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How do you divide (18n ^ { 6} + 24n ^ { 5} + 4n ^ { 4} ) \div 6n ^ { 2}?

Mar 8, 2018

Divide each term by $6 {n}^{2}$ separately in order to final answer of $3 {n}^{4} + 4 {n}^{3} + \frac{2}{3}$${n}^{2}$

Explanation:

Think of this question in terms of creating a fraction. In other words, $\frac{18 {n}^{6} + 24 {n}^{5} + 4 {n}^{4}}{6 {n}^{2}}$ . (The line in a fraction means the same thing as ÷ ... divide).

Just like any fraction, you can separate the terms in the numerator by placing each term over the denominator like this:

$\frac{18 {n}^{6}}{6 {n}^{2}} + \frac{24 {n}^{5}}{6 {n}^{2}} + \frac{4 {n}^{4}}{6 {n}^{2}}$

Now work on each term separately.

1st term : $\frac{18 {n}^{6}}{6 {n}^{2}}$ --> $\frac{18}{6} = \frac{6 \cdot 3}{6 \cdot 1} = \frac{3}{1} = 3$ and ${n}^{6} / {n}^{2} = {n}^{6 - 2} = {n}^{4}$. So the 1st term $= 3 {n}^{4}$
2nd term : $\frac{24 {n}^{5}}{6 {n}^{2}}$ --> $\frac{24}{6} = \frac{6 \cdot 4}{6 \cdot 1} = \frac{4}{1} = 4$ and ${n}^{5} / {n}^{2} = {n}^{5 - 2} = {n}^{3}$. So the 2nd term is $4 {n}^{3}$
3rd term: $\frac{4 {n}^{4}}{6 {n}^{2}}$ --> $\frac{4}{6} = \frac{2 \cdot 2}{2 \cdot 3} = \frac{2}{3}$ and ${n}^{4} / {n}^{2} = {n}^{4 - 2} = {n}^{2}$. So the 3rd term is $\frac{2}{3}$${n}^{2}$

Now put these terms back together to get the final answer

$3 {n}^{4} + 4 {n}^{3} + \frac{2}{3}$${n}^{2}$

Quick tip: Once you understand this method, try solving these problem in your head by simply taking it one step at a time.

Mar 8, 2018

Begin by dividing out the ${n}^{2}$, then the 6.
When you take out the ${n}^{2}$, you are left with $\frac{18 {n}^{2} + 24 {n}^{3} + 4 {n}^{2}}{6}$.
This results in$\frac{18 {n}^{2} + 24 {n}^{3} + 4 {n}^{2}}{3}$.