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

1 Answer
Sep 19, 2017

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

Explanation:

#(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.