Maths Ace Class 8 Solutions Chapter 7

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.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 :

Updated: June 16, 2023 — 6:23 am