# DAV Class 8 Maths Chapter 7 Worksheet 6 Solutions

The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 7 Worksheet 6 Solutions of Algebraic Identities offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 7 WS 6 Solutions

Question 1.
Factorize: x2 + z2 – 2xz.
Solution:
x2 + z2 – 2xz = (x)2 + (z)2 – 2 × x × z
[Using a2 + b2 – 2ab = (a – b)2]
= (x – z)2
= (x – z) (x – z)

Question 2.
Factorize: 4x2 + 9y2 – 12xy.
Solution:
4x2 + 9y2 – 12xy = (2x)2 + (3y)2 – 2 × 2x × 3y
[Using a2 + b2 – 2ab = (a – b)2]
= (2x – 3y)2
= (2x – 3y) (2x – 3y).

Question 3.
Factorize: 64a2 + 49b2 + 112ab.
Solution:
64a2 + 49b2 + 112ab = (8a)2 + (7b)2 + 2 × 8a × 7b
[Using a2 + b2 + 2ab = (a – b)2]
= (8a + 7b)2
= (8a + 7b) (8a + 7b).

Question 4.
Factorize: 121p2 + 16q2 – 88pq.
Solution:
121p2 + 16q2 – 88pq = (11p)2 + (4q)2 – 2 × 11p × 4q
[Using a2 + b2 – 2ab = (a – b)2]
= (11p – 4q)2
= (11p – 4q) (11p – 4q).

Question 5.
Factorize: 9x2y – 24xy2 + 16y3.
Solution:
9x2y – 24xy2 + 16y3
= y [9x2 – 24xy + 16y2]
= y [(3x)2 – 2 × 3x × 4y + (4y)2]
[Using a2 – 2ab + b2 = (a – b)2]
= y (3x – 4y)2
= y (3x – 4y) (3x – 4y).

Question 6.
Factorize: 2a3 + 4a2 b + 2ab2 .
Solution:
2a3 + 4a2b + 2ab2 = 2a[a2 + 2ab + b2]
[Using a2 + 2ab + b2 = (a + b)2]
= 2a (a + b)2
= 2a (a + b) (a + b).

Question 7.
Factorize: 50a2 + 98b2 + 140ab.
Solution:
50a2 + 98b2 + 140ab = 2[25a2 + 49b2 + 70ab]
= 2[(5a)2 + (7b)2 + 2 × 5a × 7b]
[Using a2 + b2 + 2ab = (a + b)2]
= 2 (5a + 7b)2
= 2 (5a + 7b) (5a + 7b).

Question 8.
Factorize: 64x2 – 81y2
Solution:
64x2 – 81y2 = (8x)2 – (9y)2
= (8x + 9y) (8x – 9y)
[Using a2 – b2 = (a + b) (a – b)]

Question 9.
Factorize: 25p2 – 9q2
Solution:
25p2 – 9q2 = (5p)2 – (3q)2
= (5p + 3q) (5q – 3p).
[Using a2 – b2 = (a + b) (a – b)]

Question 10.
Factorize: 16a2b – 64b3.
Solution:
16a2b – 64b3= 16b (a2 – 4b)2
= 16b [(a)2 – (2b)2]
[Using (a2 – b2) = (a + b) (a – b)]
= 16b (a + 2b) (a – 2b)

Question 11.
Factorize: 25x3y3 – 49xy
Solution:
25xy – 49xy = xy [25x2y2 – 49]
= xy [(5xy)2 – (7)2]
[Using (a2 – b2) = (a + b) (a – b)]
= xy (5xy + 7) (5xy – 7).

Question 12.
Factorize: p4 – 256.
Solution:
p4 – 256 = (p2)2 – (16)2
= (p2 – 16) (p2 + 16)
[Using (a2 – b2) = (a + b) (a – b)]
= [(p)2 – (4)2] (p2 + 16)
= (p – 4) (p + 4) (p2 + 16).

Question 13.
Factorize: a2 – (b – c)2.
Solution:
a2 – (b – c)2 = [a + (b – c)] [a – (b – c)]
= (a + b – c)(a – b + c)
[Using a2 – b2 = (a + b) (a – b)]

Question 14.
Factorize: 25m2 – (4n + 3l)2.
Solution:
25m2 – (4n + 3l)2 = (5m)2 – (4n + 3l)2
[Using a2 – b2 = (a + b) (a – b)]
= [5m + (4n + 3l)] [5m – (4n + 3l)]
= [5m + 4n + 3l] [5m – 4n – 3l].

Question 15.
Factorize: (2a + 3b)2 – 4c2.
Solution:
(2a + 3b)2 – 4c2
[Using a2 – b2 = (a + b) (a – b)]
= (2a + 3b)2 – (2c)2
= (2a + 3b + 2c) (2a + 3b – 2c).

Question 16.
Factorize: (64m2 – 144mn + 81n2) – 25p2.
Solution:
(64m2 – 144mn + 81n2) – 25p2 = [(8m)2 – 2 × 8m × 9n + (9n)2] – (5p)2
= (8m – 9n)2 – (5p)2
[Using a2 – 2ab + b2 = (a – b)2]
= (8m – 9n + 5p) (8m – 9n – 5p)
[Using a2 – b2 = (a + b) (a – b)]

Question 17.
Factorize: 16x2 + 9y2 + 4z2 + 24xy + 12yz + 16zx
Solution:
16x2 + 9y2 + 4z2 + 24xy + 12yz + 16zx = (4x)2 + (3y)2 + (2z)2 + 2 × 4x × 3y + 2 × 3y × 2z + 2 × 4x × 2z
= (4x + 3y + 2z)2
[Using a2 + b2 + c2 + 2ab + 2bc + 2ac = (a + b + c)2]
= (4x + 3y + 2z) (4x + 3y + 2z).

Question 18.
Factorize: x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx
Solution:
x2 + 4y2 + 9z2 – 4xy + 12yz – 6zx
= (- x)2 + (2y)2 + (3z)2 – 2x × 2y + 2 × 2y × 3z – 2 × 3z × x
= (- x + 2y + 3z)2
[Using a2 + b2 + c2 + 2ab + 2bc + 2ac = (a + b + c)2]
= (- x + 2y + 3z) (- x + 2y + 3z).

Question 19.
Factorize: 4a2 + b2 + 25c2 – 4ab – 10bc + 20ca
Solution:
4a2 + b2 + 25c2 – 4ab – 10bc + 20ca
= (2a)2 + (- b)2 + (5c)2 + 2 (2a) (- b) + 2 (- b) (5c) + 2 (5c) (2a)
= (2a – b + 5c)2
[Using a2 + b2 + c2 + 2ab + 2bc + 2ac = (a + b + c)2]
= (2a – b + 5c) (2a – b + 5c).

Question 20.
Factorize: a2 + $$\frac{b^2}{4}+\frac{c^2}{9}$$ + ab + $$\frac{b c}{3}+\frac{2}{3} c a$$
Solution:
a2 + $$\frac{b^2}{4}+\frac{c^2}{9}$$ + ab + $$\frac{b c}{3}+\frac{2}{3} c a$$
= (a)2 + $$\left(\frac{b}{2}\right)^2+\left(\frac{c}{3}\right)^2+2 \times a \times \frac{b}{2}+2 \times \frac{b}{2} \times \frac{c}{3}+2 \times a \times \frac{c}{3}$$
= $$\left(a+\frac{b}{2}+\frac{c}{3}\right)^2$$
[Using a2 + b2 + c2 + 2ab + 2bc + 2ca = (a + b + c)2]
= $$\left(a+\frac{b}{2}+\frac{c}{3}\right)\left(a+\frac{b}{2}+\frac{c}{3}\right)$$