## 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**

- RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7A
- RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7B
- RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7C
- RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7D
- RS Aggarwal Solutions Class 8 Chapter 7 Factorisation Ex 7E

**Question 1.**

**Solution:**

x^{2} + 8x + 16

= (x)^{2} + 2 × x × 4 + (4)^{2}

= (x + 4)^{2}

**Question 2.**

**Solution:**

x^{2} + 14x + 49

= (x)^{2} + 2 × x × 7 + (7)^{2}

= (x + 7)^{2} Ans.

**Question 3.**

**Solution:**

1 + 2x + x^{2}

= (1)^{2} + 2 × 1 × x + (x)^{2}

= (1 + x)^{2} Ans.

**Question 4.**

**Solution:**

9 + 6z + z^{2}

= (3)^{2} + 2 x 3 x z + (z)^{2}

= (3 + z)^{2} Ans.

**Question 5.**

**Solution:**

x^{2} + 6ax + 9a^{2}

= (x)^{2} + 2 × x × 3a + (3a)^{2}

= (x + 3a)^{2} Ans.

**Question 6.**

**Solution:**

4y^{2} + 20y + 25

= (2y)^{2} + 2 x 2y x 5 + (5)^{2}

= (2y^{2} + 5)^{2}

{ ∵ a^{2} + 2ab + b^{2} = (a + b)^{2}}

**Question 7.**

**Solution:**

36a^{2} + 36a + 9

= 9 [4a^{2} + 4a + 1]

= 9 [(2a)^{2} + 2 x 2a x 1 + (1)^{2}]

= 9 [2a + 1]^{2}

**Question 8.**

**Solution:**

9m^{2} + 24m + 16

= (3m)^{2} + 2 x 3m x 4 + (4)^{2}

= (3m + 4)^{2}

{ ∵ a^{2} + 2ab + b^{2} = (a + b)^{2}}

**Question 9.**

**Solution:**

z^{2} + 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:**

49a^{2} + 84ab + 36b^{2}

= (7a)^{2} + 2 x 7a x 6b + (6b)^{2}

{ ∵ a^{2} + 2ab + b^{2} = (a + b)^{2}}

= (7a + 6b)^{2}

**Question 11.**

**Solution:**

p^{2} – 10p + 25

= (p)^{2} – 2 x p x 5 + (5)^{2}

= (p – 5)^{2}

{ ∵ a^{2} – 2ab + b^{2} = (a – b)^{2}}

**Question 12.**

**Solution:**

121a^{2} – 88ab + 16b^{2}

= (11a)^{2} – 2 x 11a x 4b + 4(b)^{2}

= (11a – 4b)^{2}

**Question 13.**

**Solution:**

1 – 6x + 9x^{2}

= (1)^{2} – 2 x 1 x 3x + (3x)^{2}

= (1 – 3x)^{2}

{ ∵ a^{2} – 2ab + b^{2} = (a – b)^{2}}

**Question 14.**

**Solution:**

9y^{2} – 12y + 4

= (3y)^{2} – 2 x 3y x 2 + (2)^{2}

{ ∵ a^{2} – 2ab + b^{2} = (a – b)^{2}}

= (3y – 2)^{2}

**Question 15.**

**Solution:**

16x^{2} – 24x + 9

= (4x)^{2} – 2 x 4x x 3 + (3)^{2}

= (4x – 3)^{2} Ans.

**Question 16.**

**Solution:**

m^{2} – 4mn + 4n^{2}

= (m)^{2} -2 x m x 2n + (2n)^{2}

= (m – 2n)^{2} Ans.

**Question 17.**

**Solution:**

a^{2}b^{2} – 6abc + 9c^{2}

= (ab)^{2} – 2 x ab x 3c + (3c)^{2}

= (ab – 3c)^{2} Ans.

**Question 18.**

**Solution:**

m^{4} + 2m^{2}n^{2} + n^{4}

= (m^{2})^{2} + 2m^{2}n^{2} + (n^{2})^{2}

= (m^{2} + n^{2})^{2}

{ ∵ a^{2} + 2ab + b^{2} = (a + b)^{2}}

**Question 19.**

**Solution:**

(l + m)^{2} – 4lm

= l^{2} + m^{2} + 2lm – 4lm

= l^{2} + m^{2} – 2lm

= l^{2} – 2lm + m^{2}

= (l – m)^{2}

{ ∵ a^{2} – 2ab + b^{2} = (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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