Question

divide : 4a2+12ab+9b2-25c2 by 2a+3b+5c

Answer 1

here's your answer.,....

Answer 2

$\frac{4a^{2}+12ab+9b^{2}-25c^{2}}{2a+3b+5c}$

$= \frac{[(2a)^{2}+2\times 2a \times 3b+(3b)^{2}]-(5c)^{2}}{2a+3b+5c}$

$= \frac{(2a+3b)^{2} - (5c)^{2}}{(2a+3b+5c) }$

$\blue{ ( By \: Algebraic \:Identity )}$

$\boxed { \pink { x^{2} + 2xy + y^{2} = (x+y)^{2} }}$

$= \frac{ [(2a+3b) + 5c][(2a+3b)-5c]}{(2a+3b+5c)}$

$\blue{ ( By \: Algebraic \:Identity )}$

$\boxed { \pink { x^{2} - y^{2}= (x+y)(x-y) }}$

$= \frac{ (2a+3b + 5c)(2a+3b -5c)}{(2a+3b+5c)}$

/* After cancellation ,we get */

$= (2a+3b-5c)$

Therefore.,

$\red {\frac{4a^{2}+12ab+9b^{2}-25c^{2}}{2a+3b+5c}} \green {= (2a+3b-5c)}$

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