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

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.

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