How do you long divide $\frac{x^{2}}{2 x + 4}$ $(x^2)/(2x + 4)$ ?

Question

1756 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-08-03T18:29:20

$\frac{x^{2}}{2 x + 4} = \frac{x}{2} - 1 + \frac{2}{x + 2}$ $x^2/(2x+4)=x/2-1+2/(x+2)$

$= \frac{1}{2} \cdot \frac{x^{2}}{x + 2}$ $=1/2*x^2/(x+2)$

$= \frac{1}{2} \cdot \frac{x (x + 2) - 2 x}{x + 2}$ $=1/2*(x(x+2)-2x)/(x+2)$

$= \frac{1}{2} \cdot \frac{x (x + 2) - 2 (x + 2) + 4}{x + 2}$ $=1/2*(x(x+2)-2(x+2)+4)/(x+2)$

$= \frac{1}{2} (x - 2 + \frac{4}{x + 2})$ $=1/2(x-2+4/(x+2))$

$= \frac{x}{2} - 1 + \frac{2}{x + 2}$ $=x/2-1+2/(x+2)$

\0/ Here's our answer !

How do you long divide (x^2)/(2x + 4)x22x+4(x^2)/(2x + 4)?