How do you evaluate #(4n - 4) ( 4n ^ { 2} + 2n + 2)#?

Answer 1

#(-64n^3-32n^2-32)#

#(4n-4)(4n^2+2n+2)#

Use the distributive property! Distribute the first set of parenthesis into the second set (make sure to distribute the #4n# to all of the numbers into the second set of parenthesis AND distribute the #-4# as well:

#4n# first:

#(-4)(16n^3+8n^2+8n)#

Now the #-4#:

#(-64n^3-32n^2-32)#

You could solve for the #n#, but there would be #3# answers. You would just solve it like a cubic. If you need the values for #n#, just notify me, and I will further elaborate. But off of your question, I think that's all you needed to do for your sake.

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Answer 2

Abigail Allen

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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.