How do you divide #9x^2-16# by #3x+4#?

1 Answer
Nov 4, 2014

Treat your terms of your dividends like digits... However, you also need to consider "missing" terms (in the case of #9x^2 - 16#, the #x# term).

For your divisor, you need only divide using the first term

             3x - 4
           ____________
 (3x + 4) / (9x^2 +  0x - 16) 
          - (9x^2 + 12x)
                  - 12x - 16
               - (- 12x - 16)
                           0