Maths Ace Class 8 Solutions Chapter 7

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Maths Ace Class 8 Solutions Chapter 7 Factorisation

Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 7, Factorisation. Here students can easily find step by step solutions of all the problems for Factorisation, Exercise 7.1 and 7.2 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 7 solutions.

Factorisation Exercise 7.1 Solution :

Question no – (1)

Solution : 

(a) 12xy and 3x

= 12xy, 3x

= 12xy = 3 × 4 × x × y

= 3xy = 3 × x

∴ H.C.F is 3xy

(b) 54xyz and 12x²y²

= 54xyz = 3 × 3 × 3 × 2 × x × y × z

= 12x²y² = 3 × 2 × 2 × x × x × y × y

∴ H.C.F is = 6xy

(c) 21x²y²z and 7xyz

= 21x²y²z² = 7 × 3 × x × x × y × y × y × z × z

Now, 7xyz = 7 × x × y × z

∴ H.C.F  is 7xyz

(d) 3a²b³c³, 9a³b³c³ and 18a²b²c²

= 3a²b³c³ = 3 × a × a × b × b × b × c × c × c

9a³b³c³ = 3 × 3 × a × a × a × b × b × b × c × c × c

18a² b²c² = 3 × 2 × 3 × a × a × b  × b × c × c

∴ H.C.F = 3 × a × a × b × b × c × c

= 3ab²c²

(e) 6abc, 7ab³c and 8abc³

= 6Abc = 2 × 3 × a × b  × c

= 7ab3c = 7 × a × b × b × b × c

= 8abc³ = 2 × 2 × 2 × a × b × c × c × c

∴ H.C.F = abc

Question no – (2) 

Solution : 

(a) b²m + b²n

= b² (m + n)

(b) 54x²y + 24xy

= 6xy (9x + 4)

(c) 10xy – 5y + 8 – 16x

= (10xy – 5y) + (8 – 16x)

= 5y (2x – 1) + 8 (1 – 2x)

= 5y (2x – 1) – 8 (2x – 1)

= (2x – 1) (54 – 8)

(d) 48x³y + 36x²y² + 24x³y³

= 12x²y (4x + 3y + 2xy²)

(e) x²yz + 4x³y – 24x³

= x² (yz + 4xy – 24x)

Factorisation Exercise 7.2 Solution :

Question no – (1) 

Solution : 

(a) x² + 12x + 36

= x² + 6x + 6x + 36

= x (x + 6) + 6 (x + 6)

= (x + 6) (x + 6)

(b) p² + 8p + 16

= p² + 4p + 4p + 16

= p (p + 4) + 4 (p + 4)

= (p + 4) (p + 4)

(c) z² – 10z + 25

= z² – 5z – 5z + 25

= z (x – 5) – 5 (z – 5)

= (z – 5) (z – 5)

(d) 24S² + 30S + 9

= 25S² + 15S + 15S + 9

= 25S² (5S + 3) + 3 (5S + 3)

= (5S + 3) + 3 (5S + 3)

(e) 100b² + 60bc + 9c²

= (10b)² – 2.10b.3c (3c)²

= (10b – 3c)²

= (10b – 3c) (10b – 3c)

(f) p⁴ + 2p²q² + q²

= (p²)² + 2.p² × (q²)²

= (p² + q²)

(g) 36z² – 49x²

= (6z)² – (7x)²

= (6z + 7x) (6z – 7x)

(h) 196p² – 121q²

= (14p)² – (112)²

= (14p + 11q) (14p – 11q)

Question no – (2) 

Solution : 

(a) Given, a² – x – 2

= x² – 2x + x – 2

= x (x – 2) – 1 (x – 2)

= (x – 2) (x – 1)

(b) p² + 12p + 35

= p² + 7p + 5p + 35

= p (p + 7) + 5 (p + 7)

= (p + 7) (p + 5)

(c) z² – 3z – 54

= z² – 5z – 5² + 25

= z (z – 5) – 5 (z – 5)

= (z – 5)²

(d) s² + 11s + 30

= s² + 6s + 55 + 30

= s (s + 6) + 5 (s + 6)

= (s + 6) (s + 5)

(e) 2b² + 32b + 128

= 2(b² + 16y + 64)

= 2 {b² + 8b + 8b + 64}

= 2 {b (b + 8) + 8 (b + 8)}

= 2 {(b + 8) (b + 8)

(f) p² + 8p – 84

= p² + 14p – 6p – 84

= p (p + 14) – 6 (p + 14)

= (p – 6) (p + 14)

(g) 4x² – 56x + 196

= 4(x² – 14x + 49)

= 4{x² – 7x – 7x + 49}

= 4 {x (x – 7) – 7 (x – 7)}

= 4 (x – 7)²

(h) r² – 10r + 21

= r² – 7r – 3r + 21

= r (r – 7) – 3 (r – 7)

= (r – 7) (r – 3)

(i) p² – 17p – 38

= p² – 19p + 2p – 38

= p (p – 19) + 2 (p – 19)

= (p – 19) (p + 2)

(j) x² + 30x + 216

= x² + 18x + 12x + 216

= x (x + 18) + 12 (x + 18)

= (x + 18) (x + 12)

Next Chapter Solution : 

👉 Chapter 8 👈