myhelper: RS Aggarwal solution class 8 chapter 7 Factorisation Exercise 7B

Exercise 7B

Page-100

Q1 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 1:

Factorise:
x² − 36

Answer 1:

We have:
$x^{2} - 36 = {(x)}^{2} - {(6)}^{2} = (x + 6) (x - 6)$

∴ $x^{2} - 36 = (x + 6) (x - 6)$

Q2 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 2:

Factorise:
4a² − 9

Answer 2:

We have:
$4 a^{2} - 9 = {(2 a)}^{2} - {(3)}^{2} = (2 a + 3) (2 a - 3)$

∴ $4 a^{2} - 9 = (2 a + 3) (2 a - 3)$

Q3 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 3:

Factorise:
81 − 49x²

Answer 3:

We have:
$81 - 49 x^{2} = {(9)}^{2} - {(7 x)}^{2} = (9 + 7 x) (9 - 7 x)$

∴ $81 - 49 x^{2} = (9 + 7 x) (9 - 7 x)$

Q4 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 4:

Factorise:
4x² − 9y²

Answer 4:

We have:
$4 x^{2} - 9 y^{2} = {(2 x)}^{2} - {(3 y)}^{2} = (2 x + 3 y) (2 x - 3 y)$

∴ $4 x^{2} - 9 y^{2} = (2 x + 3 y) (2 x - 3 y)$

Q5 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 5:

Factorise:
16a² − 225b²

Answer 5:

We have:
$16 a^{2} - 225 b^{2} = {(4 a)}^{2} - {(15 b)}^{2} = (4 a + 15 b) (4 a - 15 b)$

∴ $16 a^{2} - 225 b^{2} = (4 a + 15 b) (4 a - 15 b)$

Q7 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 6:

Factorise:
9a²b² − 25

Answer 6:

We have:
$9 a^{2} b^{2} - 25 = {(3 a b)}^{2} - {(5)}^{2} = (3 a b + 5) (3 a b - 5)$

∴ $9 a^{2} b^{2} - 25 = (3 a b + 5) (3 a b - 5)$

Q7 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 7:

Factorise:
16a² − 144

Answer 7:

We have:
$16 a^{2} - 144 = {(4 a)}^{2} - {(12)}^{2} = (4 a + 12) (4 a - 12) = 4 (a + 3) 4 (a - 3) = 16 (a + 3) (a - 3)$

∴ $16 a^{2} - 144 = 16 (a + 3) (a - 3)$

Q8 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 8:

Factorise:
63a² − 112b²

Answer 8:

We have:
$63 a^{2} - 112 b^{2} = 7 (9 a^{2} - 16 b^{2}) = 7 \{{(3 a)}^{2} - {(4 b)}^{2}\} = 7 (3 a + 4 b) (3 a - 4 b)$

∴ $63 a^{2} - 112 b^{2} = 7 (3 a + 4 b) (3 a - 4 b)$

Q9 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 9:

Factorise:
20a² − 45b²

Answer 9:

We have:
$20 a^{2} - 45 b^{2} = 5 (4 a^{2} - 9 b^{2}) = 5 \{{(2 a)}^{2} - {(3 b)}^{2}\} = 5 (2 a + 3 b) (2 a - 3 b)$

∴ $20 a^{2} - 45 b^{2} = 5 (2 a + 3 b) (2 a - 3 b)$

Q10 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 10:

Factorise:
12x² − 27

Answer 10:

We have:
$12 x^{2} - 27 = 3 (4 x^{2} - 9) = 3 \{{(2 x)}^{2} - {(3)}^{2}\} = 3 (2 x + 3) (2 x - 3)$

∴ $12 x^{2} - 27 = 3 (2 x + 3) (2 x - 3)$

Q11 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 11:

Factorise:
x³ − 64x

Answer 11:

We have:
$x^{3} - 64 x = x (x^{2} - 64) = x \{{(x)}^{2} - {(8)}^{2}\} = x (x + 8) (x - 8)$

∴ $x^{3} - 64 x = x (x + 8) (x - 8)$

Q12 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 12:

Factorise:
16x⁵ − 144x³

Answer 12:

We have:
$16 x^{5} - 144 x^{3} = 16 x^{3} (x^{2} - 9) = 16 x^{3} \{{(x)}^{2} - {(3)}^{2}\} = 16 x^{3} (x + 3) (x - 3)$

∴ $16 x^{5} - 144 x^{3} = 16 x^{3} (x + 3) (x - 3)$

Q13 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 13:

Factorise:
3x⁵ − 48x³

Answer 13:

We have:
$3 x^{5} - 48 x^{3} = 3 x^{3} (x^{2} - 16) = 3 x^{3} \{{(x)}^{2} - {(4)}^{2}\} = 3 x^{3} (x + 4) (x - 4)$

∴ $3 x^{5} - 48 x^{3} = 3 x^{3} (x + 4) (x - 4)$

Q14 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 14:

Factorise:
16p³ − 4p

Answer 14:

We have:
$16 p^{3} - 4 p = 4 p (4 p^{2} - 1) = 4 p \{{(2 p)}^{2} - {(1)}^{2}\} = 4 p (2 p + 1) (2 p - 1)$

∴ $16 p^{3} - 4 p = 4 p (2 p + 1) (2 p - 1)$

Q15 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 15:

Factorise:
63a²b² − 7

Answer 15:

We have:
$63 a^{2} b^{2} - 7 = 7 (9 a^{2} b^{2} - 1) = 7 \{{(3 a b)}^{2} - {(1)}^{2}\} = 7 (3 a b + 1) (3 a b - 1)$

∴ $63 a^{2} b^{2} - 7 = 7 (3 a b + 1) (3 a b - 1)$

Q16 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 16:

Factorise:
1 − (b − c)²

Answer 16:

We have:
$1 - {(b - c)}^{2} = {(1)}^{2} - {(b - c)}^{2} = \{1 + (b - c)\} \{1 - (b - c)\} = (1 + b - c) (1 - b + c)$

∴ $1 - {(b - c)}^{2} = (1 + b - c) (1 - b + c)$

Q17 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

Question 17:

Factorise:
(2a + 3b)² − 16c²

Answer 17:

We have:
${(2 a + 3 b)}^{2} - 16 c^{2} = {(2 a + 3 b)}^{2} - {(4 c)}^{2} = \{(2 a + 3 b) + 4 c\} \{(2 a + 3 b) - 4 c\} = (2 a + 3 b + 4 c) (2 a + 3 b - 4 c)$

∴ ${(2 a + 3 b)}^{2} - 16 c^{2} = (2 a + 3 b + 4 c) (2 a + 3 b - 4 c)$

Q18 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 18:

Factorise:
(l + m)² − (l − m)²

Answer 18:

We have:
${(l + m)}^{2} - {(l - m)}^{2} = \{(l + m) + (l - m)\} \{(l + m) - (l - m)\} = (l + m + l - m) (l + m - l + m) = (2 l) (2 m)$

∴ ${(l + m)}^{2} - {(l - m)}^{2} = (2 l) (2 m)$

Q19 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 19:

Factorise:
(2x + 5y)² − 1

Answer 19:

We have:
${(2 x + 5 y)}^{2} - 1 = {(2 x + 5 y)}^{2} - {(1)}^{2} = \{(2 x + 5 y) + 1\} \{(2 x + 5 y) - 1\} = (2 x + 5 y + 1) (2 x + 5 y - 1)$

∴ ${(2 x + 5 y)}^{2} - 1 = (2 x + 5 y + 1) (2 x + 5 y - 1)$

Q20 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 20:

Factorise:
36c² − (5a + b)²

Answer 20:

We have:
$36 c^{2} - {(5 a + b)}^{2} = {(6 c)}^{2} - {(5 a + b)}^{2} = \{(6 c) + (5 a + b)\} \{(6 c) - (5 a + b)\} = (6 c + 5 a + b) (6 c - 5 a - b)$

∴ $36 c^{2} - {(5 a + b)}^{2} = (6 c + 5 a + b) (6 c - 5 a - b)$

Q21 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 21:

Factorise:
(3x − 4y)² − 25z²

Answer 21:

We have:
${(3 x - 4 y)}^{2} - 25 z^{2} = {(3 x - 4 y)}^{2} - {(5 z)}^{2} = \{(3 x - 4 y) + 5 z\} \{(3 x - 4 y) - 5 z\} = (3 x - 4 y + 5 z) (3 x - 4 y - 5 z)$

∴ ${(3 x - 4 y)}^{2} - 25 z^{2} = (3 x - 4 y + 5 z) (3 x - 4 y - 5 z)$

Q22 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 22:

Factorise:
x² − y² − 2y − 1

Answer 22:

We have:
$x^{2} - y^{2} - 2 y - 1 = x^{2} - (y^{2} + 2 y + 1)$
$= {(x)}^{2} - {(y + 1)}^{2} = \{x + (y + 1)\} \{x - (y + 1)\} = (x + y + 1) (x - y - 1)$

∴ $x^{2} - y^{2} - 2 y - 1 = (x + y + 1) (x - y - 1)$

Q23 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 23:

Factorise:
25 − a² − b² − 2ab

Answer 23:

We have:
$25 - a^{2} - b^{2} - 2 a b = 25 - (a^{2} + b^{2} + 2 a b)$
$= 25 - {(a + b)}^{2} = {(5)}^{2} - {(a + b)}^{2} = \{5 + (a + b)\} \{5 - (a + b)\} = (5 + a + b) (5 - a - b)$

∴ $25 - a^{2} - b^{2} - 2 a b = (5 + a + b) (5 - a - b)$

Q24 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 24:

Factorise:
25a² − 4b² + 28bc − 49c²

Answer 24:

We have:
$25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2} = 25 a^{2} - (4 b^{2} - 28 b c + 49 c^{2})$
$= {(5 a)}^{2} - {(2 b - 7 c)}^{2} = \{5 a + (2 b - 7 c)\} \{5 a - (2 b - 7 c)\} = (5 a + 2 b - 7 c) (5 a - 2 b + 7 c)$

∴ $25 a^{2} - 4 b^{2} + 28 b c - 49 c^{2} = (5 a + 2 b - 7 c) (5 a - 2 b + 7 c)$

Q25 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 25:

Factorise:
9a² − b² + 4b − 4

Answer 25:

We have:
$9 a^{2} - b^{2} + 4 b - 4 = 9 a^{2} - (b^{2} - 4 b + 4)$
$= {(3 a)}^{2} - {(b - 2)}^{2} = \{3 a + (b - 2)\} \{3 a - (b - 2)\} = (3 a + b - 2) (3 a - b + 2)$

∴ $9 a^{2} - b^{2} + 4 b - 4 = (3 a + b - 2) (3 a - b + 2)$

Q26 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 26:

Factorise:
100 − (x − 5)²

Answer 26:

We have:
$100 - {(x - 5)}^{2} = {(10)}^{2} - {(x - 5)}^{2}$
$= \{10 + (x - 5)\} \{10 - (x - 5)\} = (10 + x - 5) (10 - x + 5) = (5 + x) (15 - x)$

∴ $100 - {(x - 5)}^{2} = (5 + x) (15 - x)$

Q27 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 27:

Factorise:
Evaluate {(405)² − (395)²}.

Answer 27:

We have:
$\{{(405)}^{2} - {(395)}^{2}\} = (405 + 395) (405 - 395) = (800 \times 10) = 8000$

∴ $\{{(405)}^{2} - {(395)}^{2}\} = 8000$

Q28 | Ex-7B | Class 8 | RS AGGARWAL | Factorisation | Chapter 7 | myhelper

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Question 28:

Factorise:
Evaluate {(7.8)² − (2.2)²}.

Answer 28:

We have:
$\{{(7.8)}^{2} - {(2.2)}^{2}\} = (7.8 + 2.2) (7.8 - 2.2) = (10 \times 5.6) = 56$

∴ $\{{(7.8)}^{2} - {(2.2)}^{2}\} = 56$