# How do you divide (2x^4+7)/(x^2-1)?

Jul 25, 2017

Simple. You just need to figure the answer by using the basic concepts of algebra, the order of operations, and a little but synthetic division.

#### Explanation:

$\frac{2 {x}^{4} + 7}{{x}^{2} - 1}$

Use long division to figure out the answer to the question. You just need divide it like in arithmetic and simplify it the same way.

$2 {x}^{2} + 2 + \frac{5}{{x}^{2} - 1}$

Jul 25, 2017

$\frac{2 {x}^{4} + 7}{{x}^{2} - 1} = 2 {x}^{2} + 2 + \frac{9}{{x}^{2} - 1}$

#### Explanation:

Here are three ways...

We can achieve something of the effect of long division by splitting off multiples of the denominator from the numerator as follows:

$\frac{2 {x}^{4} + 7}{{x}^{2} - 1} = \frac{2 {x}^{4} - 2 {x}^{2} + 2 {x}^{2} + 7}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = \frac{2 {x}^{2} \left({x}^{2} - 1\right) + 2 {x}^{2} + 7}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = \frac{2 {x}^{2} \left({x}^{2} - 1\right) + 2 {x}^{2} - 2 + 9}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = \frac{2 {x}^{2} \left({x}^{2} - 1\right) + 2 \left({x}^{2} - 1\right) + 9}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = \frac{\left(2 {x}^{2} + 2\right) \left({x}^{2} - 1\right) + 9}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = \frac{\left(2 {x}^{2} + 2\right) \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2} - 1}}}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2} - 1}}}} + \frac{9}{{x}^{2} - 1}$

$\textcolor{w h i t e}{\frac{2 {x}^{4} + 7}{{x}^{2} - 1}} = 2 {x}^{2} + 2 + \frac{9}{{x}^{2} - 1}$

If you prefer, you can write it as long division of polynomials. One shortcut in this example is to notice that there are no terms of odd degree. So we can omit terms of odd degree in what we write:

$\textcolor{w h i t e}{{x}^{2} - 1} \textcolor{w h i t e}{X} \textcolor{w h i t e}{|} \underline{\textcolor{w h i t e}{X} 2 {x}^{2} + 2 \textcolor{w h i t e}{X X X X}}$
${x}^{2} - 1 \textcolor{w h i t e}{X} | \textcolor{w h i t e}{X} 2 {x}^{4} + 0 {x}^{2} + 7$
$\textcolor{w h i t e}{{x}^{2} - 1} \textcolor{w h i t e}{X} \textcolor{w h i t e}{|} \textcolor{w h i t e}{X} \underline{2 {x}^{4} - 2 {x}^{2}}$
$\textcolor{w h i t e}{{x}^{2} - 1} \textcolor{w h i t e}{X} \textcolor{w h i t e}{|} \textcolor{w h i t e}{X} \textcolor{w h i t e}{2 {x}^{4} -} 2 {x}^{2} + 7$
$\textcolor{w h i t e}{{x}^{2} - 1} \textcolor{w h i t e}{X} \textcolor{w h i t e}{|} \textcolor{w h i t e}{X} \textcolor{w h i t e}{2 {x}^{4} -} \underline{2 {x}^{2} - 2}$
$\textcolor{w h i t e}{{x}^{2} - 1} \textcolor{w h i t e}{X} \textcolor{w h i t e}{|} \textcolor{w h i t e}{X} \textcolor{w h i t e}{2 {x}^{4} - 2 {x}^{2} -} 9$

Alternatively, write out the answer term by term, working out what you need next...

(2x^4+7) = (x^2-1)(...

The first term must be $2 {x}^{2}$ which multiplied by ${x}^{2}$ givens the leading term $2 {x}^{4}$:

(2x^4+7) = (x^2-1)(2x^2...

Note that $\left(- 1\right) \left(2 {x}^{2}\right) = - 2 {x}^{2}$, which we need to cancel out. Choose the next term $2$ so that when multiplied by ${x}^{2}$ gives $2 {x}^{2}$:

$\left(2 {x}^{4} + 7\right) = \left({x}^{2} - 1\right) \left(2 {x}^{2} + 2\right) \ldots$

Note that $\left(- 1\right) \left(2\right) = - 2$, but we want $7$, so we need to add $9$:

$\left(2 {x}^{4} + 7\right) = \left({x}^{2} - 1\right) \left(2 {x}^{2} + 2\right) + 9$

Hence:

$\frac{2 {x}^{4} + 7}{{x}^{2} - 1} = 2 {x}^{2} + 2 + \frac{9}{{x}^{2} - 1}$