How do you factor completely #2x^3 -32x#?

Answer 1

#2x(x+4)(x-4)#

Look for a common factor between the two terms. If you focus on just the constants, #2# and #32#, it is clear that their greatest common factor is #2#. So, we can "take a #2#" out of both terms in #2x^3-32x#. We can rewrite it as #2(x^3-16x)#. We can also factor out an #x# from both terms: #2x(x^2-16)# We are not done. The term #x^2-16# is a "difference of squares". Differences of squares, like #a^2-b^2#, can be factored into #(a+b)(a-b)#. Therefore, we can factor #x^2-16# into #(x+4)(x-4)#.

So, we can factor the entire term into #2x(x+4)(x-4)#.

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Answer 2

Genesis Allen

To factor completely ( 2x^3 - 32x ), first, factor out the greatest common factor, which is ( 2x ):

[ 2x(x^2 - 16) ]

Now, factor the quadratic expression ( x^2 - 16 ) as the difference of squares:

[ 2x(x - 4)(x + 4) ]

So, ( 2x^3 - 32x ) factors completely to ( 2x(x - 4)(x + 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.