Which of the following is equivalent to
$$\frac{\left(2x+4\right)}{2x^2+8x+8}$$
For all values of x for which both expressions are defined?
$$A.\ \frac{1}{2x^2+6}$$
$$B.\ \frac{1}{9x+2}$$
$$C.\ \frac{2}{x+6}$$
$$D.\ \frac{1}{x+4}$$
$$E.\ \frac{1}{x+2}$$
The OA is E.
I think that it is a simple question, I just need to re-write the expression, then
$$\frac{2\left(x+2\right)}{2\left(x^2+4x+4\right)}=\frac{2\left(x+2\right)}{2\left(x+2\right)^2}=\frac{1}{\left(x+2\right)}$$
And that's all, right? Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
$$\frac{\left(2x+4\right)}{2x^2+8x+8}$$
For all values of x for which both expressions are defined?
$$A.\ \frac{1}{2x^2+6}$$
$$B.\ \frac{1}{9x+2}$$
$$C.\ \frac{2}{x+6}$$
$$D.\ \frac{1}{x+4}$$
$$E.\ \frac{1}{x+2}$$
The OA is E.
I think that it is a simple question, I just need to re-write the expression, then
$$\frac{2\left(x+2\right)}{2\left(x^2+4x+4\right)}=\frac{2\left(x+2\right)}{2\left(x+2\right)^2}=\frac{1}{\left(x+2\right)}$$
And that's all, right? Is there a strategic approach to this PS question? Can any experts help, please? Thanks!
