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

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Mar 8, 2018

Answer:

Begin by dividing out the #n^2#, then the 6.

Explanation:

When you take out the #n^2#, you are left with #(18n^2+24n^3+4n^2)/6#.
From here you are able to divide all terms by two, due to this being the least common denominator,
This results in#(18n^2+24n^3+4n^2)/3#.

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Sarah Share
Mar 8, 2018

Answer:

Divide each term by #6n^2# separately in order to final answer of #3n^4+4n^3+2/3##n^2#

Explanation:

Think of this question in terms of creating a fraction. In other words, #(18n^6 + 24n^5 + 4n^4)/(6n^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:

#(18n^6)/(6n^2) + (24n^5)/(6n^2) + (4n^4)/(6n^2)#

Now work on each term separately.

1st term : #(18n^6)/(6n^2)# --> #18/6=(6*3)/(6*1)=3/1=3# and #n^6/n^2=n^(6-2)=n^4#. So the 1st term #=3n^4#
2nd term : #(24n^5)/(6n^2) # --> #24/6=(6*4)/(6*1)=4/1=4# and #n^5/n^2=n^(5-2)=n^3#. So the 2nd term is #4n^3#
3rd term: #(4n^4)/(6n^2)# --> #4/6=(2*2)/(2*3)=2/3# and #n^4/n^2=n^(4-2)=n^2#. So the 3rd term is #2/3##n^2#

Now put these terms back together to get the final answer

#3n^4+4n^3+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.

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