5y^5 - 405^y plzz solve it fast
Nikki57:
is the question correct?
Tell
yess
ok then its factorisation right?
absolutely
hmm
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Hi friend,
5y^5-405y
5*y*y*y*y*y-5*9*9*y. (taking out common)
5y(y^4-81)
Using identity-3. a^2-b^2=(a+b)(a-b)
5y((y^2)^2-(9)^2)
5y(y^2-9)(y^2+9)
5y(y^2-3^2)(y^2+9)
Again using identity-3
5y(y+3)(y-3)(y^2+9)
Hope this is the answer..........
5y^5-405y
5*y*y*y*y*y-5*9*9*y. (taking out common)
5y(y^4-81)
Using identity-3. a^2-b^2=(a+b)(a-b)
5y((y^2)^2-(9)^2)
5y(y^2-9)(y^2+9)
5y(y^2-3^2)(y^2+9)
Again using identity-3
5y(y+3)(y-3)(y^2+9)
Hope this is the answer..........
ohh sry a smaall mistake
its correct now
ok?
ohhh
its soooooo easy
i understood
:)
thanks to select my answer as brainliest
my pleasure
:)
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