# How do you divide (12x^4+12x^3-5x^2+9x+3)/(x^2-4) ?

Dec 21, 2016

The answer is $= 12 {x}^{2} + 12 x + 43 + \frac{57 x + 175}{{x}^{2} - 4}$

#### Explanation:

Let's do a long division

$\textcolor{w h i t e}{a a a a}$$12 {x}^{4}$$\textcolor{w h i t e}{a a a a}$$12 {x}^{3}$$\textcolor{w h i t e}{a a a a}$$- 5 {x}^{2}$$\textcolor{w h i t e}{a a a a}$$9 x$$\textcolor{w h i t e}{a a a a}$$3$∣${x}^{2} - 4$

$\textcolor{w h i t e}{a a a a}$$12 {x}^{4}$$\textcolor{w h i t e}{a a a a a a a a}$color(white)(aaaa)-48x^2$\textcolor{w h i t e}{a a a a}$color(white)(aaaaaa)∣$12 {x}^{2} + 12 x + 43$

$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$12 {x}^{3}$$\textcolor{w h i t e}{a a a}$color(white)(aaaaaaaaaa)9x$\textcolor{w h i t e}{a a a a a a}$

$\textcolor{w h i t e}{a a a a a a}$color(white)(aaaaaa)12x^3$\textcolor{w h i t e}{a a a a}$$+ 43 {x}^{2}$$\textcolor{w h i t e}{a a}$$- 48 x$$\textcolor{w h i t e}{a a}$$+ 3$

$\textcolor{w h i t e}{a a a a a a}$color(white)(aaaaaa)0$\textcolor{w h i t e}{a a a a a a a}$$+ 43 {x}^{2}$$\textcolor{w h i t e}{a a}$$+ 57 x - 172$$\textcolor{w h i t e}{a a a a a a}$

$\textcolor{w h i t e}{a a a a a a}$color(white)(aaaaaa)$\textcolor{w h i t e}{a a a a a a a a a a a a}$$0$$\textcolor{w h i t e}{a a}$$+ 57 x + 175$$\textcolor{w h i t e}{a a a a a a}$

Therefore,

$\frac{12 {x}^{4} + 12 {x}^{3} - 5 {x}^{2} + 9 x + 3}{{x}^{2} - 4}$

$= 12 {x}^{2} + 12 x + 43 + \frac{57 x + 175}{{x}^{2} - 4}$