Question

factorise
$factorise 3y^3-48y$
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Answer 1

Answer :

• We have to factorise these numbers.

= 3y³ - 48y

• Take 3y as common

= 3y ( y² - 16 )

= 3y ( y² - 4² )

• Use the identity a² - b² = (a - b) (a + b)

= 3y ( y - 4)( y + 4) $\:$

•°• The factors are 3y ( y - 4)( y + 4).

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Answer 2

3y^3- 48y
Here, we can write 48y as 3 x 16 x y
= 3 x y x y x y - 3 x 16 x y
So, we should take 3y as common.
= 3y( y^2 - 16)
so, 16 is also known as 4^2
= 3y(y^2 - 4^2)
Now, we should do a^2-b^2= (a+b)(a-b)
Here, a= y ; b= 4
Therefore, (y+4)(y-4).

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