Question

Long Division: (3x^4-8x^3-x^2+9x+5) / (x-2) pleaseee answer this

Answer 1

3x⁴ = 3x³ • x, and

3x³ (x - 2) = 3x⁴ - 6x³

Subtract this from the numerator to get a remainder of

(3x⁴ - 8x³ - x² + 9x + 5) - (3x⁴ - 6x³) = -2x³ - x² + 9x + 5

-2x³ = -2x² • x, and

-2x² (x - 2) = -2x³ + 4x²

Subtract this from the previous remainder to get a new remainder of

(-2x³ - x² + 9x + 5) - (-2x³ + 4x²) = -5x² + 9x + 5

-5x² = -5x • x, and

-5x (x - 2) = -5x² + 10x

Subtract this from the last remainder to get a new one of

(-5x² + 9x + 5) - (-5x² + 10x) = -x + 5

-x = -1 • x, and

-1 (x - 2) = -x + 2

This gives a new remainder of

(-x + 5) - (-x + 2) = 3

3 does not divide x, so we're done.

So, we have

(3x⁴ - 8x³ - x² + 9x + 5) / (x - 2) = 3x³ - 2x² - 5x - 1 + 3/(x - 2)