Question 202748
Here's the long division:
<pre>
    3x^2 + (-10x) + 40
    ___________________________
x+3/3x^3 + (-x^2) + 10x +  (-4)
    3x^3 +  9x^2
    ------------
          -10x^2 +  10x 
          -10x^2 + -30x
          -------------
                    40x + (-4)
                    40x + 120
                    ----------
                         -124
</pre>
As usual, when the remainder is not 0, you put it into the numerator of a fraction with the divisor in the denominator. So:
(3x^3 – x^2 + 10^x – 4) ÷ (x + 3) = {{{3x^2 - 10x + 40 + (-124)/(x+3)}}}