Question

divide. if there is a remainder, include it as a\\((4n^{8} + 24n^{6} + 12n^{5}) \div 4n^{4}\\)

Explanation:

Step1: Divide each term by \(4n^4\)

We use the rule of dividing monomials: \(\frac{a^m}{a^n}=a^{m - n}\) and \(\frac{ac}{bc}=\frac{a}{b}\) (when \(c
eq0\)).
For the first term: \(\frac{4n^8}{4n^4}\)
The coefficient \(4\div4 = 1\), and for the variable part \(n^{8-4}=n^4\), so this term becomes \(n^4\).
For the second term: \(\frac{24n^6}{4n^4}\)
The coefficient \(24\div4 = 6\), and for the variable part \(n^{6 - 4}=n^2\), so this term becomes \(6n^2\).
For the third term: \(\frac{12n^5}{4n^4}\)
The coefficient \(12\div4=3\), and for the variable part \(n^{5 - 4}=n\), so this term becomes \(3n\).

Answer:

\(n^4 + 6n^2+3n\)