How do you divide #\frac { 13u ^ { 4} + 6u ^ { 3} + 4u ^ { 2} } { 2u ^ { 2} }#?

2 Answers
Jacob
Jun 6, 2017

#\frac{13}{2}u^2 + 3u + 2#

Explanation:

The denominator is only a single term so you can split the fractions up and divide them individually: #\frac{13u^4 + 6u^3 + 4u^2}{2u^2} = \frac{13u^4}{2u^2} + \frac{6u^3}{2u^2} + \frac{4u^2}{2u^2}#, and the "#u#"s cancel out to give: #\frac{13}{2}u^2 + 3u + 2#. I hope I have explained it in enough detail.

Jim G.
Jun 6, 2017

#13/2u^2+3u+2#

Explanation:

#"'split' up the terms in the numerator dividing each one by"#
#"the denominator"#

#rArr(13u^4+6u^3+4u^2)/(2u^2)#

#=(13u^4)/(2u^2)+(6u^3)/(2u^2)+(4u^2)/(2u^2)#

#=13/2u^2+3u+2#

Related questions
Impact of this question
1113 views around the world
You can reuse this answer
Creative Commons License