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

RS Aggarwal solution class 8 chapter 7 Factorisation Exercise 7A

Exercise 7A

Page-99

Q1 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 1:

Factorise:
(i) 12x + 15
(ii) 14m − 21
(iii) 9n − 12n²

Answer 1:

(i) $12 x + 15 = 3 (4 x + 5)$
(ii) $14 m - 21 = 7 (2 m - 3)$
(iii) $9 n - 12 n^{2} = 3 n (3 - 4 n)$

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Question 2:

Factorise:
(i) 16a² − 24ab
(ii) 15ab² − 20a²b
(iii) 12x²y³ − 21x³y²

Answer 2:

(i) H.C.F. of $16 a^{2}$ and $24 a b$ is $8 a$ .

∴ $16 a^{2} - 24 a b = 8 a (2 a - 3 b)$

(ii) H.C.F. of $15 a b^{2}$ and $20 a^{2} b$ is $5 a b$ .

∴ $15 a b^{2} - 20 a^{2} b = 5 a b (3 b - 4 a)$

(iii) H.C.F. of $12 x^{2} y^{3}$ and $21 x^{3} y^{2}$ is $3 x^{2} y^{2}$ .

∴ $12 x^{2} y^{3} - 21 x^{3} y^{2} = 3 x^{2} y^{2} (4 y - 7 x)$

Q3 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 3:

Factorise:
(i) 24x³ − 36x²y
(ii) 10x³ − 15x²
(iii) 36x³y − 60x²y³z

Answer 3:

(i) H.C.F. of $24 x^{3}$ and $36 x^{2} y$ is $12 x^{2}$ .

∴ $24 x^{3} - 36 x^{2} y = 12 x^{2} (2x - 3 y)$

(ii) H.C.F. of $10 x^{3}$ and $15 x^{2}$ is $5 x^{3}$ .

∴ $10 x^{3} - 15 x^{2} = 5 x^{2} (2 x - 3)$

(iii) H.C.F. of $36 x^{3} y$ and $60 x^{2} y^{3} z$ is $12 x^{2} y$ .

∴ $36 x^{3} y - 60 x^{2} y^{3} z = 12 x^{2} y (3 x - 5 y^{2} z)$

Q4 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 4:

Factorise:
(i) 9x³ − 6x² + 12x
(ii) 8x² − 72xy + 12x
(iii) 18a³b³ − 27a²b³ + 36a³b²

Answer 4:

(i) H.C.F. of $9 x^{3}$ , $6 x^{2}$ and $12 x$ is $3 x$ .

∴ $9 x^{3} - 6 x^{2} + 12 x = 3 x (3 x^{2} - 2 x + 4)$

(ii) H.C.F. of $8 x^{3}$ , $72 x y$ and $12 x$ is $4 x$ .

∴ $8 x^{3} - 72 x y + 12 x = 4 x (2 x^{2} - 18 y + 3)$

(iii) H.C.F. of $18 a^{3} b^{3}$ , $27 a^{2} b^{3}$ and $36 a^{3} b^{2}$ is $9 a^{2} b^{2}$ .

∴ $18 a^{3} b^{3} - 27 a^{2} b^{3} + 36 a^{3} b^{2} = 9 a^{2} b^{2} (2 a b - 3 b + 4 a)$

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Question 5:

Factorise:
(i) 14x³ + 21x⁴y − 28x²y²
(ii) −5 − 10t + 20t²

Answer 5:

(i) H.C.F. of $14 x^{3}$ , $21 x^{4} y$ and $28 x^{2} y^{2}$ is $7 x^{2}$ .

∴ $14 x^{3} + 21 x^{4} y - 28 x^{2} y^{2} = 7 x^{2} (2 x + 3 x^{2} y - 4 y^{2})$

(ii) H.C.F. of $- 5$ , $- 10 t$ and $20 t^{2}$ is 5.

∴ $- 5 - 10 t + 20 t^{2} = 5 (- 1 - 2 t + 4 t^{2})$

Q6 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 6:

Factorise:
(i) x(x + 3) + 5(x + 3)
(ii) 5x(x − 4) − 7(x − 4)
(iii) 2m(1 − n) + 3(1 − n)

Answer 6:

(i) $x (x + 3) + 5 (x + 3) = (x + 3) (x + 5)$

(ii) $5 x (x - 4) - 7 (x - 4) = (x - 4) (5 x - 7)$

(iii) $2 m (1 - n) + 3 (1 - n) = (1 - n) (2 m + 3)$

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Question 7:

Factorise:
6a(a − 2b) + 5b(a − 2b)

Answer 7:

We have:
$6 a (a - 2 b) + 5 b (a - 2 b) = (a - 2 b) (6 a + 5 b)$

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Question 8:

Factorise:
x³(2a − b) + x²(2a − b)

Answer 8:

We have:
$x^{3} (2 a - b) + x^{2} (2 a - b) = (2 a - b) (x^{3} + x^{2}) = x^{2} (x + 1) (2 a - b)$

Q9 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 9:

Factorise:
9a(3a − 5b) − 12a²(3a − 5b)

Answer 9:

We have:
$9 a (3 a - 5 b) - 12 a^{2} (3 a - 5 b) = (3 a - 5 b) (9 a - 12 a^{2}) = 3 a (3 a - 5 b) (3 - 4 a)$

Q10 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 10:

Factorise:
(x + 5)² − 4(x + 5)

Answer 10:

We have:
${(x + 5)}^{2} - 4 (x + 5) = (x + 5) \{(x + 5) - 4\}$
$= (x + 5) (x + 5 - 4) = (x + 5) (x + 1)$

∴ ${(x + 5)}^{2} - 4 (x + 5) = (x + 5) (x + 1)$

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Question 11:

Factorise:
3(a − 2b)² − 5(a − 2b)

Answer 11:

$3 {(a - 2 b)}^{2} - 5 (a - 2 b) = (a - 2 b) \{3 (a - 2 b) - 5\}$
$= (a - 2 b) (3 a - 6 b - 5)$

∴ $3 {(a - 2 b)}^{2} - 5 (a - 2 b) = (a - 2 b) (3 a - 6 b - 5)$

Q12 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 12:

Factorise:
2a + 6b − 3(a + 3b)²

Answer 12:

We have:
$2 a + 6 b - 3 {(a + 3 b)}^{2} = 2 (a + 3 b) - 3 {(a + 3 b)}^{2} = (a + 3 b) \{2 - 3 (a + 3 b)\} = (a + 3 b) (2 - 3 a - 9 b)$

∴ $2 a + 6 b - 3 {(a + 3 b)}^{2} = (a + 3 b) (2 - 3 a - 9 b)$

Q13 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 13:

Factorise:
16(2p − 3q)² − 4(2p − 3q)

Answer 13:

We have:
$16 {(2 p - 3 q)}^{2} - 4 (2 p - 3 q) = (2 p - 3 q) \{16 (2 p - 3 q) - 4\} = (2 p - 3 q) (32 p - 48 q - 4)$

∴ $16 {(2 p - 3 q)}^{2} - 4 (2 p - 3 q) = (2 p - 3 q) (32 p - 48 q - 4)$

Q14 | Ex-7A | Class 8 | RS AGGARWAL | chapter 7 | Factorisation | myhelper

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Question 14:

Factorise:
x(a − 3) + y(3 − a)

Answer 14:

We have:
$x (a - 3) + y (3 - a) = x (a - 3) - y (a - 3) = (a - 3) (x - y)$

∴ $x (a - 3) + y (3 - a) = (a - 3) (x - y)$

Q15 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 15:

Factorise:
12(2x − 3y)² − 16(3y − 2x)

Answer 15:

We have:
$12 {(2 x - 3 y)}^{2} - 16 (3 y - 2 x) = 12 {(2 x - 3 y)}^{2} + 16 (2 x - 3 y) = (2 x - 3 y) \{12 (2 x - 3 y) + 16\} = (2 x - 3 y) (24 x - 36 y + 16)$

$∴ 12 {(2 x - 3 y)}^{2} - 16 (3 y - 2 x) = (2 x - 3 y) (24 x - 36 y + 16)$

Q16 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 16:

Factorise:
(x + y)(2x + 5) − (x + y)(x + 3)

Answer 16:

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

Q17 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 17:

Factorise:
ar + br + at + bt

Answer 17:

By grouping the terms:
$a r + b r + a t + b t = (a r + b r) + (a t + b t) = r (a + b) + t (a + b) = (a + b) (r + t)$

∴ $a r + b r + a t + b t = (a + b) (r + t)$

Q18 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 18:

Factorise:
x² − ax − bx + ab

Answer 18:

By suitably arranging the terms:
$x^{2} - a x - b x + a b = x^{2} - b x - a x + a b = (x^{2} - b x) - (a x - a b) = x (x - b) - a (x - b) = (x - b) (x - a)$

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

Q19 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 19:

Factorise:
ab² − bc² − ab + c²

Answer 19:

By suitably arranging the terms:
$a b^{2} - b c^{2} - a b + c^{2} = a b^{2} - a b - b c^{2} + c^{2} = (a b^{2} - a b) - (b c^{2} - c^{2}) = a b (b - 1) - c^{2} (b - 1) = (b - 1) (a b - c^{2})$

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

Q20 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 20:

Factorise:
x² − xz + xy − yz

Answer 20:

By suitably arranging the terms:
$x^{2} - x z + x y - y z = x^{2} + x y - x z - y z = (x^{2} + x y) - (x z + y z) = x (x + y) - z (x + y) = (x + y) (x - z)$

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

Q21 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 21:

Factorise:
6ab − b² + 12ac − 2bc

Answer 21:

By suitably arranging the terms:
$6 a b - b^{2} + 12 a c - 2 b c = 6 a b + 12 a c - b^{2} - 2 b c = (6 a b + 12 a c) - (b^{2} + 2 b c) = 6 a (b + 2 c) - b (b + 2 c) = (b + 2 c) (6 a - b)$

∴ $6 a b - b^{2} + 12 a c - 2 b c = (b + 2 c) (6 a - b)$

Q22 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 22:

Factorise:
(x − 2y)² + 4x − 8y

Answer 22:

We have:
${(x - 2 y)}^{2} + 4 x - 8 y = {(x - 2 y)}^{2} + 4 (x - 2 y) = (x - 2 y) (x - 2 y) + 4 (x - 2 y) = (x - 2 y) \{(x - 2 y) + 4\} = (x - 2 y) (x - 2 y + 4)$

∴ ${(x - 2 y)}^{2} + 4 x - 8 y = (x - 2 y) (x - 2 y + 4)$

Q23 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 23:

Factorise:
y² − xy(1 − x) − x³

Answer 23:

We have:
$y^{2} - x y (1 - x) - x^{3} = y^{2} - x y + x^{2} y - x^{3} = (y^{2} - x y) + (x^{2} y - x^{3}) = y (y - x) + x^{2} (y - x) = (y - x) (y + x^{2})$

∴ $y^{2} - x y (1 - x) - x^{3} = (y - x) (y + x^{2})$

Q24 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 24:

Factorise:
(ax + by)² + (bx − ay)²

Answer 24:

We have:
${(a x + b y)}^{2} + {(b x - a y)}^{2} = (a^{2} x^{2} + b^{2} y^{2} + 2 a x b y) + (b^{2} x^{2} + a^{2} y^{2} - 2 b x a y) = a^{2} x^{2} + a^{2} y^{2} + b^{2} y^{2} + b^{2} x^{2} + 2 a x b y - 2 b x a y = a^{2} (x^{2} + y^{2}) + b^{2} x^{2} + b^{2} y^{2} + 2 a x b y - 2 a x b y = a^{2} (x^{2} + y^{2}) + b^{2} (x^{2} + y^{2}) = (x^{2} + y^{2}) (a^{2} + b^{2})$

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

Q25 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 25:

Factorise:
ab² + (a − 1)b − 1

Answer 25:

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

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

Q26 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 26:

Factorise:
x³ − 3x² + x − 3

Answer 26:

We have:
$x^{3} - 3 x^{2} + x - 3 = (x^{3} - 3 x^{2}) + (x - 3) = x^{2} (x - 3) + 1 (x - 3) = (x - 3) (x^{2} + 1)$

∴ $x^{3} - 3 x^{2} + x - 3 = (x - 3) (x^{2} + 1)$

Q27 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 27:

Factorise:
ab(x² + y²) − xy(a² + b²)

Answer 27:

We have:
$a b (x^{2} + y^{2}) - x y (a^{2} + b^{2}) = a b x^{2} + a b y^{2} - a^{2} x y - b^{2} x y = a b x^{2} - a^{2} x y + a b y^{2} - b^{2} x y = a x (b x - a y) + b y (a y - b x) = a x (b x - a y) - b y (b x - a y) = (b x - a y) (a x - b y)$

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

Q28 Ex-7A Class 8 RS AGGARWAL chapter 7 Factorisation

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Question 28:

Factorise:
x² − x(a + 2b) + 2ab

Answer 28:

We have:
$x^{2} - x (a + 2 b) + 2 a b = x^{2} - a x - 2 b x + 2 a b$
$= x^{2} - 2 b x - a x + 2 a b = (x^{2} - 2 b x) - (a x - 2 a b) = x (x - 2 b) - a (x - 2 b) = (x - 2 b) (x - a)$

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

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