Question

Factorize using suitable identities:
a. 36a²+120ab + 100b²​

Answer 1

a. = (6a)²+2×6a×10b+(10b)²

= (6a)²+(10b)²

= (6a+10b)²

hope it helps uh!!! Ãmî ❤️

Answer 2

$36 {a}^{2} + 120ab + 100 {b}^{2} \\ \\ = > {(6a)}^{2} + 2.(6a).(10b) + {(10b)}^{2}$

As we know that :

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

Hence,

$= > {(6a + 10b)}^{2}$

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Hope it helps ☺

Fóllòw Më ❤