Sorry for the mistake in previous question.
Factorise :
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Answers
Answered by
8
Hola there..
Given
=> 9x²y - 24xy² + 16y³
=> 9x²y - 12xy² - 12xy² + 16y³
=> 3xy(3x - 4y) - 4y²(3x - 4y)
=> (3x - 4y)(3xy - 4y²)
=> (3x - 4y)y(3x - 4y)
=> y(3x - 4y)². ....Ans
Hope this helps....:)
Given
=> 9x²y - 24xy² + 16y³
=> 9x²y - 12xy² - 12xy² + 16y³
=> 3xy(3x - 4y) - 4y²(3x - 4y)
=> (3x - 4y)(3xy - 4y²)
=> (3x - 4y)y(3x - 4y)
=> y(3x - 4y)². ....Ans
Hope this helps....:)
AkshithaZayn:
so, splitting the middle term is used. right?can we do it by any identity?
Answered by
11
Hi friend
---------------
Your answer
--------------------
★ FACTORISATION
-------------------------------
9x²y - 24xy + 16y³
Now ,
---------
Taking y common from the three terms , we get,
y(9x² - 24xy + 16y²)
Now, we can factorise by splitting the middle term.
y(9x² - 12xy - 12xy + 16y²)
=> y[3x(3x - 4y) - 4y(3x - 4y)]
=> y(3x - 4y)(3x - 4y)
=> y(3x - 4y)²
Alternately : -
We can solve by using the identity => (a - b)² = a² - 2ab + b².
=> y(9x² - 24xy + 16y²)
=> y[(3x)² - 2 × (3x) × (4y) + (4y)²]
=> y(3x - 4y)²
HOPE IT HELPS
---------------
Your answer
--------------------
★ FACTORISATION
-------------------------------
9x²y - 24xy + 16y³
Now ,
---------
Taking y common from the three terms , we get,
y(9x² - 24xy + 16y²)
Now, we can factorise by splitting the middle term.
y(9x² - 12xy - 12xy + 16y²)
=> y[3x(3x - 4y) - 4y(3x - 4y)]
=> y(3x - 4y)(3x - 4y)
=> y(3x - 4y)²
Alternately : -
We can solve by using the identity => (a - b)² = a² - 2ab + b².
=> y(9x² - 24xy + 16y²)
=> y[(3x)² - 2 × (3x) × (4y) + (4y)²]
=> y(3x - 4y)²
HOPE IT HELPS
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