Answer with explanation:
[tex]\frac{x^4-2}{x+1}\\\\=\frac{x^4-1-1}{x+1}\\\\=\frac{x^4-1^4}{x+1}-\frac{1}{x+1}\\\\=\frac{(x^2-1^2)(x^2+1^2)}{x+1}-\frac{1}{x+1}\\\\=\frac{(x-1)(x+1)(x^2+1)}{x+1}-\frac{1}{x+1}\\\\=(x-1)(x^2+1)-\frac{1}{x+1}\\\\=x \times (x^2+1)-1\times (x^2+1)-\frac{1}{x+1}\\\\=x^3+x-x^2-1-\frac{1}{x+1}\\\\=x^3-x^2+x-1-\frac{1}{x+1}[/tex]
Or
Writing the coefficient of [tex]x^4,x^3,x^2, x[/tex]
and made the table to solve by synthetic Division method.
Option A
