# How do you divide (x^4+x^2+2) / (x-2) ?

Apr 27, 2017

${x}^{3} + 2 {x}^{2} + 5 x + 10 + \setminus \frac{22}{x - 2}$

#### Explanation:

We have to find what multiplied by $x - 2$ cancels ${x}^{4}$. In this case it is ${x}^{3}$. Multiplying by ${x}^{3}$ we get ${x}^{4} - 2 {x}^{3}$. We can then subtract this from the numerator to get
${x}^{4} + {x}^{2} + 2 - \left({x}^{4} - 2 {x}^{3}\right) = 2 {x}^{3} + {x}^{2} + 2$

Now we need to find a value that will cancel $2 {x}^{3}$. This value is $2 {x}^{2}$. Multiplying by this we get $2 {x}^{3} - 4 {x}^{2}$. Subtracting we get
$2 {x}^{3} + {x}^{2} + 2 - \left(2 {x}^{3} - 4 {x}^{2}\right) = 5 {x}^{2} + 2$

The value that will cancel out $5 {x}^{2}$ is $5 x$. Subtracting we get
$5 {x}^{2} + 2 - \left(5 {x}^{2} - 10 x\right) = 10 x + 2$

To cancel out $10 x$ we can us $10$.
$10 x + 2 - \left(10 x - 20\right) = 22$

Since we can't divide anymore, 22 is the remainder. We add the remainder divided by the denominator to the rest of the terms we got to get
${x}^{3} + 2 {x}^{2} + 5 x + 10 + \setminus \frac{22}{x - 2}$