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Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.3 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

9th Standard Maths 1 Practice Set 3.3 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.3 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
Divide each of the following polynomials by synthetic division method and also by linear division method. Write the quotient and the remainder.
i. (2m² – 3m + 10) ÷ (m – 5)
ii. (x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
iii. (y³ – 216) ÷ (y – 6)
iv. (2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
v. (x⁴ – 3x² – 8) ÷ (x + 4)
vi. (y³ – 3y² + 5y – 1) ÷ (y – 1)
Solution:
i. Synthetic division:
(2m² – 3m + 10) ÷ (m – 5)
Dividend = 2m² – 3m + 10
∴ Coefficient form of dividend = (2, -3, 10)
Divisor = m – 5
∴ Opposite of -5 is 5.

Coefficient form of quotient = (2, 7)
∴ Quotient = 2m + 7,
Remainder = 45
Linear division method:
2m² – 3m + 10
To get the term 2m², multiply (m – 5) by 2m and add 10m,
= 2m(m – 5) + 10m- 3m + 10
= 2m(m – 5) + 7m + 10
To get the term 7m, multiply (m – 5) by 7 and add 35
= 2m(m – 5) + 7(m- 5) + 35+ 10
= (m – 5) (2m + 7) + 45
∴ Quotient = 2m + 7,
Remainder = 45

ii. Synthetic division:
(x⁴ + 2x³ + 3x² + 4x + 5) ÷ (x + 2)
Dividend = x⁴ + 2x³ + 3x² + 4x + 5
∴ Coefficient form of dividend = (1, 2, 3, 4, 5)
Divisor = x + 2
∴ Opposite of + 2 is -2.

Coefficient form of quotient = (1, 0, 3, -2)
∴ Quotient = x³ + 3x – 2,
Remainder = 9

Linear division method:
x⁴ + 2x³ + 3x² + 4x + 5
To get the term x⁴, multiply (x + 2) by x³ and subtract 2x³,
= x³(x + 2) – 2x³ + 2x³ + 3x² + 4x + 5
= x³(x + 2) + 3x² + 4x + 5
To get the term 3x², multiply (x + 2) by 3x and subtract 6x,
= x³(x + 2) + 3x(x + 2) – 6x + 4x + 5
= x³(x + 2) + 3x(x + 2) – 2x + 5
To get the term -2x, multiply (x + 2) by -2 and add 4,
= x³(x + 2) + 3x(x + 2) – 2(x + 2) + 4 + 5
= (x + 2) (x3 + 3x – 2) + 9
∴ Quotient = x³ + 3x – 2,
Remainder – 9

iii. Synthetic division:
(y³ – 216) ÷ (y – 6)
Dividend = y³ – 216
∴ Index form = y³ + 0y³ + 0y – 216
∴ Coefficient form of dividend = (1, 0, 0, -216)
Divisor = y – 6
∴ Opposite of – 6 is 6.

Coefficient form of quotient = (1, 6, 36)
∴ Quotient = y² + 6y + 36,
Remainder = 0

Linear division method:
y³ – 216
To get the term y³, multiply (y – 6) by y² and add 6y²,
= y²(y – 6) + 6y² – 216
= y²(y – 6) + 6ysup>2 – 216
To get the, term 6 y² multiply (y – 6) by 6y and add 36y,
= y²(y – 6) + 6y(y – 6) + 36y – 216
= y²(y – 6) + 6y(y – 6) + 36y – 216
To get the term 36y, multiply (y- 6) by 36 and add 216,
= y² (y – 6) + 6y(y – 6) + 36(y – 6) + 216 – 216
= (y – 6) (y² + 6y + 36) + 0
Quotient = y² + 6y + 36
Remainder = 0

iv. Synthetic division:
(2x⁴ + 3x³ + 4x – 2x²) ÷ (x + 3)
Dividend = 2x⁴ + 3x³ + 4x – 2x²
∴ Index form = 2x⁴ + 3x³ – 2x² + 4x + 0
∴ Coefficient form of the dividend = (2,3, -2,4,0)
Divisor = x + 3
∴ Opposite of + 3 is -3

Coefficient form of quotient = (2, -3, 7, -17)
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

Linear division method:
2x⁴ + 3x³ + 4x – 2x² = 2x² + 3x³ – 2x² + 4x
To get the term 2x⁴, multiply (x + 3) by 2x³ and subtract 6x³,
= 2x³(x + 31 – 6x³ + 3x³ – 2x² + 4x
= 2x³(x + 3) – 3x³ – 2x² + 4x

To get the term – 3x³, multiply (x + 3) by -3x² and add 9x²,
= 2x³(x + 3) – 3x²(x + 3) + 9x² – 2x² + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x² + 4x

To get the term 7x², multiply (x + 3) by 7x and subtract 21x,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 21x + 4x
= 2x³(x + 3) – 3x²(x + 3) + 7x(x + 3) – 17x

To get the term -17x, multiply (x + 3) by -17 and add 51,
= 2x³(x + 3) – 3x²(x + 3) + 7x(x+3) – 17(x + 3) + 51
= (x + 3) (2x³ – 3x² + 7x- 17) + 51
∴ Quotient = 2x³ – 3x² + 7x – 17,
Remainder = 51

v. Synthetic division:
(x⁴ – 3x² – 8) + (x + 4)
Dividend = x⁴ – 3x² – 8
∴ Index form = x⁴ + 0x³ – 3x² + 0x – 8
∴ Coefficient form of the dividend = (1,0, -3,0, -8)
Divisor = x + 4
∴ Opposite of + 4 is -4

∴ Coefficient form of quotient = (1, -4, 13, -52)
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder = 200

Linear division method:
x⁴ – 3x² – 8
To get the term x⁴, multiply (x + 4) by x³ and subtract 4x³,
= x³(x + 4) – 4x³ – 3x² – 8
= x³(x + 4) – 4x³ – 3x² – 8
To get the term – 4x³, multiply (x + 4) by -4x² and add 16x²,
= x³(x + 4) – 4x² (x + 4) + 16x² – 3x² – 8
= x³(x + 4) – 4x² (x + 4) + 13x² – 8
To get the term 13x², multiply (x + 4) by 13x and subtract 52x,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52x – 8
To get the term -52x, multiply (x + 4) by – 52 and add 208,
= x³(x + 4) – 4x²(x + 4) + 13x(x + 4) – 52(x + 4) + 208 – 8
= (x + 4) (x³ – 4x² + 13x – 52) + 200
∴ Quotient = x³ – 4x² + 13x – 52,
Remainder 200

vi. Synthetic division:
(y³ – 3y² + 5y – 1) ÷ (y – 1)
Dividend = y³ – 3y² + 5y – 1
Coefficient form of the dividend = (1, -3, 5, -1)
Divisor = y – 1
∴Opposite of -1 is 1.

∴ Coefficient form of quotient = (1, -2, 3)
∴ Quotient = y² – 2y + 3,
Remainder = 2

Linear division method:
y³ -3y² + 5y – 1
To get the term y³ , multiply (y – 1) by y² and add y²
= y² (y – 1) + y² – 3y² + 5y – 1
= y² (y – 1) – 2y² + 5y – 1
To get the term -2y², multiply (y – 1) by -2y and subtract 2y,
= y² (y – 1) – 2y(y – 1) – 2y + 5y – 1
= y² (y – 1) – 2y(y – 1) + 3y – 1
To get the term 3y, multiply (y – 1) by 3 and add 3,
= y² (y – 1) – 2y(y – 1) + 3(y- 1) + 3 – 1
= (y – 1)(y² – 2y + 3) + 2
∴ Quotient = y² – 2y + 3,
Remainder = 2.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

9th Standard Maths 1 Practice Set 3.2 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.2 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
Use the given letters to write the answer.
i. There are ‘a’ trees in the village Lat. If the number of trees increases every year by ’b‘. then how many trees will there be after ‘x’ years?
ii. For the parade there are y students in each row and x such row are formed. Then, how many students are there for the parade in all ?
iii. The tens and units place of a two digit number is m and n respectively. Write the polynomial which represents the two digit number.
Solution:
i. Number of trees in the village Lat = a
Number of trees increasing each year = b
∴ Number of trees after x years = a + bx
∴ There will be a + bx trees in the village Lat after x years.

ii. Total rows = x
Number of students in each row = y
∴ Total students = Total rows × Number of students in each row
= x × y
= xy
∴ There are in all xy students for the parade.

iii. Digit in units place = n
Digit in tens place = m
∴ The two digit number = 10 x digit in tens place + digit in units place
= 10m + n
∴ The polynomial representing the two digit number is 10m + n.

Question 2.
Add the given polynomials.
i. x³ – 2x² – 9; 5x³ + 2x + 9
ii. -7m⁴+ 5m³ + √2 ; 5m⁴ – 3m³ + 2m² + 3m – 6
iii. 2y² + 7y + 5; 3y + 9; 3y² – 4y – 3
Solution:
i. (x³ – 2x² – 9) + (5x³ + 2x + 9)
= x³ – 2x² – 9 + 5x³ + 2x + 9
= x³ + 5x³ – 2x² + 2x – 9 + 9
= 6x³ – 2x² + 2x

ii. (-7m⁴ + 5m³ + √2 ) + (5m⁴ – 3m³ + 2m² + 3m – 6)
= -7m⁴ + 5m³ + √2 + 5m⁴ – 3m³ + 2m² + 3m – 6
= -7m⁴ + 5m⁴ + 5m³ – 3m³ + 2m² + 3m +√2 – 6
= -2m⁴ + 2m³ + 2m² + 3m + √2 – 6

iii. (2y² + 7y + 5) + (3y + 9) + (3y² – 4y – 3)
= 2y² + 7y + 5 + 3y + 9 + 3y² – 4y – 3
= 2y² + 3y² + 7y + 3y – 4y + 5 + 9 – 3
= 5y² + 6y + 11

Question 3.
Subtract the second polynomial from the first.
i. x² – 9x + √3 ; – 19x + √3 + 7x²
ii. 2ab² + 3a²b – 4ab; 3ab – 8ab² + 2a²b
Solution:
i. x² – 9x + √3 -(- 19x + √3 + 7x²)
= x² – 9x + √3 + 19x – √ 3 – 7x²
= x² – 7x² – 9x + 19x + √3 – √3
= – 6x² + 10x

ii. (2ab² + 3a²b – 4ab) – (3ab – 8ab² + 2a²b)
= 2ab² + 3a²b – 4ab – 3ab + 8ab² – 2a²b
= 2ab² + 8ab² + 3a²b – 2a²b – 4ab – 3ab
= 10ab² + a²b – 7ab

Question 4.
Multiply the given polynomials.
i. 2x; x² – 2x – 1
ii. x⁵ – 1; x³ + 2x² + 2
iii. 2y +1; y² – 2y + 3y
Solution:
i. (2x) x (x² – 2x – 1) = 2x³ – 4x² – 2x

ii. (x⁵ – 1) × (x³ + 2x² + 2)
= x⁵ (x³ + 2x² + 2) -1(x³ + 2x² + 2)
= x⁸ + 2x⁷ + 2x⁵ – x³ – 2x² – 2

iii. (2y + 1) × (y² – 2y³ + 3y)
= 2y(y² – 2y³ + 3y) + 1(y² – 2y³ + 3y)
= 2y³ – 4y⁴ + 6y² + y² – 2y³ + 3y
= -4y⁴ + 2y³ – 2y³ + 6y² + y² + 3y
= -4y⁴ + 7y² + 3y

Question 5.
Divide first polynomial by second polynomial and write the answer in the form ‘Dividend = Divisor x Quotient + Remainder’.
i. x³ – 64; x – 4
ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 ; x² – x
Solution:
i. x³ – 64 = x3 + 0x² + 0x – 64

∴ Quotient = x² + 4x + 16, Remainder = 0
Now, Dividend = Divisor x Quotient + Remainder
∴ x³ – 64 = (x – 4)(x² + 4x + 16) + 0

ii. 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = 5x⁵ + 4x⁴ – 3x³ + 2x + 0x + 2

∴ Quotient = 5x³ + 9x² + 6x + 8,
Remainder = 8x + 2
Now, Dividend = Divisor x Quotient + Remainder
∴ 5x⁵ + 4x⁴ – 3x³ + 2x² + 2 = (x² – x)(5x³ + 9x² + 6x + 8) + (8x + 2)

Question 6.
Write down the information in the form of algebraic expression and simplify.
There is a rectangular farm with length (2a² + 3b²) metre and breadth (a² + b²) metre. The farmer used a square shaped plot of the farm to build a house. The side of the plot was (a2 – b2) metre. What is the area of the remaining part of the farm? [4 Marks]
Solution:
Length of the rectangular farm = (2a² + 3b²) m
Breadth of the rectangular farm = (a² + b²) m
Area of the farm = length x breadth = (2a² + 3b²) x (a² + b²)
= 2a²(a² + b²) + 3b²(a² + b²)
= 2a² + 2a²b² + 3a²b² + 3b⁴
= (2a⁴ + 5a²b² + 3b⁴) sq. m … (i)
The farmer used a square shaped plot of the farm to build a house.
Side of the square shaped plot = (a² – b²) m
∴ Area of the plot = (side)²
= (a² – b²)²
= (a⁴ – 2a²b² + b⁴) sq m… .(ii)

∴ Area of the remaining farm = Area of the farm – Area of the plot
= (2a⁴ + 5a²b² + 3b⁴) – (a⁴ – 2a²b² + b⁴) … [From (i) and (ii)]
= 2a⁴ + 5a²b² + 3b⁴ – a⁴ + 2a²b² – b⁴
= 2a⁴ – a⁴ + 5a²b² + 2a²b² + 3b⁴ – b⁴
= a⁴ + 7a²b² + 2b⁴
∴ The area of the remaining farm is (a⁴ + 7a²b² + 2b⁴) sq. m.

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.4 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.4 Solutions Maharashtra Board

Std 8 Maths Practice Set 6.4 Chapter 6 Solutions Answers

Question 1.
Simplify:
i. \(\frac{m^{2}-n^{2}}{(m+n)^{2}} \times \frac{m^{2}+m n+n^{2}}{m^{3}-n^{3}}\)
ii. \(\frac{a^{2}+10 a+21}{a^{2}+6 a-7} \times \frac{a^{2}-1}{a+3}\)
iii. \(\frac{8 x^{3}-27 y^{3}}{4 x^{2}-9 y^{2}}\)
iv. \(\frac{x^{2}-5 x-24}{(x+3)(x+8)} \times \frac{x^{2}-64}{(x-8)^{2}}\)
v. \(\frac{3 x^{2}-x-2}{x^{2}-7 x+12} \div \frac{3 x^{2}-7 x-6}{x^{2}-4}\)
vi. \(\frac{4 x^{2}-11 x+6}{16 x^{2}-9}\)
vii. \(\frac{a^{3}-27}{5 a^{2}-16 a+3} \div \frac{a^{2}+3 a+9}{25 a^{2}-1}\)
viii. \(\frac{1-2 x+x^{2}}{1-x^{3}} \times \frac{1+x+x^{2}}{1+x}\)
Solution:
i. \(\frac{m^{2}-n^{2}}{(m+n)^{2}} \times \frac{m^{2}+m n+n^{2}}{m^{3}-n^{3}}\)

ii. \(\frac{a^{2}+10 a+21}{a^{2}+6 a-7} \times \frac{a^{2}-1}{a+3}\)

iii. \(\frac{8 x^{3}-27 y^{3}}{4 x^{2}-9 y^{2}}\)

iv. \(\frac{x^{2}-5 x-24}{(x+3)(x+8)} \times \frac{x^{2}-64}{(x-8)^{2}}\)

v. \(\frac{3 x^{2}-x-2}{x^{2}-7 x+12} \div \frac{3 x^{2}-7 x-6}{x^{2}-4}\)

vi. \(\frac{4 x^{2}-11 x+6}{16 x^{2}-9}\)

vii. \(\frac{a^{3}-27}{5 a^{2}-16 a+3} \div \frac{a^{2}+3 a+9}{25 a^{2}-1}\)

viii. \(\frac{1-2 x+x^{2}}{1-x^{3}} \times \frac{1+x+x^{2}}{1+x}\)

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 3.1 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 3 Polynomials.

9th Standard Maths 1 Practice Set 3.1 Chapter 3 Polynomials Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Practice Set 3.1 Chapter 3 Polynomials Questions With Answers Maharashtra Board

Question 1.
State whether the given algebraic expressions are polynomials? Justify.
i. y + \(\frac { 1 }{ y }\)
ii. 2 – 5√x
iii. x² + 7x + 9
iv. 2m^-2 + 7m – 5
v. 10
Answer:
i. No, because power of v in the term 5√x is -1 (negative number).
ii. No, because the power of x in the term 5√x is
i. e. 0.5 (decimal number).
iii. Yes. All the coefficients are real numbers. Also, the power of each term is a whole number.
iv. No, because the power of m in the term 2m^-2 is -2 (negative number).
v. Yes, because 10 is a constant polynomial.

Question 2.
Write the coefficient of m³ in each of the given polynomial.
i. m³
ii. \(\sqrt [ -3 ]{ 2 }\) + m – √3m³
iii. \(\sqrt [ -2 ]{ 3 }\)m³ + 5m² – 7m -1
Answer:
i. 1
ii. -√3
iii. – \(\frac { 2 }{ 3 }\)

Question 3.
Write the polynomial in x using the given information. [1 Mark each]
i. Monomial with degree 7
ii. Binomial with degree 35
iii. Trinomial with degree 8
Answer:
i. 5x⁷
ii. x³⁵ – 1
iii. 3x⁸ + 2x⁶ + x⁵

Question 4.
Write the degree of the given polynomials.
i. √5
ii. x°
iii. x²
iv. √2m¹⁰ – 7
v. 2p – √7
vi. 7y – y³ + y⁵
vii. xyz +xy-z
viii. m³n⁷ – 3m⁵n + mn
Answer:
i. √5 = √5 x°
∴ Degree of the polynomial = 0

ii. x°
∴Degree of the polynomial = 0

iii. x²
∴Degree of the polynomial = 2

iv. √2m¹⁰ – 7
Here, the highest power of m is 10.
∴Degree of the polynomial = 10

v. 2p – √7
Here, the highest power of p is 1.
∴ Degree of the polynomial = 1

vi. 7y – y³ + y⁵
Here, the highest power of y is 5.
∴Degree of the polynomial = 5

vii. xyz + xy – z
Here, the sum of the powers of x, y and z in the term xyz is 1 + 1 + 1= 3,
which is the highest sum of powers in the given polynomial.
∴Degree of the polynomial = 3

viii. m³n⁷ – 3m⁵n + mn
Here, the sum of the powers of m and n in the term m³n⁷ is 3 + 7 = 10,
which is the highest sum of powers in the given polynomial.
∴ Degree of the polynomial = 10

Question 5.
Classify the following polynomials as linear, quadratic and cubic polynomial. [2 Marks]
i. 2x² + 3x +1
ii. 5p
iii. √2 – \(\frac { 1 }{ 2 }\)
iv. m³ + 7m² + \(\sqrt [ 5 ]{ 2 }\)m – √7
v. a²
vi. 3r³
Answer:
Linear polynomials: ii, iii
Quadratic polynomials: i, v
Cubic polynomials: iv, vi

Question 6.
Write the following polynomials in standard form.
i. m³ + 3 + 5m
ii. – 7y + y⁵ + 3y³ – \(\frac { 1 }{ 2 }\)+ 2y⁴ – y²
Answer:
i. m³ + 5m + 3
ii. y⁵ + 2y⁴ + 3y³ – y² – 7y – \(\frac { 1 }{ 2 }\)

Question 7.
Write the following polynomials in coefficient form.
i. x³ – 2
ii. 5y
iii. 2m⁴ – 3m² + 7
iv. – \(\frac { 2 }{ 3 }\)
Answer:
i. x³ – 2 = x³ + 0x² + 0x – 2
∴ Coefficient form of the given polynomial = (1, 0, 0, -2)

ii. 5y = 5y + 0
∴Coefficient form of the given polynomial = (5,0)

iii. 2m⁴ – 3m² + 7
= 2m⁴ + Om³ – 3m² + 0m + 7
∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7)

iv. – \(\frac { 2 }{ 3 }\)
∴Coefficient form of the given polynomial = (- \(\frac { 2 }{ 3 }\))

Question 8.
Write the polynomials in index form.
i. (1, 2, 3)
ii. (5, 0, 0, 0 ,-1)
iii. (-2, 2, -2, 2)
Answer:
i. Number of coefficients = 3
∴ Degree = 3 – 1 = 2
∴ Taking x as variable, the index form is x² + 2x + 3

ii. Number of coefficients = 5
∴ Degree = 5 – 1=4
∴ Taking x as variable, the index form is 5x⁴ + 0x³ + 0x² + 0x – 1

iii. Number of coefficients = 4
∴Degree = 4 – 1 = 3
∴Taking x as variable, the index form is -2x³ + 2x² – 2x + 2

Question 9.
Write the appropriate polynomials in the boxes.

Answer:
i. Quadratic polynomial: x²; 2x² + 5x + 10; 3x² + 5x
ii. Cubic polynomial: x³ + x² + x + 5; x³ + 9
iii. Linear polynomial: x + 7
iv. Binomial: x + 7; x³ + 9; 3x² + 5x
v. Trinomial: 2x² + 5x + 10
vi. Monomial: x²

Question 1.
Write an example of a monomial, a binomial and a trinomial having variable x and degree 5. ( Textbook pg. no. 3)
Answer:
Monomial: x⁵
Binomial: x⁵ + x
Trinomial: 2x⁵ – x² + 5

Question 2.
Give example of a binomial in two variables having degree 5. (Textbook pg. no. 38)
Answer:
x³y² + xy

Class 9 Maths Digest

• Practice Set 3.1 Class 9 Answers
• Practice Set 3.2 Class 9 Answers
• Practice Set 3.3 Class 9 Answers
• Practice Set 3.4 Class 9 Answers
• Practice Set 3.5 Class 9 Answers
• Practice Set 3.6 Class 9 Answers
• Practice Set 3 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.3 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 6.3 Chapter 6 Solutions Answers

Question 1.
Factorize
i. y³ – 27
ii. x³ – 64y³
iii. 27m³ – 216n³
iv. 125y³ – 1
v. \(8 p^{3}-\frac{27}{p^{3}}\)
vi. 343a³ – 512b³
vii. 64x³ – 729y³
viii. \(16 a^{3}-\frac{128}{b^{3}}\)
Solution:
i. y³ – 27
= y³ – (3)³
Here, a = y and b = 3
∴ y³ – 27 = (y – 3)[y² + y(3) + (3)2]
…[∵ a³ – b³ = (a – b) (a² + ab + b²)]
= (y – 3)(y² + 3y + 9)

ii. x³ – 64y³
= x³ – (4y)³
Here, a = x and b = 4y
∴ x³ – 64y³ = (x – 4y)[x² + x(4y) + (4y)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (x – 4y)(x² + 4xy + 16y²)

iii. 27m³ – 216n³
= 27 (m³ – 8n³)
… [Taking out the common factor 27]
= 27 [m³ – (2n)³]
Here, a = m and b = 2n
∴ 27m³ – 216n³
= 27 {(m – 2n) [m² + m(2n) + (2n)²]}
….[∵ a³ – b³ = (a – b) (a² + ab + b²)]
= 27 (m – 2n)(m² + 2mn + 4n²)

iv. 125y³ – 1
= (5y)³ – 1³
Here, a = 5y and b = 1
∴ 125y³ – 1 = (5y – 1) [(5y)² + (5y)(1) + (1)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (5y – 1) (25y² + 5y + 1)

v. \(8 p^{3}-\frac{27}{p^{3}}\)

vi. 343a³ – 512b³
= (7a)³ – (8b)³
Here, A = 7a and B = 8b
∴ 343a³ – 512b³
= (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]
…[∵ A³ – B³ = (A – B)(A² + AB + B²)]
= (7a – 8b) (49a² + 56ab + 64b²)

vii. 64x³ – 729y³
= (4x)³ – (9y)³
Here, a = 4x and b = 9y
∴ 64x³ – 729y³
= (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]
…[∵ a³ – b³ = (a – b)(a² + ab + b²)]
= (4x – 9y) (16x² + 36xy + 81y²)

viii. \(16 a^{3}-\frac{128}{b^{3}}\)

Question 2.
Simplify:
i. (x + y)³ – (x – y)³
ii. (3a + 5b)³ – (3a – 5b)³
iii. (a + b)³ – a³ – b³
iv. p³ – (p + 1)³
v. (3xy – 2ab)³ – (3xy + 2ab)³
Solution:
i. (x + y)³ – (x – y)³
Here, a = x + y and b = x – y
(x + y)³ – (x – y)³
= [(x + y) – (x – y)] [(x + y)² + (x + y) (x – y) + (x – y)]
…[a³ – b³ = (a – b)(a² + ab + b²)]
= (x + y – x + y) [(x² + 2xy + y²) + (x² – y²) + (x² – 2xy + y²)]
= 2y(x² + x² + x² + 2xy – 2xy + y² – y² + y²)
= 2y (3x² + y²)
= 6x²y + 2y³

ii. (3a + 5b)³ – (3a – 5b)³
Here, A = 3a + 5b and B = 3a – 5b
= [(3a + 5b) – (3a – 5b)] [(3a + 5b)² + (3a + 5b) (3a – 5b) + (3a – 5b)²]
…[∵ A³ – B³ = (A – B)(A² + AB + B²)]
= (3a + 5b – 3a + 5b) [(9a² + 30ab + 25b²) + (9a² – 25b²) + (9a² – 30ab + 25b²)]
= 10b (9a² + 9a² + 9a² + 30ab – 30ab + 25b² – 25b² + 25b²)
= 10b (27a² + 25b²)
= 270a²b + 250b³

iii. (a + b)³ – a³ – b³
= a³ + 3a²b + 3ab² + b³ – a³ – b³
= 3a²b + 3ab²

iv. p³ – (p + 1)³
= p³ – (p³ + 3p² + 3p + 1) …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= p³ – p³ – 3p² – 3p – 1
= – 3p² – 3p – 1

v. (3xy – 2ab)³ – (3xy + 2ab)³
Here, A = 3xy – 2ab and B = 3xy + 2ab
∴ (3xy – 2ab)³ – (3xy + 2ab)³
= [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]
…[∵ A³ – B³ = (A – B) (A² + AB + B²)]
= (3xy – 2ab – 3xy – 2ab) [(9x²y² – 12xyab + 4a²b²) + (9x²y² – 4a²b²) + (9x²y² + 12xyab + 4a²b²)]
= (- 4ab) (9x²y² + 9x²y² + 9x²y² – 12xyab + 12xyab + 4a²b² – 4a²b² + 4a²b²)
= (- 4ab) (27 xy² + 4a²b²)
= -108x²y²ab – 16a³b³

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.2 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 6.2 Chapter 6 Solutions Answers

Question 1.
Factorise:
i. x³ + 64y³
ii. 125p³ + q³
iii. 125k³ + 27m³
iv. 2l³ + 432m³
v. 24a³ + 81b³
vi. \(y^{3}+\frac{1}{8 y^{3}}\)
vii. \(\mathrm{a}^{3}+\frac{8}{\mathrm{a}^{3}}\)
viii. \(1+\frac{\mathrm{q}^{3}}{125}\)
Solution:
i. x³ + 64y³
= x³ + (4y)³
Here, a = x and b = 4y
∴ x³ + 64y³ = (x + 4y) [x² – x(4y) + (4y)²]
….[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (x + 4y)(x² – 4xy + 16y²)

ii. 125p³ + q³
= (5p)³ + q³
Here, a = 5p and b = q
∴ 125p³ + q³ = (5p + q)[(5p)² – (5p)(q) + q²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5p + q)(25p² – 5pq + q²)

iii. 125k³ + 27m³
= (5k)³ + (3m)³
Here, a = 5k and b = 3m
∴ 125k³ + 27m³
= (5k + 3m) [(5k)² – (5k)(3m) + (3m)²]
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= (5k + 3m)(25k² – 15km + 9m²)

iv. 2l³ + 432m³
= 2 (l³ + 216m³)
… [Taking out the common factor 2]
= 2[l³ + (6m)³]
Here, a = l and b = 6m
2l³ + 432m³ = 2 {(l + 6m)[l² – l(6m) + (6m)²]}
…[∵ a³ + b³ = (a + b)(a² – ab + b²)]
= 2(l + 6m)(l² – 6lm + 36m²)

v. 24a³ + 81b³
…[Taking out the common factor 3]
= 3 [(2a)³ + (3b)³]
Here, A = 2a and B = 3b
∴ 24a³ + 81b³
= 3 {(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]}
…[∵ A³ + B³ = (A + B) (A² – AB + B²)]
= 3(2a + 3b)(4a² – 6ab + 9b²)

vi. \(y^{3}+\frac{1}{8 y^{3}}\)

vii. \(\mathrm{a}^{3}+\frac{8}{\mathrm{a}^{3}}\)

viii. \(1+\frac{\mathrm{q}^{3}}{125}\)

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 6.1 8th Std Maths Answers Solutions Chapter 6 Factorisation of Algebraic Expressions.

Factorisation of Algebraic Expressions Class 8 Maths Chapter 6 Practice Set 6.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 6.1 Chapter 6 Solutions Answers

Question 1.
Factorize:
i. x² + 9x + 18
ii. x² – 10x + 9
iii. y² + 24y + 144
iv. 5y² + 5y – 10
v. p² – 2p – 35
vi. p² – 7p – 44
vii. m² – 23m + 120
viii. m² – 25m + 100
ix. 3x² + 14x + 15
x. 2x² + x – 45
xi. 20x² – 26x + 8
xii. 44x² – x – 3
Solution:
i. x² + 9x + 18
= x² + 6x + 3x + 18
= x (x + 6) + 3(x + 6)
= (x + 6) (x + 3)

ii. x² – 10x + 9
= x² – 9x – x + 9
= x (x – 9) – 1(x – 9)
= (x – 9)(x – 1)

iii. y² + 24y + 144
= y² + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12)(y + 12)

iv. 5y² + 5y – 10
= 5(y² + y – 2)
… [Taking out the common factor 5]
= 5(y² + 2y – y – 2)
= 5[y(y + 2) – 1(y + 2)]
= 5 (p + 2)(y- 1)

v. p² – 2p – 35
= p² – 7p + 5p – 35
= p(p – 7) + 5(p – 7)
= (p – 7)(p + 5)

vi. p² – 7p – 44
= p² – 11p + 4p – 44
= p(p – 11) + 4(p – 11)
= (p – 11)(p + 4)

vii. m² – 23m + 120
= m² – 15m – 8m + 120
= m (m – 15) – 8 (m – 15)
= (m – 15) (m – 8)

viii. m² – 25m + 100
= m² – 20m – 5m + 100
= m(m – 20) – 5(m – 20)
= (m – 20) (m – 5)

ix. 3x² + 14x + 15 3 × 15 = 45
= 3x² + 9x + 5x + 15
= 3x(x + 3) + 5(x + 3)
= (x + 3) (3x + 5)

x. 2x² + x – 45 2 × (- 45) = -90
= 2x² + 10x – 9x – 45
= 2x(x + 5) – 9 (x + 5)
= (x + 5) (2x – 9)

xi. 20x² – 26x + 8
= 2(10x² – 13x + 4) 10 × 4 = 40
… [Taking out the common factor 2]
= 2(10x² – 8x – 5x + 4)
= 2[2x(5x – 4) – 1(5x – 4)]
= 2 (5x – 4) (2x – 1)

xii. 44x² – x – 3 44 × (-3) = -132
= 44x² – 12x + 11x – 3
= 4x(11x – 3) + 1(11x – 3)
= (11x – 3) (4x + 1)

Std 8 Maths Digest

• Practice Set 6.1 Class 8 Answers
• Practice Set 6.2 Class 8 Answers
• Practice Set 6.3 Class 8 Answers
• Practice Set 6.4 Class 8 Answers
• Practice Set 7.1 Class 8 Answers
• Practice Set 7.2 Class 8 Answers
• Practice Set 7.3 Class 8 Answers

Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Problem Set 2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 2 Real Numbers.

9th Standard Maths 1 Problem Set 2.6 Chapter 2 Real Numbers Textbook Answers Maharashtra Board

Class 9 Maths Part 1 Problem Set 2.6 Chapter 2 Real Numbers Questions With Answers Maharashtra Board

Question 1.
Choose the correct alternative answer for the questions given below. [1 Mark each]

i. Which one of the following is an irrational number?

Answer:
√5

ii. Which of the following is an irrational number?
(A) 0.17
(B) \(1.\overline { 513 }\)
(C) \(0.27\overline { 46 }\)
(D) 0.101001000……..
Answer:
(D) 0.101001000……..

iii. Decimal expansion of which of the following is non-terminating recurring?

Answer:
(C) \(\frac { 3 }{ 11 }\)

iv. Every point on the number line represents which of the following numbers?
(A) Natural numbers
(B) Irrational numbers
(C) Rational numbers
(D) Real numbers
Answer:
(D) Real numbers

v. The number [/latex]0.\dot { 4 }[/latex] in \(\frac { p }{ q }\) form is ……

Answer:
(A) \(\frac { 4 }{ 9 }\)

vi. What is √n , if n is not a perfect square number ?
(A) Natural number
(B) Rational number
(C) Irrational number
(D) Options A, B, C all are correct.
Answer:
(C) Irrational number

vii. Which of the following is not a surd ?

Answer:
(C) \(\sqrt [ 3 ]{ \sqrt { 64 } }\)

viii. What is the order of the surd \(\sqrt [ 3 ]{ \sqrt { 5 } }\) ?
(A) 3
(B) 2
(C) 6
(D) 5
Answer:
(C) 6

ix. Which one is the conjugate pair of 2√5 + √3 ?
(A) -2√5 + √3
(B) -2√5 – √3
(C) 2√3 – √5
(D) √3 + 2√5
Answer:
(A) -2√5 + √3

x. The value of |12 – (13 + 7) x 4| is ____ .
(A) – 68
(B) 68
(C) – 32
(D) 32
Answer:
(B) 68

Hints:
ii. Since the decimal expansion is neither terminating nor recurring, 0.101001000…. is an irrational number.

iii. \(\frac { 3 }{ 11 }\)
Denominator =11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal expansion of \(\frac { 3 }{ 11 }\) will be non terminating recurring.

v. Let x = [/latex]0.\dot { 4 }[/latex]
∴10 x = [/latex]0.\dot { 4 }[/latex]
∴10 – x = [/latex]4.\dot { 4 }[/latex] – [/latex]0.\dot { 4 }[/latex]
∴9x = 4
∴ x = \(\frac { 4 }{ 9 }\)

vii. \(\sqrt[3]{61}\) = 4, which is not an irrational number.

viii. \(\sqrt[3]{\sqrt{5}}=\sqrt[3 \times 2]{5}=\sqrt[6]{5}\)
∴ Order = 6

ix. The conjugate of 2√5 + √3 is 2√5 – √3 or -2√5 + √3

x. |12 – (13+7) x 4| = |12 – 20 x 4|
= |12 – 80|
= |-68|
= 68

Question 2.
Write the following numbers in \(\frac { p }{ q }\) form.
i. 0.555
ii. \(29.\overline { 568 }\)
iii. 9.315315…..
iv. 357.417417…..
v . \(30.\overline { 219 }\)
Solution:

ii. Let x = \(29.\overline { 568 }\) …(i)
x = 29.568568…
Since, three numbers i.e. 5, 6 and 8 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 29568.568568…
1000 x= \(29568.\overline { 568 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(29568.\overline { 568 }\) – \(29.\overline { 568 }\)
∴ 999x = 29539

iii. Let x = 9.315315 … = \(9.\overline { 315 }\) …(i)
Since, three numbers i.e. 3, 1 and 5 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 9315.315315…
∴1000x = \(9315.\overline { 315 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(9315.\overline { 315 }\) – \(9.\overline { 315 }\)
∴ 999x = 9306

iv. Let x = 357.417417… = \(357.\overline { 417 }\) …(i)
Since, three numbers i.e. 4, 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 357417.417417…
∴ 1000x = 357417.417 …(ii)
Subtracting (i) from (ii),
1000x – x = \(357417.\overline { 417 }\) – \(357.\overline { 417 }\)
∴ 999x = 357060

v. Let x = \(30.\overline { 219 }\) …(i)
∴ x = 30.219219
Since, three numbers i.e. 2, 1 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x= 30219.219219…
∴ 1000x = \(30219.\overline { 219 }\) …(ii)
Subtracting (i) from (ii),
1000x – x = \(30219.\overline { 219 }\) – \(30.\overline { 219 }\)
∴ 999x = 30189

Question 3.
Write the following numbers in its decimal form.

Solution:
i. \(\frac { -5 }{ 7 }\)

ii. \(\frac { 9 }{ 11 }\)

iii. √5

iv. \(\frac { 121 }{ 13 }\)

v. \(\frac { 29 }{ 8 }\)

Question 4.
Show that 5 + √7 is an irrational number. [3 Marks]
Solution:
Let us assume that 5 + √7 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, ‘a’ and ‘b’ are integers, \(\sqrt [ a ]{ b }\) – 5 is a rational number and so √7 is a rational number.
∴ But this contradicts the fact that √7 is an irrational number.
Our assumption that 5 + √7 is a rational number is wrong.
∴ 5 + √7 is an irrational number.

Question 5.
Write the following surds in simplest form.

Solution:

Question 6.
Write the simplest form of rationalising factor for the given surds.

Solution:

Now, 4√2 x √2 = 4 x 2 = 8, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √32 .

Now, 5√2 x √2 = 5 x 2 = 10, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √50 .

Now, 3√3 x √3 = 3 x 3 = 9, which is a rational number.
∴ √ 3 is the simplest form of the rationalising factor of √27 .

= 6, which is a rational number.
∴ √10 is the simplest form of the rationalising factor of \(\sqrt [ 3 ]{ 5 }\) √10 .

Now, 18√2 x √2 = 18 x 2 = 36, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of 3√72.

vi. 4√11
4√11 x √11 = 4 x 11 = 44, which is a rational number.
∴ √11 is the simplest form of the rationalising factor of 4√11.

Question 7.
Simplify.

Solution:

Question 8.
Rationalize the denominator.

Solution:

Question 1.
Draw three or four circles of different radii on a card board. Cut these circles. Take a thread and measure the length of circumference and diameter of each of the circles. Note down the readings in the given table. (Textbook pg.no.23 )

Solution:
i. 14,44,3.1
ii. 16,50.3,3.1
iii. 11,34.6,3.1
From table, we observe that the ratio \(\sqrt [ c ]{ d }\) is nearly 3.1 which is constant. This ratio is denoted by π (pi).

Question 2.
To find the approximate value of π, take the wire of length 11 cm, 22 cm and 33 cm each. Make a circle from the wire. Measure the diameter and complete the following table.

Verify that the ratio of circumference to the diameter of a circle is approximately \(\sqrt [ 22 ]{ 7 }\). (Textbook pg. no. 24)
Solution:
i. 3.5, \(\sqrt [ 22 ]{ 7 }\)
ii. 7, \(\sqrt [ 22 ]{ 7 }\)
iii. 10.5, \(\sqrt [ 22 ]{ 7 }\)
∴ The ratio of circumference to the diameter of each circle is \(\sqrt [ 22 ]{ 7 }\).

Class 9 Maths Digest

• Practice Set 1.1 Class 9 Answers
• Practice Set 1.2 Class 9 Answers
• Practice Set 1.3 Class 9 Answers
• Practice Set 1.4 Class 9 Answers
• Problem Set 1 Class 9 Answers
• Practice Set 2.1 Class 9 Answers
• Practice Set 2.2 Class 9 Answers
• Practice Set 2.3 Class 9 Answers
• Practice Set 2.4 Class 9 Answers
• Practice Set 2.5 Class 9 Answers
• Problem Set 2 Class 9 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.4 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.4 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.4 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (2p + q + 5)²
ii. (m + 2n + 3r)²
iii. (3x + 4y – 5p)²
iv. (7m – 3n – 4k)²
Solution:
i. (2p + q + 5)² = (2p)² + (q)² + (5)² + 2(2p) (q) + 2(q) (5) + 2(2p) (5)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 4p² + q² + 25 + 4pq + 10q + 20p

ii. (m + 2n + 3r)² = (m)² + (2n)² + (3r)² + 2(m) (2n) + 2(2n) (3r) + 2(m) (3r)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= m² + 4n² + 9r² + 4mn + 12nr + 6mr

iii. (3x + 4y – 5p)² = (3x)² + (4y)² + (- 5p)² + 2(3x) (4y) + 2(4y) (- 5p) + 2(3x) (- 5p)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 9x + 16y² + 25p² + 24xy – 40py – 30px

iv. (7m – 3n – 4k)² = (7m)² + (- 3n)² + (- 4k)² + 2(7m) (- 3n) + 2 (- 3n) (- 4k) + 2 (7m) (- 4k)
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49m² + 9n² + 16k² – 42mn + 24nk – 56km

Question 2.
Simplify:
i. (x – 2y + 3)² + (x + 2y – 3)²
ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
Solution:
i. (x – 2y + 3)² + (x + 2y – 3)²
= [(x)² + (- 2y)² + (3)² + 2 (x) (- 2y) + 2 (- 2y) (3) + 2 (x) (3)] + [(x)² + (2y)² + (- 3)² + 2 (x) (2y) + 2 (2y) (- 3) + 2 (x) (- 3)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= x² + 4y² + 9 – 4xy – 12y + 6x + x² + 4y² + 9 + 4xy – 12y – 6x
= x + x² + 4y² + 4y² + 9 + 9 – 4xy + 4xy – 12y – 12y + 6x – 6x
= 2x² + 8y² + 18 – 24y

ii. (3k – 4r – 2m)² – (3k + 4r – 2m)²
= [(3k)² + (- 4r)² + (- 2m)² + 2 (3k) (- 4r) + 2 (- 4r) (- 2m) + 2 (3k) (- 2m)] – [(3k)² + (4r)² + (- 2m)² + 2 (3k) (4r) + 2 (4r) (- 2m) + 2 (3k) (- 2m)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= (9k² + 16r² + 4m² – 24kr + 16rm – 12km) – (9k² + 16r² + 4m² + 24kr – 16rm – 12km)
= 9k² + 16r² + 4m² – 24kr + 16rm – 12km – 9k² – 16r² – 4m² – 24kr + 16rm + 12km
= 9k² – 9k² + 16r² – 16r² + 4m² – 4m² – 24kr – 24kr + 16rm + 16rm – 12km + 12km
= 32rm – 48kr

iii. (7a – 6b + 5c)² + (7a + 6b – 5c)²
= [(7a)² + (- 6b)² + (5c)² + 2(7a) (-6b) + 2(-6b) (5c) + 2(7a) (5c)] + [(7a)² + (6b)² + (- 5c)² + 2 (7a) (6b) + 2 (6b) (- 5c) + 2 (7a) (- 5c)]
… [(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac]
= 49a² + 36b² + 25c² – 84ab – 60bc + 70ac + 49a² + 36b² + 25c² + 84ab – 60bc – 70ac
= 49a² + 49a² + 36b² + 36b² + 25c² + 25c² – 84ab + 84ab – 60bc – 60bc + 70ac – 70ac
= 98a² + 72b² + 50c² – 120bc

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.4 Intext Questions and Activities

Question 1.
Fill in the boxes with appropriate terms in the steps of expansion. (Textbook pg. no. 27)
(2p + 3m + 4n)²
= (2p)² + (3m)² + __ + 2 × 2p × 3m + 2 × __ × 4n + 2 × 2p × __
= __ + 9m² + __ + 12pm + __ + __
Solution:
(2p + 3m + 4n)²
= (2p)² + (3m)² + (4n)² + 2 x 2p x 3m + 2 x 3m x 4n + 2 x 2p x 4n
= 4p² + 9m² + 16n² + 12pm + 24mn + 16pn

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 5.3 8th Std Maths Answers Solutions Chapter 5 Expansion Formulae.

Expansion Formulae Class 8 Maths Chapter 5 Practice Set 5.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 5.3 Chapter 5 Solutions Answers

Question 1.
Expand:
i. (2m – 5)³
ii. (4 – p)³
iii. (7x – 9y)³
iv. (58)³
v. (198)³
vi. \(\left(2 p-\frac{1}{2 p}\right)^{3}\)
vii. \(\left(1-\frac{1}{a}\right)^{3}\)
viii. \(\left(\frac{x}{3}-\frac{3}{x}\right)^{3}\)
Solution:
i. Here, a = 2m and b = 5
(2m – 5)³
= (2m)³ – 3(2m)² (5) + 3(2m) (5)² – (5)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8m³ – 3(4m²)(5) + 3(2m)(25) – 125
= 8m³ – 60m² + 150m – 125

ii. Here, a = 4 and b = p
(4 – p)³ = (4)³ – 3(4)²(p) + 3(4)(p)² – (p)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 64 – 3(16)(p) + 3(4)(p²) – p³
= 64 – 48p + 12p² – p³

iii. Here, a = 7x and b = 9y
(7x – 9y)³
= (7x)³ – 3(7x)² (9y) + 3 (7x)(9y)² – (9y)³
…[(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 343x³ – 3(49x²)(9y) + 3(7x)(81y²) – 729y³
= 343x³ – 1323x²y + 1701xy² – 729y³

iv. (58)³ = (60 – 2)³
Here, a = 60 and b = 2
(58)³ = (60)³ – 3(60)²(2) + 3(60)(2)² – (2)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 216000 – 3(3600)(2) + 3(60)(4) – 8
= 216000 – 21600 + 720 – 8
=195112

v. (198)³ = (200 – 2)³
Here, a = 200 and b = 2
(198)³ = (200)³ – 3(200)²(2) + 3(200)(2)² – (2)³
… [(a – b)³ = a³ – 3a²b + 3ab² – b³]
= 8000000 – 3(40000)(2) + 3(200)(4) – 8
= 8000000 – 240000 + 2400 – 8
= 7762392

vi. Here, a = 2p and b = \(\frac { 1 }{ 2p }\)

vii. Here, A = 1 and B = \(\frac { 1 }{ a }\)

viii. Here, a = \(\frac { x }{ 3 }\) and b = \(\frac { 3 }{ x }\)

Question 2.
Simplify:
i. (2a + b)³ – (2a – b)³
ii. (3r – 2k)³ + (3r + 2k)³
iii. (4a – 3)³ – (4a + 3)³
iv. (5x – 7y)³ + (5x + 7y)³
Solution:
i. (2a + b)³ – (2a – b)³
= [(2a)³ + 3(2a)²(b) + 3 (2a)(b)² + (b)³] – [(2a)³ – 3(2a)²(b) + 3 (2a)(b)² – (b)³]
… [(a + b)³ = a³ + 3a²b + 3ab² + b³, (a – b)³ = a³ – 3a²b + 3ab² – b³]
= (8a³ + 12a²b + 6ab² + b³) – (8a³ – 12a²b + 6ab² – b³)
= 8a³ + 12a²b + 6ab² + b³ – 8a³ + 12a²b – 6ab² + b³
= 8a³ – 8a³ + 12a²b + 12a²b + 6ab² – 6ab² + b³ + b³
= 24a²b + 2b³

ii. (3r – 2k)³ + (3r + 2k)³
= [(3r)³ – 3(3r)²(2k) + 3(3r)(2k)² – (2k)³] + [(3r)³ + 3(3r)²(2k) + 3(3r)(2k)² + (2k)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (27r³ – 54r²k + 36rk² – 8k³) + (27r³ + 54r²k + 36rk² + 8k³)
= 27r³ – 54r²k + 36rk² – 8k³ + 27r³ + 54r²k + 36rk² + 8k³
= 27r³ + 27r³ – 54r²k + 54r²k + 36rk² + 36rk² – 8k³ + 8k³
= 54r³ + 72rk²

iii. (4a – 3)³ – (4a + 3)³
= [(4a)³ – 3(4a)² (3) + 3(4a)(3)² – (3)³] – [(4a)³ + 3(4a)²(3) + 3(4a)(3)² + (3)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (64a³ – 144a² + 108a – 27) – (64a³ + 144a² + 108a + 27)
= 64a³ – 144a² + 108a – 27 – 64a³ -144a² – 108a – 27
= 64a³ – 64a³ – 144a² – 144a² + 108a – 108a – 27 – 27
= -288a² – 54

iv. (5x – 7y)³ + (5x + 7y)³
= [(5x)³ – 3(5x)²(7y) + 3(5x)(7y)² – (7y)³] + [(5x)³ + 3(5x)² (7y) + 3(5x) (7y)² +(7y)³]
… [(a – b)³ = a³ – 3a²b + 3ab² – b³, (a + b)³ = a³ + 3a²b + 3ab² + b³]
= (125x³ – 525x²y + 735xy² – 343y³) + (125x³ + 525x²y + 735xy² + 343y³)
= 125x³ – 525x²y + 735xy² – 343y³ + 125x³ + 525x²y + 735xy² + 343y³
= 125x³ + 125x³ – 525x²y + 525x²y + 735xy² + 735xy² – 343y³ + 343y³
= 250x³ + 1470xy²

Maharashtra Board Class 8 Maths Chapter 5 Expansion Formulae Practice Set 5.3 Intext Questions and Activities

Question 1.
Make two cubes of side a and of side b each. Make six parallelopipeds; three of them measuring a × a × b and the remaining three measuring b × b × a. Arrange all these solid figures properly and make a cube of side (a + b). (Textbook pg. no. 25)
Solution:
(a + b)³ = a³ + 3a²b + 3ab² + b³
= a × a × a + 3 × a × a × b + 3 × a × b × b + b × b × b

Std 8 Maths Digest

• Practice Set 3.1 Class 8 Answers
• Practice Set 3.2 Class 8 Answers
• Practice Set 3.3 Class 8 Answers
• Practice Set 4.1 Class 8 Answers
• Practice Set 5.1 Class 8 Answers
• Practice Set 5.2 Class 8 Answers
• Practice Set 5.3 Class 8 Answers
• Practice Set 5.4 Class 8 Answers