after waste management in towns?
Answer:
Municipality

Waste and its Management Short Answer Type Questions

Question 1:
What is the effect of sewage on water bodies?
Answer:
Sewage from households pollutes water bodies like rivers and ponds. It is harmful for aquatic animals and plants. It also pollutes groundwater.

Question 2:
What is the need for waste management?
Answer:
Waste can keep on accumulating if not managed in time. This will make ugly site around us. Moreover, heaps of garbage can trigger many epidemics. Hence, there is a need for waste management.

Question 3:
Waste management is not efficient in small towns. Why?
Answer:
Municipalities in small towns suffer from shortage of funds. So, the contractors do not get timely payment. Even sanitation workers do not get regular salary. Due to these reasons, waste management is not efficient in small towns.

Question 4:
What were the suggestions of Shivraman Committee?
Answer:
Following are the suggestions of Shivaraman Committee:

• Installation of large garbage bins.
• Proper arrangement of management of human excreta.
• Proper system for picking up garbage in towns.
• Using incinerators for disposing off the waste.

Waste and its Management Long Answer Type Questions

Question 1:
What are the harms of waste?
Answer:
Following are the harms of waste:

• Accumulation of waste around us is not good from the point of aesthetics.
• Decomposition of waste produces methane and other harmful gases. Methane causes global warming.
• A heap of waste is the breeding ground for harmful germs, houseflies and mosquitoes. These agents can cause many epidemics.
• Hospital waste can cause serious infection to a person. HIV and Hepatitis B can spread from coming in contact with infected needle.
• Plastic waste turns the soil infertile. It chokes the drainage. Stray animals often swallow plastic and die due to choking. Prolonged exposure to plastic increases thalatus in blood which hampers the development of foetus.
• Sewage from households pollutes water bodies like rivers and ponds. It is harmful for aquatic animals and plants. It also pollutes groundwater.

We hope the given RBSE Solutions for Class 10 Science Chapter 13 Waste and its Management will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Science Solutions Chapter 13 Waste and its Management, drop a comment below and we will get back to you at the earliest.

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions.

Board	RBSE
Class	Class 9
Subject	Maths
Chapter	Chapter 3
Chapter Name	Polynomial
Exercise	Additional Questions
Number of Questions Solved	36
Category	RBSE Solutions

RBSE Solutions for Class 9 Maths Chapter 3 Polynomial Additional Questions

Multiple Choice Questions

Question 1.
A polynomial (RBSESolutions.com) of degree n in x has atmost:
(A) n terms
(B) (n – 1) terms
(C) (n + 1) terms
(D) \(\frac { n }{ 2 }\) terms
Solution
C

Question 2.
The zeroes of the (RBSESolutions.com) polynomial p(x) = x(x² – 1) are:
(A) 0, 1
(B) 0, – 1
(C) 0, – 1, 1
(D) ± 1
Solution
C

Question 3.
If

is equal to

Solution
B

Question 4.
√7 is a polynomial (RBSESolutions.com) of degree:
(A) 0
(B) 7
(C) \(\frac { 1 }{ 2 }\)
(D) 2
Solution
A

Question 5.
The coefficient of x in (x – 3)(x – 4) is:
(A) 7
(B) 1
(C) – 7
(D) 12
Solution
C

Question 6.
If p(x) = x² -3√2x + 1, then p(3√2) is equal to:
(A) 3√2
(B) 3√2 – 1
(C) 6√2 – 1
(D) 1
Solution
D

Question 7.
Which of the following is (RBSESolutions.com) a polynomial in one variable:

Solution
A

Question 8.
The degree of the zero polynomial is:
(A) 0
(B) 1
(C) any real number
(D) not exist
Solution
D

Question 9.
If x⁹¹ + 91 is divided by x + 1, then (RBSESolutions.com) the remainder is:
(A) 0
(B) 90
(C) 92
(D) None of these
Solution
B

Question 10.
Zero of the polynomial p(x) = √3x + 3 is

Solution
A

Very Short Answer Type Questions

Question 1.
Show that \(\frac { 1 }{ 2 }\) is zero (RBSESolutions.com) of the polynomial 2x² + 7x – 4.
Solution.
Let p(x) = 2x² + 7x – 4

Question 2.
Find the (RBSESolutions.com) remainder when
x⁴ + x³ – 2x² + x + 1 is divided by x – 1.
Solution.
Let p(x) = x⁴ + x³ – 2x² + x + 1
zero of x – 1 is 1
So, P(1) = (1)⁴ + (1)³ – 2(1)² + (1) + 1 = 1 + 1 – 2 + 2 = 2
Hence, 2 is the remainder, when x⁴ + x³ – 2x² + x + 1 is divided by x – 1.

Question 3.
If x – 1 is a factor of p(x) = 2x² + kx + √2 then find (RBSESolutions.com) the value of k.
Solution.
x – 1 is a factor of p(x) = 2x² + kx + √2
p(1) = 0 (by factor theorem)
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
k = -(2 + √2)

Question 4.
Find the (RBSESolutions.com) value of m, if (x + 3) is a factor of 3x² + mx + 6.
Solution.
Let p(x) = 3x² + mx + 6
If x + 3 is a factor of p(x), then
p(-3) = 0
p(-3) = 3(- 3)² + m(-3) + 6 = 0
⇒ 3 x 9 – 3m + 6 = 0
⇒ 27 – 3m + 6 = 0
⇒ 3m = 33
⇒ m = 11

Question 5.
Verify that 1 is not a zero (RBSESolutions.com) of the polynomial 4x⁴ – 3x³ + 2x² – 5x + 1.
Solution.
Let p(x) = 4x⁴ – 3x³ + 2x² – 5x + 1
Now
P(1) = 4(1)⁴ – 3(1)³ + 2(1)² – 5(1) + 1 = 4 – 3 + 2 – 5 + 1 = -1
p(1) ≠ 0
1 is not a zero of p(x).

Remainder Theorem Calculator. Various types of online remainder theorem calculators are available online.

Question 6.
Using remainder theorem, find the (RBSESolutions.com) value of k so that (4x² + kx – 1) leaves the remainder 2 when divided by (x – 3).
Solution.
Let p(x) = 4x² + kx – 1
Here we are given that p(3) = 2
i.e. 4(3)² + k(3) -1 = 2
⇒ 4 x 9 + 3k – 1 = 2
⇒ 36 + 3k = 3
⇒ 3k = -33
⇒ k = -11

Question 7.
If (x – 2) is a factor of the (RBSESolutions.com) polynomial x⁴ – 2x³ + ax – 1, then find the value of a.
Solution.
(x – 2) is a factor of the polynomial p(x) = x⁴ – 2x³ + ax – 1, then according to factor theorem p(2) = 0
⇒ (2)⁴ – 2(2)³ + a(2) – 1 = 0
⇒ 2a – 1 = 0
⇒ a = \(\frac { 1 }{ 2 }\)

Question 8.
Find the value of
(x – y)³ + (y – z)³ + (z – x)³.
Solution.
Let a = x – y, b = y – z, c = z – x
Here a + b + c = x – y + y – z + z – x = 0
a³ + b³ + c³ = 3abc
⇒ (x – y)³ + (y – z)³ + (z – x)³ = 3(x – y)(y – z)(z – x).

Question 9.
If a, b, c are all non-zero and a + b + c = 0 then find the (RBSESolutions.com) value of \(\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }\)
Solution.
We know that a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
a + b + c then a³ + b³ + c³ = 3abc …(1)
Dividing equation (1) by abc, we get

Question 10.

Solution.

Short Answer Type Questions

Question 1.
Find the remainder (RBSESolutions.com) obtained on dividing p(x) = x³ + 1 by x + 1.
Solution.
By long division method

Here we find that remainder is zero.
Also p(-1) = (-1)³ + 1 = -1 + 1 = 0,
which is equal to the (RBSESolutions.com) remainder obtained by actual division.

Question 2.
Find the value of
4a² + b² + 25c² + 4ab – 10bc – 20ca when a = 1, b = 2 and c = 3.
Solution.
We have,
4a² + b² + 25c² + 4ab – 10bc – 20ca = (2a)² + (b)² + (-5c)² – 2(2a)(b) + 2(b)(-5c) + 2(-5c)(2a) = (2a + b – 5c)²
It is given that a = 1, b = 2 and c = 3,
so (2a + b – 5c)² = (2 x 1 + 2 – 5 x 3)² = (2 + 2 – 15)² = (4 – 15)² = (-11)² = 121

Question 3.
If 3a – 2b = 11 and ab = 12, then find (RBSESolutions.com) the value of 27a³ – 8b³.
Solution.
We know that
(a – b)³ = a³ – b³ – 3ab(a – b)
Using this identity,
(3a – 2b)³ = (3a)³ – (2b)³ -3 x 3a x 2b (3a – 2b)
⇒ (3a – 2b)³ = 27a³ – 8b³ – 18ab(3a – 2b)
Now substituting 3a – 2b = 11 and ab = 12,
we get
(11)² = 27a³ – 8b³ – 18 x 12 x 11
⇒ 1331 = 27a³ – 8b³ – 2376
⇒ 27a³ – 8b² = 1331 + 2376
⇒ 27a³ – 8b³ = 3707

Question 4.
Use factor (RBSESolutions.com) theorem, show that (x + √2) is a factor of (2√2x² + 5x + √2).
Solution.
Let p(x) = (2√2x² + 5x + √2)
If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.
p(-√2) = 2√2(-√2)² + 5(-√2) + √2 = 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0
p(-√2) = 0
(x + √2) is a factor p(x) i.e. (2√2 x² + 5x + √2).

Question 5.
If a + b + c = 9 and ab + bc + ca = 23 then find (RBSESolutions.com) the value of a² + b² + c².
Solution.
We have, a + b + c = 9 Squaring both sides, we get
(a + b + c)² = (9)²
⇒ a² + b² + c² + 2ab + 2bc + 2ca = 81
⇒ a² + b² + c² + 2(ab + bc + ca) = 81
⇒ a² + b² + c² + 2 x 23 = 81
⇒ a² + b² + c² = 81 – 46
⇒ a² + b² + c² = 35 Thus, a² + b² + c² = 35.

Question 6.
Simplify by using (RBSESolutions.com) suitable identity

Solution.

Question 7.
Factorise
(i) 216a³ – 125
(ii) a³ – b³ – a + b
Solution.
(i) we have,
216a³ – 125 = (6a)³ – (5)³
[a³ – b³ = (a – b)(a² + ab + b²)]
= (6a – 5) {(6a)² + 6a x 5 + (5)²}
= (6a – 5)(36a² + 30a + 25)
(ii) We have,
a³ – b³ – a + b = a³ – b³ – (a – b)
= (a – b)(a² + ab + b²) – (a – b)
[a³ – b³ = (a – b)(a² + ab + b²)]
= (a – b)(a² + ab + b² – 1)

Question 8.
If p = 4 – q show that p³ + q³ + 12pq = 64.
Solution.
We have,
p = 4 – q ⇒ p + q = 4 …(i)
Cubing both sides, (RBSESolutions.com) we get
(p + q)³ = (4)³
⇒ p³ + 3p2q + 3pq² + q³ = 64
⇒ p³ + q³ + 3pq(p + q) = 64
⇒ p³ + q³ + 3pq x 4 = 64 [using (i)]
⇒ p³ + q³ + 12pq = 64

Question 9.

Solution.

Long Answer Type Questions

Question 1.
If (x – 3) and (x – \(\frac { 1 }{ 2 }\)) are both (RBSESolutions.com) factors of ax² + 5x + b, show that a = b.
Solution.
Let p(x) = ax² + 5x + b
x – 3 is a factor of p(x).
p(3) = 0

Question 2.
For what values of a and b so that (RBSESolutions.com) the polynomial x³ + 10x² + ax – 6 is exactly divisible by (x – 1) and (x + 2).
Solution.
Let p(x) = x³ + 10x² + ax + b
p(x) i.e. x³ + 10x² + ax + 6 is exactly divisible by (x – 1) and (x + 2)
Therefore, p(1) and p(-2) must equal to zero.
p(1) = (1)³ + 10(1)² + a x 1 + b = 0
⇒ 1 + 10 + a + b = 0
⇒ a + b = – 11 …(i)
Also, p(-2) = 0
(-2)³ + 10(-2)² + a x (- 2) + b = 0
-8 + 40 – 2a + b = 0
⇒ – 2a + b = – 32 …(ii)
Solving (i) and (ii), we get
a = 7 and b = -18.

Question 3.
If ax³ + bx² + x – 6 has x + 2 as a factor and leaves (RBSESolutions.com) a remainder 4 when divided by (x – 2), find the values of a and b.
Solution.
Let p(x) = ax³ + bx² + x – 6
(x + 2) is a factor of p(x)
⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]
⇒ a(-2)³ + b(-2)² + (-2) – 6 = 0
⇒ -8a + 4b – 2 – 6 = 0
⇒ -8a + 4b = 8
⇒ -2a + b = 2 …(i)
It is given that p(x) leaves the (RBSESolutions.com) remainder 4 when it divided by (x – 2) i.e. p(2) = 4.
⇒ a(2)³ + b(2)² + (2) – 6 = 4
⇒ 8a + 4b – 4 = 4
⇒ 8a + 46 = 8
⇒ 2a + b = 2 …(ii)
Solving (i) and (ii), we get a = 0 and b = 2.

Question 4.
Evaluate

Solution.
In the given expression, we see that both numerator and denominator (RBSESolutions.com) is in the form a³ + b³ + c³ = 3abc because a + b + c = 0.
From numerator, we see that a² – b² + b² – c² + c² – a² = 0
⇒ (a² – b²)³ + (b² – c²)³ + (c² – a²)³ = 3 (a² – b²)(b² – c²)(c² – a²)
Similarly, from denominator,
a – b + b – c + c – a = 0
(a – b)³ + (b – c)³ + (c – a)³ = 3(a – b)(b – c)(c – a)
Value of the expression

Question 5.
If the polynomials (3x³ + ax² + 3x + 5) and (4x³ + x² – 2x + a) leaves the same (RBSESolutions.com) remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.
Solution.
Let the given polynomials are p(x) = 3x³ + ax² + 3x + 5 and f(x) = 4x³ + x² – 2x + a
According to question p(2) = f(2)
3(2)³ + a(2)² + 3(2) + 5 = 4(2)³ + (2)² – 2(2) + a
⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a
⇒ 24 + 4a + 11 = 32 + a
⇒ 4a – a = 32 – 35
⇒ 3a = -3
⇒ a = -1
As the remainder is same, so p(2) or f(2) would (RBSESolutions.com) be same when a = – 1
p(2) = 3(2)³ + (-1)(2)² + 3 x 2 + 5 [∴ a = -1]
= 24 – 4 + 6 + 5
= 35 – 4 = 31
Thus, a = – 1 and p(2) or f(2) = 31

Question 6.
Prove that (a + b + c)³ – a³ – b³ – c³ = 3(a + b)(b + c)(c + a).
Solution.
We have,
L.H.S.
= (a + b + c)³ – a³ -b³ – c³
= {(a + b + c)³ – (a)³} – (b³ + c³)
= (a + b + c – a){(a + b + c)² + a(a + b + c) + a²} – (b + c){b² – bc + c²)
[∵ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)]
= (b + c){a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²} – (b + c)(b² – bc + c²)
= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²]
= (b + c)[3a² + 3ab + 3bc + 3ca]
= 3(b + c)[a² + ab + bc + ca]
= 3(b + c)[a(a + b) + c(a + b)]
= 3(b + c)(a + b)(a + c)
= 3(a + b)(b + c)(c + a)
= R.H.S.

Question 7.
(i) For what (RBSESolutions.com) value of m is x³ – 2mn² + 16 divisible by (x + 2)?
(ii) Show that (2x – 3) is a factor of x + 2x³ – 9x + 12.
Solution.
(i) Let p(x) = x³ – 2mn² + 16
p(x) will be divisible by (x + 2) if
p(-2)= 0
p(-2) = (-2)³ – 2m(-2)² + 16 = -8 – 8m + 16 = – 8m + 8
p(-2) = 0
⇒ -8m + 8 = 0
⇒ 8m = 8
⇒ m = 1
(ii) We know that (2x – 3) will be a factor of x + 2x³ – 9x + 12 if p(x) on dividing by 2x – 3, leaves (RBSESolutions.com) a remainder zero

So, the remainder obtained (RBSESolutions.com) on dividing p(x) by 2x – 3 is zero.
Hence, (2x – 3) is a factor of x + 2x³ – 9x + 12

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