## RS Aggarwal Class 8 Math Seventh Chapter Factorisation Exercise 7C Solution

## EXERCISE 7C

#### Factorise:

### FORMULA:

### (i) a^{2} + 2ab + b^{2} = (a + b)^{2}

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

**(1) x ^{2} + 8x + 16**

= (x)^{2} + (2 × x × 4) + (4)^{2}

= (x + 4)^{2}

**(2) x ^{2} + 14x + 49**

= (x)^{2} + (2 × x × 7) + (7)^{2}

= (x + 7)^{2}

**(3) 1 + 2x + x ^{2}**

= (1)^{2} + (2 × 1 × x) + (x)^{2}

= (1 + x)^{2}

**(4) 9 + 6z + z ^{2}**

= (3)^{2} + (2 × 3 × z) + (z)^{2}

= (3 + z)^{2}

**(5) x ^{2} + 6ax + 9a^{2}**

= (x)^{2} + (2 × x × 3a) + (3a)^{2}

= (x + 3a)^{2}

**(6) 4y ^{2} + 20y + 25**

= (2y)^{2} + (2 × 2y × 5) + (5)^{2}

= (2y + 5)^{2}

**(7) 36a ^{2} + 36a + 9**

= (6a)^{2} + (2 × 6a × 3) + (3)^{2}

= (6a + 3)^{2}

**(8) 9m ^{2} + 24m + 16**

= (3m)^{2} + (2 × 3m × 4) + (4)^{2}

= (3m + 4)^{2}

**(10) 49a ^{2} + 84ab + 36b^{2}**

= (7a)^{2} + (2 × 7a + 6b) + (6b)^{2}

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

**(11) p ^{2} – 10p + 25**

= (p)^{2} – (2 × p × 5) + (5)^{2}

= (p – 5)^{2}

**(12) 121a ^{2} – 88ab + 16b^{2}**

= (11a)^{2} – (2 × 11a × 4b) + (4b)^{2}

= (11a – 4b)^{2}

**(13) 1 – 6x + 9x ^{2}**

= (1)^{2} – (2 × 1 × 3x) + (3x)^{2}

= (1 – 3x)^{2}

**(14) 9y ^{2} – 12y + 4**

= (3y)^{2} – (2 × 3y × 2) + (2)^{2}

= (3y – 2)^{2}

**(15) 16x ^{2} – 24x + 9**

= (4x)^{2} – (2 × 4x × 3) + (3)^{2}

= (4x – 3)^{2}

**(16) m ^{2} – 4mn + 4n^{2}**

= (m)^{2} – (2 × m × 2n) + (2n)^{2}

= (m – 2n)^{2}

**(17) a ^{2}b^{2} – 6abc + 9c^{2}**

= (ab)^{2} – (2 × ab × 3c) + (3c)^{2}

= (ab – 3c)^{2}

**(18) m ^{4} + 2m^{2}n^{2} + n^{4}**

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

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

**(19) (l + m) ^{2} – 4lm**

= l^{2} + 2lm + m^{2} – 4lm

= l^{2} – 2lm + m^{2}

= (l – m)^{2}