12n^3-48 = 12(n^3-4)
I'm stuck on the second one.
well 3 is a factor of the coefficients and I might be incorrect but I'd pull 3c^2 out of each to make:
3c^2(c^2+8C+16)
3c^2(c+4)(c+4)
I may be wrong but that is the best I could come up with.
First factor out the 3c^2 and you get:
3c^2(c^2 + 8c + 16)
This can then be factored to:
3c^2(c+4)(c+4)
Could be simplified to:
3c^2(c+4)^2

Any input on these??
12n^3-48 and 3d^3+4d-2
Thanks for the real help on these!! My brain isn't what it once was!
Could you pull d for the second one and get d(3d^2+4)-2?
Make sure the math teacher will accept this answer, since (c+4)^2 can also be still factored to (c+4)(c+4). Some math teachers can be really particular.![]()
The c^2 can also be factored to (c)(c) but no teacher would prefer that as a final answer. The best answer is left in the squared form.
It depends on what level of math one is teaching. I am a math teacher and I would prefer for my beginning Algebra students to factor out the (c+4)^2 to the (c+4)(c+4) since many will ignore the ^2 if a number is to be substituted in for the c and was asked to solve.
Yes c^2 can be (c)(c) and no I wouldn't want that factored out but you are only dealing with the one term and not another term to go along with it.