Question

$\frac{2 {x}^{3} + 4 {x}^{2} - 5 }{x + 3}$
I need help with this one.​

Answer 1

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

$\dfrac{2x^3+4x^2-5}{x+3}=2x^2-2x+6-\dfrac{23}{x+3}$