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

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

2 Answers

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Answer 1 · 2017-06-06T18:16:17

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

Explanation:

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

Answer 2 · 2017-06-06T18:36:49

$\frac{13}{2} u^{2} + 3 u + 2$ $13/2u^2+3u+2$

Explanation:

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

$\Rightarrow \frac{13 u^{4} + 6 u^{3} + 4 u^{2}}{2 u^{2}}$ $rArr(13u^4+6u^3+4u^2)/(2u^2)$

$= \frac{13 u^{4}}{2 u^{2}} + \frac{6 u^{3}}{2 u^{2}} + \frac{4 u^{2}}{2 u^{2}}$ $=(13u^4)/(2u^2)+(6u^3)/(2u^2)+(4u^2)/(2u^2)$

$= \frac{13}{2} u^{2} + 3 u + 2$ $=13/2u^2+3u+2$

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

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