2x³ = x • 2x², and
2x² (x + 3) = 2x³ + 6x²
Subtract this from the dividend to get a remainder of
(2x³ + 4x² - 5) - (2x³ + 6x²) = -2x² - 5
-2x² = x • (-2x), and
-2x (x + 3) = -2x² - 6x
Subtract this from the last remainder to get a new one of
(-2x² - 5) - (-2x² - 6x) = 6x - 5
6x = x • 6, and
6 (x + 3) = 6x + 18
This gives a new remainder of
(6x - 5) - (6x + 18) = -23
x does not divide -23, so we're done, leaving us with
[tex]\dfrac{2x^3+4x^2-5}{x+3}=2x^2-2x+6-\dfrac{23}{x+3}[/tex]
