Question 60411
Well, here's a short cut that works when you're dividing two polynomials where one is a linear expression, as is (x+2). The zero value, or the value of x that would give us zero for the expression, for (x+2) would be (-2). So, what we do is we line up the -2 on the left, which should be followed by the coefficients. Hence we have the following:
<br>
-2 ||....2...3...-3...4

__________________________
.............2..-1...-1...6

In the above, we brought the two down. Then we multiplied this by (-2) which gives us -4. We add 3 and -4, which gives us -1. Then -1 is multiplied with -2, which gives us 2. This 2 is added to -3, which gives us -1. -2 is then multiplied to -1, which gives us 2. This 2 is added to 4 which gives us 6. 
<br>
Now all we have to do is bring back the 'x' values and reduce the degree by 1, so we get:
<br>
{{{2x^2-x-1+(6/(x+2))}}}
<br>
Hope this helped and wasn't too confusing. If it was, go to:
<br>
www.purplemath.com/modules/polydiv2.htm for further help.
<br>
They use a clearer method which works for more cases then the above.