Polynomials by Monomials?

Question

Alright. So I asked this yesterday and no one could get it.

^= to the power of.

7-3[(n^3+ 8n) divided by (-n) + 9n^2]

I got -24n^4 - 17 but they said that was wrong.

Answer 1

7 - 3[(n^3 + 8n) / (-n) + 9n^2] =
7 - 3[-n^2 - 8 + 9n^2] = { The (-n) cancels out one n each on top }
7 - 3[8n^2 - 8] = {Combine like terms }
7 - 24n^2 + 24 = { Multiply through by 3 }
-24n^2 + 31 { Again, combine like terms }

Answer 2

7-3[(n^3+ 8n) divided by (-n) + 9n^2]
Case 1: You mean the following:
7-3[n^3+8n)/-n +9n^2]
=7-3[-n^2-8 +9n^2]
=7-3[8n^2-8]
= 7-24n^2 +56
=63-24n^2
Case 2: You mean the following:
7-3[n^3+8n)/(-n +9n^2)]
=7- 3[n(n^2+8)/n(-1+9n)]
=7 - 3[n^2+8)/(9n-1)]
= 7 -(3n^2 -24)/(9n-1)
= (63n -7 -3n^2 +24)/(9n-1)
= (-3n^2 +63n +17)(9n-1)

Answer 3

I assume you mean:

7 -3[((n^3+8n)/-n) +9n^2]
=7 -3[-n^2 -8 +9n^2
=7 -3[8n^2-8]
=-24n^2 +31

but if you mean 7-3[(n^3+8n)/(-n +9n^2)]
=7 -3[(n^3 + 8n)/n(-1+9n)]
=7 -3[(n^2+8)/(9n -1)]
=(63n -7 -3n^2 -24)/(9n-1)
=(-3n^2 +63n -31)/(9n -1)
=