Divide $x^{4}-5x+6$ by $2-x^{2}$.

(N/A) To divide $x^{4}-5x+6$ by $2-x^{2}$,we first rewrite the polynomials in descending order of their degrees:
Dividend: $x^{4}+0x^{3}+0x^{2}-5x+6$
Divisor: $-x^{2}+2$
Step $1$: Divide the first term of the dividend $(x^{4})$ by the first term of the divisor $(-x^{2})$: $x^{4} / (-x^{2}) = -x^{2}$. This is the first term of the quotient.
Step $2$: Multiply $-x^{2}$ by the divisor $(-x^{2}+2)$: $-x^{2}(-x^{2}+2) = x^{4}-2x^{2}$.
Step $3$: Subtract this from the dividend: $(x^{4}+0x^{3}+0x^{2}-5x+6) - (x^{4}-2x^{2}) = 2x^{2}-5x+6$.
Step $4$: Divide the first term of the new dividend $(2x^{2})$ by the first term of the divisor $(-x^{2})$: $2x^{2} / (-x^{2}) = -2$. This is the second term of the quotient.
Step $5$: Multiply $-2$ by the divisor $(-x^{2}+2)$: $-2(-x^{2}+2) = 2x^{2}-4$.
Step $6$: Subtract this from the current dividend: $(2x^{2}-5x+6) - (2x^{2}-4) = -5x+10$.
Thus,the quotient is $-x^{2}-2$ and the remainder is $-5x+10$.