+561 How would I express (2x - 1) / (x + 2) as A + B / (x+2), where A and B are integers?
I can do 2x / (x + 2) - 1 / (x+2) of course, but that doesn't work...
Any ideas?
EDIT: Sorry if I wasn't clear, but it's NOT:
$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}} = {\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
But instead I need to change:
$${\frac{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
Into something that looks like:
$${\mathtt{A}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{B}}}{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}}$$
Thank you!
+118608 I think anon's logic should work but I would do it much more simply.
$$\frac{2x-1 }{ x+2}\\\\
=\frac{2(x+2)-1-4}{x+2}\\\\
=\frac{2(x+2)-5}{x+2}\\\\
=\frac{2(x+2)}{x+2}+\frac{-5}{x+2}\\\\
=2+\frac{-5}{x+2}$$
Don't give anon negative points for trying to help you!!!
+118608 YES anon understood what you wanted and gave you a correct answer!
I shall show you
$$\\\frac{2x-1}{x+2}=A+\frac{B}{x+2}\\\\
$multiply both sides by (x+2)$\\
2x-1=A(x+2)+B\\
2x-1=Ax+2A+B\\
2x-1=Ax+(2A+B)\\
$ equating co-efficients$\\
2=A\\
-1=2A+B\\
-1=4+B\\
B=-5\\
so\\
\frac{2x-1}{x+2}=2+\frac{-5}{x+2}\\\\$$
so maybe you owe anon an appology 
