Factorise the following :
`(i) 4x^(2)+20x+25 " " (ii) 9y^(2)-66yz+121z^(2) " " (iii) (2x+(1)/(3))^(2)-(x-(1)/(2))^(2)`

Explanation

(i) We have,
`4x^(2)+20x+25=(2x)^(2)+2(2x)(5)+(5)^(2)`
`=(2x+5)^(2) " " [because a^(2)+2ab+b^(2)=(a+b)^(2)]`
=(2x+5)(2x+5)
We have,
`9y^(2)-66yz+121z^(2)=(-3y)^(2)+2(-3y)(11z)+(11z)^(2)`
`=(-3y+11z)^(2) " " [because a^(2)+2ab+b^(2)=(a+b)^(2)]`
=(-3y+11z)(-3y+11z)
={-(3y-11z)}{-(3y-11z)}=(3y-11z)(3y-11z)
(iii) Using identity `a^(2)-b^(2)=(a+b)(a-b)`
`(2x+(1)/(3))^(2)-(x-(1)/(2))^(2)=[(2x+(1)/(3))+(x-(1)/(2))][(2x+(1)/(3))-(x-(1)/(2))]`
`=(2x+(1)/(3)+x-(1)/(2))(2x+(1)/(3)-x+(1)/(2))=(3x-(1)/(6))(x+(5)/(6))`