Question

(х^3-4х):(х-4)
(х^4-х^3+3х^2-2х+2):(х^2-х+1)​

Answer 1

(х^3-4х):(х-4)

$\frac{x^3-4x}{x-4}$

(х^4-х^3+3х^2-2х+2):(х^2-х+1)​

$\frac{x^4-x^3+3x^2-2x+2}{x^2-x+1}$

$\frac{x^4-x^3+x^2+2x^2-2x+2}{x^2-x+1}$

$\frac{x^2*(x^2-x+1) + 2(x^2-x+1)}{x^2-x+1}$

$\frac{(x^2-x+1)*(x^2+2)}{x^2-x+1}$

$x^2+2$