How do you factor completely #3y^3 - 48y #?

Answer 1

Isaiah Anthony

#3y(y-4)(y+4)#

Notice that 3 is a factor of 3 and 48 Also y is a factor of #y^2# and #y#

Factoring out #3y# giving

#3y(y^2-16)#

But 16 is #4^2#

#3y(y^2-4^2)#

Compare this to the example: #a^2+b^2=(a-b)(a+b)#

Using the same principle we have:

#3y(y-4)(y+4)#

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Answer 2

Abigail Allen

To factor completely (3y^3 - 48y), you first need to factor out the greatest common factor, which is (3y). After factoring out (3y), the expression becomes (3y(y^2 - 16)). Then, factor the quadratic expression (y^2 - 16) using the difference of squares formula, which results in ((y - 4)(y + 4)). Therefore, the completely factored form of (3y^3 - 48y) is (3y(y - 4)(y + 4)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.