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Frank Learning Maths Class 8 Solutions Chapter 7 Factorisation
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Frank Learning Maths Class 8 Book, Chapter 7 Factorisation. Here students can easily find step by step solutions of all the problems for Factorisation. Here students will find solutions for Exercise 7.1, 7.2 and 7.3. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here in this post all Mathematic solutions are based on the CBSE latest curriculum.
Factorisation Exercise 7.1 Solution
Question no – (1)
Solution :
(a) 9x2 + 24x + 16
= from of identity, x = a
∴ 9a2 + 24a + 16
= (3a)2 + 2.3a.4 + (4)2
= (3a + 4)2
∴ (3x + 4)2
(b) 4x2y2 – 4xyz + z2
= From of identity, xy = a, z = b
4a2 – 4ab + b2
= (2a)2 – 2.2a.b (b)2
= (2a – b)2
∴ (2xy – z)2
(c) x2 + 2 + 1/x2
= x2 + 2.x.1/x + 1/x2
= from of identify, x = a, 1/x = b
∴ a2 + 2ab + b2
= (a + b)2
∴ (x + 1/x)2
(d) x4 – 10x2y2 + 25y4
= from of identity – x2 = a, y2 = b
∴ a2 – 10ab + 25b2
= a2 – 2.a.5b +(5b)2
= (a – 5b)2
∴ (x2 – 5y2)2
(e) 64a2 – 25b2
∴ (6a)2 – (5b)2
= (6a – 5b) (6a + 5b)
(f) 9x2 + 42x + 49 – y4
= from of identity, x = a, y2 = b
= 9a2 + 42a + 49 – b2
= (3a)2 + 2.3a.7 + (7)2 – b2
= (3a + 7)2 – b2
= (3a + 7 + b) (3a + 7 – b)
= (3a + b + 7) (3a – b + 7)
∴ (3x + y2 + 7) (3x – y2 + 7)
(g) x – y – x2 + y2
Let x = a, y = b
∴ a – b – a2 + b2
= (a – b) – (a2 – b2)
= (a – b) – (a – b) (a + b)
= (a – b) (1 – a – b)
∴ (x – y) (1 – x – y)
Question no – (2)
Solution :
(a) x4 – 16
= (x2)2 – (4)2
= (x2 – 4) (x2 + 4)
= (x2 – 22) (x2 + 22)
= (x + 2) (x – 2) (x2 + 4)
(b) (x + 2y)4 – 1
= {(x + 2y)2}2 – 12
= [(x + 2y)2 – 12] [(x + 2)2 + 1]
= (x + 2y – 1) (x + 2y + 1) [(x + 2)2 + 1]
(c) x8 – 1
= (x4)2 – 12
= (x4 – 12) (x4 + 1)
= (x2 – 1) (x2 +1) (x4 + 1)
= (x – 1) (x + 1) (x2 + 1) (x4 + 1)
(d) (7a – 6b)2 – 16y2
= (7a – 6a + 4y) (7a – 6b – 4y)
(e) 49x2 – 1/144
= (7x)2 – (1/12)2
= (7x + 1/12) (7x – 1/12)
(f) 49 (x + y)2 – 64 (x – y)2
= [ 7 (x + y)]2 – [ 8 (x – y)]2
= ( 7x + 7y + 8x – 8y ) (7x + 7y – 8x + 8y)
= ( 15x – y ) ( -x + 15y )
(g) 24a2 – 54b2
= 6 (4a2 – 9b2)
= 6 [(2a)2 – (3b)2]
= 6 [(2a – 3b) (2a + 3b)]
Factorisation Exercise 7.2 Solution
Question no – (1)
Solution :
(a) x2 + 11x + 24
= x2 + (8 + 3) x + 24
= x2 + 8x + 3x + 24
= x(x + 8) +3 (x + 8)
= (x + 3) (x + 8)
(b) a2 + 5a – 36
= a2 + 9a – 4a – 36
= a(a + 9) -4 (a + 9)
= (a – 4) (a + 9)
(c) x2 + 14x + 48
= x2 + 8x + 6x + 48
= x(x + 8) +6 (x + 8)
= (x + 8) (x + 6)
(d) x2 – 4x – 21
= x2 – 7x + 3x – 21
= x(x – 7) +3 (x – 7)
= (x – 7) (x + 3)
(e) x2 – 11x – 42
= x2 – 14x + 3x – 42
= x(x – 14) +3 (x – 14)
= (x + 3) (x – 14)
(f) x2 + 2x – 24
= x2 + 6x – 4x – 24
= x(x + 6) -4 (x + 6)
= (x – 4) (x + 6)
(g) a2 – 8a + 15
= a2 – 5a – 3a + 15
= a(a – 5) -3 (a – 5)
= (a – 3) (a – 5)
(h) x2 + 8x – 48
= x2 + 12x – 4x – 48
= x (x + 12) -4 (x + 12)
= (x + 12) (x – 4)
(i) 9p2 + 24p + 16
= (3p)2 + 2.3p.4 + (4)2
= (3p + 4)2
= (3p + 4) (3p + 4)
Question no – (2)
Solution :
(a) 6a2 – 13ab + 2b2
= 6a2 – (12 + 1) ab + 2b2
= 6a2 – 12ab – ab + 2b2
= 6a(a – 2b) – b (a – 2b)
= (6a – b) (a – 2b)
(b) 3x2 + 22x + 35
= 3x2 + 15x + 7x + 35
= 3x(x + 5) +7 (x + 5)
= (3x + 7) (x + 5)
(c) 11m2 – 54mn + 63n2
= 11m2 – 33mn – 21mn + 63n2
= 11m(m – 3n) -21n (m – 3n)
= (11m – 2n) (m – 3n)
(d) 2x2 + 19xy + 24y2
= 2x2 + 16xy + 3xy + 24y2
= 2x (x + 8y) +3y (x + 8y)
= (2x + 3y) (x + 8y)
(e) x2 + 3xy – 18y2
= x2 + 6xy – 3xy – 18y2
= x (x + 6y) -3y (x + 6y)
= (x + 6y) (x – 3y)
(f) 6x2 – 7x – 20
= 6x2 – 15x + 8x – 20
= 3x (2x – 5) +4 (2x – 5)
= (3x + 4) (2x – 5)
(g) 2a2 – 7a + 6
= 2a2 – 4a – 3a + 6
= 2a (a -2) -3 (a – 2)
= (2a – 3) (a – 2)
(h) a2 + a + 1/4
= 4a2 + 4a + 1/4
= 4a2 + 2a + 2a + 1/4
= (2a (2a + 1) +1 (2a + 1)/4
= (2a + 1) (2a + 1)/4
= (2a + 1/2) (2a + 1/2)
= (a + 1/2) (a + 1/2)
(i) 49a6 – 28a3b3 + 4b6
= 49a6 – 14a3b3 – 14a3b3 + 4b6
= 7a3 (7a2 – 2b3) -2b3 (7a2 – 2b3)
= (7a3 – 2b3) (7a2 – 2b3)
Factorisation Exercise 7.3 Solution
Question no – (1)
Solution :
(a) Given, y² + 5y – 36 by y + 9
= y² + 9y – 4y – 36
= y(y + 9) -4 (y + 9)
= (y – 4) (y + 9)
Divide by (y + 9)
∴ (y – 4) (y + 9) / (y + 9)
= (y – 4)
(b) In the question, x2 + 14x + 48 by x + 6
= x2 + 6x + 8x + 48
= x(x + 6) + 8 (x + 6)
= (x + 6) (x + 8)
Divide by (x + 6)
∴ (x + 6) (x + 8) / (x + 6)
= (x + 8)
(c) Given, 36a2 + 12ab – 15b2 by 2a – b
= 3 (12a2 + 4ab – 5b2)
= 3 (12a2 + 10ab – 6ab – 5b2)
= 3 [2a(6a + 5b) –b (6a + 5b)]
= 3 [(2a – b) (6a + 5b)]
Divide by (2a – b)
∴ 3 (2a – b) (6a + 5b) / (2a – b)
= 3 (6a + 5b)
= 18a + 15b
(d) Given,5x2 + 31x – 28 by 5x – 4
= 5x2 + 35x – 4x – 28
= 5x(x + 7) – 4 (x + 7)
= (5x – 4) (x + 7)
Divided by (5x – 4)
∴ (5x – 4) (x+ 7)/(5x – 4)
= x + 7
(e) Given, 3x2 + 10x + 3 by 3x + 1
= 3×2 + 9x + x + 3
= 3x(x + 3) +1 (x + 3)
= (3x + 1) (x + 3)
Divided by (3x + 1)
∴ (3x + 1) (x + 3) / (3x + 1)
= (x + 3)
(f) Given, x2 – 14x – 51 by x + 3
= x2 – 17x + 3x – 51
= x(x – 17) +3 (x – 17)
= (x – 17) (x + 3)
Divided by (x + 3)
∴ (x + 7) (x + 3) / (x + 3)
= (x + 7)
Question no – (2)
Solution :
(a) 6x^2 – 31x + 47 by 2x – 5
∴ quotient = 3x + 8
∴ remainder = 87
(b) x^2 + 12x + 38 by x + 7
∴ quotient = x + 5
∴ remainder = 3
(c) 15x^2 + x – 6 by 3x + 2
∴ quotient = 5x – 3
(d) 4a^3 + 8a^2 + 24 by a + 4
∴ quotient = 4a^2 – 8a + 32
∴ remainder = 104
(e) 2x^3 – 5x^2 + 8x – 5 by 2×2 – 3x + 5
∴ quotient = x – 1
(f) x^5 + 3x^4 – 5x^3 + 14x^2 + 30x – 16 by 2x – 5
∴ quotient = 1/2x^4 + 11/4x^3 + 35/8x^2 – 63/16x + 165/32
∴ remainder = 313/32
Question no – (3)
Solution :
(a) x^2 – 4x + 8, x – 2
∴ No
(b) 2x^3 + x^2 – 5x – 3, 2x + 3
∴ Yes
(c) 8x^4 + 10x^3 – 5x^2 – 4x + 1, 2x^2 + x – 1
∴ No
Previous Chapter Solution :
