Question 473314
{{{(8y^2-2y-1)/(2y+2)}}}
<pre>
It doesn't help to factor as {{{((4y+1)(2y-1))/(2(y+1))}}} because 
nothing cancels when you do.  So we must use long division:

      <u>       4y -  5</u>
2y + 2)8y² - 2y -  1
       <u>8y² + 8y</u>
           -10y -  1
           <u>-10y - 10</u>
                   9


       4y - 5 + {{{9/(2y+2)}}}
 
Edwin</pre>