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factorise:50a5-8a3b2
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take \(a^3\) common @Faris_Waleed What do you get?
a3(50a2-8ab2)
good , now take 2 common from 50a2 - 8ab2
2(50a-8ab)
sorry
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i meant 2(25a2-4ab2)
i think it is \(50a^2-8ab^2\)
\[\large{(a^3)(50a^2-8b^2)}\] \[\large{\color{blue}{a^3 (2)(25a^2-4b^2)}}\] \[\large{\color{green}{2a^3(25a^2-4b^2)}}\]
now we have : \[\large{a^2-b^2 = \color{blue}{(a+b)(a-b)}}\] \[\large{(2a^3)\color{green}{(5a)^2-(2b)^2})}\] \[\large{(2a^3)\color{blue}{(5a+2b)(5a-2b)}}\]
yes thats the answer i got in my notebook
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@mathslover is right , forever ;)
thanks @jiteshmeghwal9
you deserved that ;)
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