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

Jun 6, 2017

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

#### Explanation:

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

Jun 6, 2017

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

#### Explanation:

$\text{'split' up the terms in the numerator dividing each one by}$
$\text{the denominator}$

$\Rightarrow \frac{13 {u}^{4} + 6 {u}^{3} + 4 {u}^{2}}{2 {u}^{2}}$

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

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