CBSE Class 8

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7E

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

OBJECTIVE QUESTIONS.
Tick the correct answer in each of the following :

Question 1.
Solution:
7a² – 63b² = 7 (a² – 9b²)
= 7 {(a)² – (3b)²}
= 7 (a – 3b) (a + 36) (d)

Question 2.
Solution:
2x – 32x³ = 2x (1 – 16x²)
= 2x {(1)² – (4x)²}
= 2x (1 – 4x) (1 + 4x) (d)

Question 3.
Solution:
x³ – 144x
= x (x² – 144)
= x {(x)² – (12)²}
= x (x – 12) (x + 12) (c)

Question 4.
Solution:
2 – 50x² = 2 (1 – 25x²)
= 2 {(1)² – (5x)²}
= 2 (1 – 5x) (1 + 5x) (d)

Question 5.
Solution:
a² + bc + ab + ac
= a2 + ab + ac + bc
= a (a + b) + c (a + b)
= (a + b) (a + c)   (a)

Question 6.
Solution:
pq² + q (p – 1) – 1
= pq² + pq – q – 1
= pq (q + 1) – 1 (q + 1)
= (q + 1) (pq – 1)
= (pq – 1) (q + 1) (d)

Question 7.
Solution:
ab – mn + an – bm
= ab + an – bm – mn
= a (b + n) – m (b + n)
= (b + n) (a – m) (b)

Question 8.
Solution:
ab – a – b + 1
= a (b – 1) – 1 (b – 1)
= b – 1) (a – 1) (a)

Question 9.
Solution:
x² – xz + xy – yz
= x (x – z) + y (x – z)
= (x – z) (x + y) (c)

Question 10.
Solution:
12m² – 21
= 3 (4m² – 9)
= 3 {(2m)² – (3)²}
= 3 (2m – 3) (2m + 3) (c)

Question 11.
Solution:
x³ – x
= x (x² – 1)
= x (x – 1) (x + 1) (d)

Question 12.
Solution:
1 – 2ab – (a² + b²)
= 1 – 2ab – a² – b²
= 1 – (a² + b² + 2ab)
= 1 – (a + b)²
= (1 + a + b) (1 – a – b) (c)

Question 13.
Solution:
x² + 6x + 8
= x² + 4x + 2x + 8
{8 = 4 x 2, 6 = 4 + 2}
= x (x + 4) + 2 (x + 4)
= (x + 4) (x + 2) (c)

Question 14.
Solution:
x² + 4x – 21
= x² + 7x – 3x – 21
{ – 21 = + 7 x ( – 3), 4 = 7 – 3}
= x (x + 7) – 3 (x + 7)
= (x + 7) (x – 3) (b)

Question 15.
Solution:
y² + 2y – 3
{ – 3 = 3 x ( – 1), 2 = 3 – 1}
= y² + 3y – y – 3
= y (y + 3) – 1 (y + 3)
= (y + 3) (y – 1) (a)

Question 16.
Solution:
40 + 3x – x²
= 40 + 8x – 5x -x²
{40 = 8 x ( – 5), 3 = 8 – 5}
= 8(5 + x) – x(5 + x)
= (5 + x)(8 – x) (c)

Question 17.
Solution:
2x² + 5x + 3
= 2x² + 2x + 3x + 3
{2 x 3 = 6, 6 = 2 x 3, 5 = 2 + 3}
= 2x(x + 1) + 3(x + 1)
= (x + 1)(2x + 3) (b)

Question 18.
Solution:
6a² – 13a + 6
= 6a² – 9a – 4a + 6
{6 x 6 = 36, 36 = ( – 9)x( – 4), – 13 = – 9 – 4}
= 3a (2a – 3) – 2 (2a – 3)
= (2a – 3) (3a – 2) (c)

Question 19.
Solution:
4z² – 8z + 3
= 4z² – 6z – 2z + 3
{4 x 3 = 12,12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2z (2z – 3) – 1 (2z – 3)
= (2z – 3) (2z – 1) (a)

Question 20.
Solution:
3 + 23y – 8y²
= 3 + 24y – y – 8y²
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3(1 + 8y) – y(1 + 8y)
= (1 + 8y) (3 – y) (b)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7D

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² + 5x + 6
= x² + 2x + 3x + 6
{6 = 2 x 3, 5 = 2 + 3}
= x (x + 2) + 3 (x + 2)
= (x + 2) (x + 3) Ans.

Question 2.
Solution:
y² + 10y + 24
= y² + 6y + 4y + 24
{24 = 6 x 4, 10 = 6 + 4}
= y (y + 6) + 4 (y + 6)
= (y + 6) (y + 4)

Question 3.
Solution:
z² + 12x + 27
{27 = 9 x 3, 12 = 9 + 3}
= z² + 9z + 3z + 27
= z (z + 9) + 3 (z + 9)
= (z + 9) (z + 3)

Question 4.
Solution:
p² + 6p + 8
= p² + 4p + 2p + 8
{ 8 = 4 x 2, 6 = 4 + 2}
= p (p + 4) + 2 (p + 4)
= (p + 4) (p + 2)

Question 5.
Solution:
x² + 15x + 56
= x² + 8x + 7x + 56
{56 = 8 x 7, 15 = 8 x 7}
= x(x + 8) + 7(x + 8)
= (x + 8) (x + 7) Ans.

Question 6.
Solution:
y² + 19y + 60
= y² + 15y + 4y + 60
{60 = 15 x 4, 19 = 15 + 4}
= y (y + 15) + 4 (y + 15)
= (y + 15) (y + 4)

Question 7.
Solution:
x² + 13x + 40
= x² + 5x + 8x + 40
{40 = 5 x 8, 13 = 5 + 8}
= x(x + 5) + 8(x + 5)
= (x + 5) (x + 8) Ans.

Question 8.
Solution:
q² – 10q + 21
= q² – 7q – 3q + 21
{21 = ( – 7)x( – 3), – 10 = – 7 – 3}
= q {q – 7) – 3 (q – 7)
= (q – 7)(q – 3)

Question 9.
Solution:
p² + 6p – 16
= p² + 8p – 2p – 16
{ – 16 = 8x( – 2),6 = 8 – 2}
= p (p + 8) – 2 (p + 8)
= (p + 8) (p – 2)

Question 10.
Solution:
x² – 10x + 24
= x² – 6x – 4x + 24
{24 = ( – 6) x ( – 4), – 10 = – 6 – 4 }
= x(x – 6) – 4(x – 6)
= (x – 6) (x – 4) Ans.

Question 11.
Solution:
x² – 23x + 42
= x² – 2x – 21x + 42
{42 = ( – 2) x ( – 21), – 23 = – 2 – 21}
= x(x – 2) – 21(x – 2)
= (x – 2) (x – 21) Ans.

Question 12.
Solution:
x² – 17x + 16
= x² – x – 16x + 16
{ 16 = ( – 1) x ( – 16), 17 = – 1 – 16}
= x (x – 1) – 16(x – 1)
= (x – 1) (x – 16) Ans.

Question 13.
Solution:
y² – 21y + 90
= y² – 15y – 6y + 90
{90 = ( – 15)x ( – 6), – 21 = – 15 – 6}
= y(y – 15) – 6 (y – 15)
= (y – 15) (y – 6)

Question 14.
Solution:
x² – 22x + 117
= x² – 13x – 9x + 117
{117 = ( – 13) x ( – 9), – 22 = – 13 – 9}
= x(x – 13) – 9(x – 13)
= (x – 9) (x – 13) Ans.

Question 15.
Solution:
x² – 9x + 20
= x² – 5x – 4x + 20
{ 20 = ( – 5) x ( – 4), – 9 = – 5 – 4}
= x(x – 5) – 4(x – 5)
= (x – 5) (x – 4) Ans.

Question 16.
Solution:
x² + x – 132
= x² + 12x – 11x – 132
{ – 132 = 12 x ( – 11), 1 = 12 – 11}
= x(x + 12) – 11(x + 12)
= (x + 12) (x – 11) Ans.

Question 17.
Solution:
x² + 5x – 104
= x² + 13x – 8x – 104
{ – 104 = 13 x ( – 8), 5 = 13 – 8}
= x(x + 13) – 8(x + 13)
= (x + 13) (x – 8) Ans.

Question 18.
Solution:
y² + 7y – 144
= y² + 16y – 9y – 144
{144 = – 16 x 9, 7 = 16 – 9}
= y(y +16) – 9(y + 16)
= (y + 16) (y – 9) Ans.

Question 19.
Solution:
z² + 19z – 150
= z² + 25z – 6z – 150
{ – 150 = 25 x ( – 6), 19 = 25 – 6}
= z(z + 25) – 6(z + 25)
= (z + 25) (z – 6) Ans.

Question 20.
Solution:
y² + y – 72
= y² + 9y – 8y – 72
{ – 72 = 9x( – 8), 1 = 9 – 8}
= y(y + 9) – 8(y + 9)
= (y + 9) (y – 8) Ans

Question 21.
Solution:
a² + 6a – 91
= a² + 13a – 7a – 91
{ – 91 = 13x( – 7), 6 = 13 – 7}
= a (a + 13) – 7 (a + 13)
= (a + 13) (a – 7)

Question 22.
Solution:
p² – 4p – 11
= p² – 11p + 7p – 77
{ – 77 = – 11 x 7, – 4 = – 11 + 7}
= p(p – 11) + 7 (p – 11)
= (p – 11)(p + 7)

Question 23.
Solution:
x² – 7x – 30
= x² – 10x + 3x – 30
{ – 30 = – 10 x 3, – 7 = – 10 + 3}
= x(x – 10) + 3(x – 10)
= (x – 10) (x + 3) Ans.

Question 24.
Solution:
x² – 14 x + 3 x – 42
{ – 11 = – 14 + 3], – 42 = – 14 x 3}
= x (x – 14) + 3 (x – 14)
= (x – 14) (x + 3) Ans.

Question 25.
Solution:
x² – 5x – 24
= x² – 8x + 3x – 24
{ – 24 = – 8 x 3, – 5 = – 8 + 3}
= x(x – 8) + 3(x – 8)
= (x – 8) (x + 3) Ans.

Question 26.
Solution:
y² – 6y – 135
= y² – 15y + 9y – 135
{ – 135 = – 15 x 9, – 6 = – 15 + 9}
= y (y – 15) + 9 (y – 15)
= (y – 15) (y + 9)

Question 27.
Solution:
z² – 12z – 45
= z² – 15z + 3z – 45
= z (z – 15) + 3 (z – 15)
{ – 45 = – 15 x 3, – 12 = – 15 + 3}
= (z – 15)(z + 3)

Question 28.
Solution:
x² – 4x – 12
= x² – 6x + 2x – 12
{ – 12 = – 6 x 2, – 4 = – 6 + 2}
= x (x – 6) + 2 (x – 6)
= (x – 6) (x + 2)

Question 29.
Solution:
3x² + 10x + 8
= 3x² + 6x + 4x + 8
{ 3 x 8 = 24, 24 = 6 x 4,10 = 6 + 4}
= 3x(x + 2) + 4(x + 2)
= (x + 2) (3x + 4) Ans.

Question 30.
Solution:
3y² + 14y + 8
= 3y² + 12y + 2y + 8
{3 x 8 = 24, 24 = 12 x 2, 14 = 12 + 2}
= 3y (y + 4) + 2 (y + 4)
= (y + 4) (3y + 2)

Question 31.
Solution:
3z² – 10z + 8
= 3z² – 6z – 4z + 8
{ 3 x 8 = 24, 24 = ( – 6)x( – 4), – 10 = – 6 – 4}
= 3z (z – 2) – 4 (z – 2)
= (z – 2) (3z – 4)

Question 32.
Solution:
2x² + x – 45
= 2x² + 10x – 9x – 45
{2 x ( – 45) = – 90,- 90 = (10)x( – 9), 1 = 10 – 9}
= 2x (x – 5) + 9 (x – 5)
= (x – 5) (2x + 9)

Question 33.
Solution:
6p² + 11p – 10
= 6x² + 15x – 4x – 10
{6x ( – 10) = 60, – 60 = 15 x ( – 4),11 = 15 – 4}
= 3x (2x + 5) – 2 (2x + 5)
= (3x – 2) (2x + 5) Ans.

Question 34.
Solution:
2x² – 20x + 3x – 30
{2 x ( – 30) = – 60, – 17 = – 20 + 3, – 60 = – 20 x 3}
= 2x (x – 10) + 3 (x – 10)
= (x – 10) (2x + 3) Ans.

Question 35.
Solution:
7y² -19y – 6
= 7y² – 21 y + 2y – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7y(y – 3) + 2 (y – 3)
= (y – 3) (7y + 2)

Question 36.
Solution:
28 – 31x – 5x²
= 28 – 35x + 4x – 5x²
{28 x ( – 5)= – 140, -140 = – 35 x 4, – 31 = – 35 + 4}
= 7 (4 – 5x) + x (4 – 5x)
= (4 – 5x) (7 + x)

Question 37.
Solution:
3 + 23z – 8z²
= 3 + 24z – z – 8z²
{3 x ( – 8) = – 24, – 24 = 24 x ( – 1), 23 = 24 – 1}
= 3 (1 + 8z) – z (1 + 8z)
= (1 + 8z) (3 – z)

Question 38.
Solution:
6x² – 5x – 6
= 6x² – 9x + 4x – 6
{ 6 x ( – 6) = – 36,- 36 = – 9 x 4, – 5 = – 9 + 4}
= 3x (2x – 3) + 2 (2x – 3)
= (2x – 3) (3x + 2)

Question 39.
Solution:
3m² + 24m + 36
= 3 {m² + 8m + 12}
= 3 {m² + 6m + 2m + 12)
{12 = 6 x 2, 8 = 6 + 2}
= 3 {m (m + 6) + 2 (m + 6)}
= 3 (m + 6) (m + 2)

Question 40.
Solution:
4n² – 8n + 3
= 4n² – 6n – 2n + 3
{4 x 3 = 12, 12 = ( – 6)x( – 2), – 8 = – 6 – 2}
= 2n (2n – 3) – 1 (2n- 3)
= (2n – 3) (2n – 1)

Question 41.
Solution:
6x² – 17x – 3
= 6x² – 18x + x – 3
{6 x ( – 3)= – 18, – 18 = – 18 x 1, – 17 = – 18 + 1}
= 6x (x – 3) + 1 (x – 3)
= (x – 3) (6x + 1)

Question 42.
Solution:
7x² – 19x – 6
7x² – 19x – 6
= 7x² – 21x + 2x – 6
{7 x ( – 6) = – 42, – 42 = – 21 x 2, – 19 = – 21 + 2}
= 7x (x – 3) + 2 (x – 3)
= (x – 3) (7x + 2)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7C

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² + 8x + 16
= (x)² + 2 × x × 4 + (4)²
= (x + 4)²

Question 2.
Solution:
x² + 14x + 49
= (x)² + 2 × x × 7 + (7)²
= (x + 7)² Ans.

Question 3.
Solution:
1 + 2x + x²
= (1)² + 2 × 1 × x + (x)²
= (1 + x)² Ans.

Question 4.
Solution:
9 + 6z + z²
= (3)² + 2 x 3 x z + (z)²
= (3 + z)² Ans.

Question 5.
Solution:
x² + 6ax + 9a²
= (x)² + 2 × x × 3a + (3a)²
= (x + 3a)² Ans.

Question 6.
Solution:
4y² + 20y + 25
= (2y)² + 2 x 2y x 5 + (5)²
= (2y² + 5)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 7.
Solution:
36a² + 36a + 9
= 9 [4a² + 4a + 1]
= 9 [(2a)² + 2 x 2a x 1 + (1)²]
= 9 [2a + 1]²

Question 8.
Solution:
9m² + 24m + 16
= (3m)² + 2 x 3m x 4 + (4)²
= (3m + 4)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 9.
Solution:
z² + z + \(\\ \frac { 1 }{ 4 } \)
= (z)² + 2 x z x \(\\ \frac { 1 }{ 2 } \) + \({ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\)
= \({ \left( z+\frac { 1 }{ 2 } \right) }^{ 2 }\)

Question 10.
Solution:
49a² + 84ab + 36b²
= (7a)² + 2 x 7a x 6b + (6b)²
{ ∵ a² + 2ab + b² = (a + b)²}
= (7a + 6b)²

Question 11.
Solution:
p² – 10p + 25
= (p)² – 2 x p x 5 + (5)²
= (p – 5)²
{ ∵ a² – 2ab + b² = (a – b)²}

Question 12.
Solution:
121a² – 88ab + 16b²
= (11a)² – 2 x 11a x 4b + 4(b)²
= (11a – 4b)²

Question 13.
Solution:
1 – 6x + 9x²
= (1)² – 2 x 1 x 3x + (3x)²
= (1 – 3x)²
{ ∵ a² – 2ab + b² = (a – b)²}

Question 14.
Solution:
9y² – 12y + 4
= (3y)² – 2 x 3y x 2 + (2)²
{ ∵ a² – 2ab + b² = (a – b)²}
= (3y – 2)²

Question 15.
Solution:
16x² – 24x + 9
= (4x)² – 2 x 4x x 3 + (3)²
= (4x – 3)² Ans.

Question 16.
Solution:
m² – 4mn + 4n²
= (m)² -2 x m x 2n + (2n)²
= (m – 2n)² Ans.

Question 17.
Solution:
a²b² – 6abc + 9c²
= (ab)² – 2 x ab x 3c + (3c)²
= (ab – 3c)² Ans.

Question 18.
Solution:
m⁴ + 2m²n² + n⁴
= (m²)² + 2m²n² + (n²)²
= (m² + n²)²
{ ∵ a² + 2ab + b² = (a + b)²}

Question 19.
Solution:
(l + m)² – 4lm
= l² + m² + 2lm – 4lm
= l² + m² – 2lm
= l² – 2lm + m²
= (l – m)²
{ ∵ a² – 2ab + b² = (a – b)}

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7B

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Question 1.
Solution:
x² – 36
= (x)² – (6)² { ∵ a² – b² = (a + b) (a – b)}
= (x + 6) (x – 6) Ans.

Question 2.
Solution:
4a² – 9
= (2a)² – (3)²
= (2a + 3) (2a – 3)
{ ∵ a² – b² = (a + b) (a – b)}

You can also Download NCERT Solutions for Class 8 English to help you to revise complete Syllabus and score more marks in your examinations.

Question 3.
Solution:
81 – 49x²
= (9)² – (7x)²
= (9 + 7x) (9 – 7x) { ∵ a² – b² = (a + b) (a – b)}

Question 4.
Solution:
= (2x)² – (3y)²
= (2x + 3y)(2x-3y)
{∵ a² – b² = (a + b) (a – b)}

Question 5.
Solution:
Using a² – b²
= (a + b) (a – b)
= 16a² – 225b²
= (4a)² – (15b)²
= (4a + 15b) (4b – 5b)

Question 6.
Solution:
Using a² – b²
= (a + b) (a – b)
= (3ab)² – (5)²
= (3ab + 5) (3ab – 5)

Question 7.
Solution:
Using a² – b² = (a + b) (a – b)
16a² – 144 = (4a)² = (12)²
= (4a + 12) (4a – 12)
= 4 (a + 3) x 4 (a – 3)
= 16 (a + 3) (a – 3)

Question 8.
Solution:
63a² – 112b²
= 7 (9a² – 16b²)
= 7 [(3a)² – (4b)²
= 7 (3a + 4b) (3a – 4b)

Question 9.
Solution:
20a² – 45b²
= 5 {4a² – 9b²}
= 5{(2a)² – (3b)²)
= 5(2a + 3b) (2a – 3b) Ans.

Question 10.
Solution:
12x² – 27
= 3(4x² – 9)
= 3{(2x)² – (3)²}
= 3(2x + 3) (2x – 3) Ans.

Question 11.
Solution:
x³ – 64x
= x(x² – 64)
= x{(x)² – (8)²}
= x(x + 8) (x – 8) Ans.

Question 12.
Solution:
16x⁵ – 144x³
= 16x³ [x² – 9]
= 16x³ [(x)² – (3)²]
= 16x³ (x + 3) (x – 3)

Question 13.
Solution:
3x⁵ – 48x³
= 3x³ {x² – 16}
= 3x³{(x)² – (4)²}
= 3x³ (x + 4) (x – 4) Ans.

Question 14.
Solution:
16p³ – 4p
= 4p [4p² – 1]
= 4p ((2p)² – (1)²]
= 4p(2p + 1)(2p – 1)

Question 15.
Solution:
63a²b² – 7
= 7(9a²b² – 1)
= 7{(3ab)² – (1)²)
= 7(3ab + 1) (3ab – 1) Ans.

Question 16.
Solution:
1 – (b – c)²
= (1)² – (b – c)²
= (1 + b + c) (1 – b + c) Ans.
{ ∵ a² – b² = (a + b) (a – b)}

Question 17.
Solution:
(2a + 3b)² – 16c²
= (2a + 3b)² – (4c)²
=(2a + 3b + 4c)(2a + 3b – 4c)Ans.
{ ∵ a² – b² = (a + b)(a – b)}

Question 18.
Solution:
(l + m)² – (l – m)²
= (l + m + l – m)(l + m – l + m)
{ ∵ a² – b² = (a + b)(a – b)}
= 2l x 2m = 4lm

Question 19.
Solution:
(2x + 5y)² – (1)²
=(2x + 5y + 1)(2x + 5y – 1)
{ ∵ a² – b² = (a + b)(a – b)}

Question 20.
Solution:
36c² – (5a + b)²
= (6c)² – (5a + b)²
{ ∵ a² – b² = (a + b)(a – b)}
= (6c + 5a + b)(6c – 5a – b)

Question 21.
Solution:
(3x – 4y)² – 25z²
= (3x – 4y)² – (5z)²
= (3x – 4y + 5z) (3x – 4y – 5z) Ans.

Question 22.
Solution:
x² – y² – 2y – 1
= x² – (y² + 2y + 1)
= (x)² – (y + 1)²
= (x + y + 1)(x – y – 1)Ans.

Question 23.
Solution:
25 – a² – b² – 2ab
= 25 – (a² + b² + 2ab)
= (5)² – (a + b)²
= (5 + a + b)(5 – a – b)Ans.

Question 24.
Solution:
25a² – 4b² + 28bc – 49c²
= 25a² – [4b² – 28bc + 49c²]
{ ∵ a² – 2ab + b² = (a – b)²}
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
{ ∵ (a² – b² = (a + b)(a – b)}
= (5a + 2b – 7c) (5a – 2b + 7c)

Question 25.
Solution:
9a² – b² + 4b – 4
= 9a² – (b² – 4b + 4)
= (3a)² – [(b)² – 2 x b x 2 + (2)²]
= (3a)² – (b – 2)²
{ ∵ a² – 2ab + b² = (a – b)²}
= (3a + b – 2)(3a – b + 2)
{ ∵ a² – b² = (a + b)(a – b)}

Question 26.
Solution:
(10)² – (x – 5)²
= (10)² – (x – 5)²
= (10 + x – 5)(10 – x + 5)
= (5 + x) (15 – x) Ans.

Question 27.
Solution:
{(405)² – (395)²}
= (405)² – (395)²
= (405 + 395) (405 – 395)
{ ∵ a² – b²(a + b) (a – b)}
= 800 x 10 = 8000

Question 28.
Solution:
(7.8)² – (2.2)²
= (7.8 + 2.2) (7.8 – 2.2)
= 10.0 x 5.6
= 56 Ans.

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 8 Solutions Chapter 7 Factorisation Ex 7A

These Solutions are part of RS Aggarwal Solutions Class 8. Here we have given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A.

Other Exercises

• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
• RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

Factorize :

Question 1.
Solution:
(i) 12x + 15 = 3(4x + 5) {HCF of 12 and 15 = 3)
(ii) 14m – 21 = 7(2m – 3) {HCF of 14 and 21 = 7)
(iii) 9n – 12n² = 3n(3 – 4n) Ans.{HCF of 9 and 12 = 3}

Question 2.
Solution:
(i) 16a² – 24ab = 8a(2a – 3b) { HCF of 16 and 24 = 8)
(ii) 15ab² – 20a²b = 5ab(3b – 4a) {HCF of 15 and 20 5}
(iii) 12x²y³ – 21x³y²= 3x²y²(4y – 7x) Ans. {HCF of 12 and 21 = 3)

Question 3.
Solution:
(i) 24x³ – 36x²y = 12x²(2x – 3y) {HCF of 24 and 36 = 12}
(ii) 10x³ – 15x² = 5x²(2x – 3) {HCF of 10, 15 = 5}
(iii) 36x³y – 60x²y³z = 12x²y (3x – 5y²z) {HCF of 36 and 60 = 12}

Question 4.
Solution:
(i)9x³ – 6x² + 12x
= 3x(3x² – 2x + 4) {HCF of 9, 6, 12 = 3)
(ii)8x² – 72xy + 12x
= 4x(2x – 18y + 3) (HCF of 8, 72, 124)
(iii)18a³b³ – 27a²b³ + 36a³b²
= 9a²b²(2ab – 3b + 4a) Ans.{HCF of 10, 27, 36 = 9)

Question 5.
Solution:
(i) 14x³ + 21x⁴y – 28x²y²
= 7x² (2x + 3x²y – 4y²) {HCF of 14, 21 28 = 7)
(ii) – 5 – 10t + 20t²
= – 5(1 + 2t – 4t²) Ans. {HCF of 5, 10, 20=5)

Question 6.
Solution:
(i) x(x + 3) + 5(x + 3)
=(x + 3)(x + 5)
(ii) 5x(x – 4) – 7(x – 4)
= (x – 4) (5x – 7)
(iii) 2m (1 – n) + 3(1 – n)
= (1 – n) (2m + 3) Ans.

Question 7.
Solution:
6a(a – 2b) + 5b(a – 2b)
= (a – 2b) (6a + 5b) Ans.

Question 8.
Solution:
x³(2a – b) + x²(2a – b)
=x²(2a – b)(x + 1)Ans.

Question 9.
Solution:
9a(3a – 5b) – 12a²(3a – 5b)
= 3a (3a – 5b) (3 – 4a) Ans.

Question 10.
Solution:
(x + 5)² – 4(x + 5)
= (x + 5)(x + 5 – 4)
=(x + 5)(x + 1)Ans.

Question 11.
Solution:
3(a – 2b)² – 5 (a – 2b)
= (a – 2b) {3(a – 2b) – 5}
= (a – 2b) (3a – 6b – 5) Ans.

Question 12.
Solution:
2a + 6b – 3 (a + 3b)²
= 2(a + 3b) – 3(a + 3b)²
= (a + 3b){ 2 – 3 (a + 3b)}
= (a + 3b) (2 – 3a – 9b) Ans.

Question 13.
Solution:
16(2p – 3q)² – 4 (2p – 3q)
= 4(2p – 3q) {4(2p – 3q) – 1}
= 4(2p – 3q) (8p – 12q – 1) Ans.

Question 14.
Solution:
x(a – 3) + y (3 – a)
= x(a – 3) – y(a – 3)
= (a – 3) (x – y) Ans.

Question 15.
Solution:
12(2x – 3y)² – 16(3y – 2x)
= 12(2x – 3y)² + 16(2x – 3y)
= 4(2x – 3y) {3(2x – 3y) + 4}
= 4(2x – 3y) (6x – 9y + 4) Ans.

Question 16.
Solution:
(x + y)(2x + 5) – (x + y)(x + 3)
= (x + y)(2 + 5 – x – 3)
= (x + y)(x + 2) Ans.

Question 17.
Solution:
ar + br + at + bt
= r(a + b) + t(a + b)
= (a + b) (r + t) Ans.

Question 18.
Solution:
x² – ax – bx + ab
= x(x – a) – b(x – a)
= (x – a)(x – b)Ans.

Question 19.
Solution:
ab² – bc² – ab + c²
= ab² – ab – bc² + c²
= ab(b – 1) – c²(b – 1)
= (b – 1) (ab – c²) Ans.

Question 20.
Solution:
x² – xz + xy – yz
= x(x – z) + y(x – z)
= (x – z)(x + y)Ans.

Question 21.
Solution:
6ab – b² + 12ac – 2bc
6ab + 12ac – b² – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b) Ans.

Question 22.
Solution:
(x – 2y)² + 4x – 8y
= (x – 2y)² + 4(x – 2y)
= (x – 2y)(x – 2y + 4)Ans.

Question 23.
Solution:
y² – xy(1 – x) – x³
y² – xy + x²y – x²
= y(y – x) + x² (y – x)
= (y – x)(y + x²)Ans.

Question 24.
Solution:
(ax + by)² + (bx – ay)²
= a²x² + b²y² + 2abxy + b²x² + a²y² – 2abxy
= a²x² + b²y² + b²x² + a²y²
= a²x² + b²x² + a²y² + b²y²
= x²(a² + b²) + y²(a² + b²)
= (a² + b²) (x² + y²) Ans.

Question 25.
Solution:
ab² + (a – 1)b – 1
= ab² + ab – b – 1
= ab(b + 1) – 1(b + 1)
= (b + 1) (ab – 1) Ans.

Question 26.
Solution:
x³ – 3x² + x – 3
= x²(x – 3) + 1(x – 3)
(x – 3)(x² + 1)Ans.

Question 27.
Solution:
ab (x² + y²) – xy (a² + b²)
= abx² + aby² + xya² – xyb²
= abx² – xya² – xyb² + aby2
=ax(bx – ay) – by(bx – ay)
= (bx – ay) (ax – by) Ans.

Question 28.
Solution:
x² – x (a + 2b) + 2ab
= x² – xa – 2bx + 2ab
= x(x – a) – 2b(x – a)
= (x – a)(x – 2b)

Hope given RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.