Answer the following and justify: Can $x^{2}-1$ be the quotient on division of $x^{6}+2 x^{3}+x-1$ by a polynomial in $x$ of degree $5$?

(B) No. According to the division algorithm for polynomials,if $p(x)$ is the dividend and $g(x)$ is the divisor,then $p(x) = g(x) \cdot q(x) + r(x)$,where $\text{deg}(r(x)) < \text{deg}(g(x))$.
Given,$\text{deg}(p(x)) = 6$ and $\text{deg}(g(x)) = 5$.
If the quotient $q(x) = x^{2}-1$,then $\text{deg}(q(x)) = 2$.
The degree of the product $g(x) \cdot q(x)$ would be $\text{deg}(g(x)) + \text{deg}(q(x)) = 5 + 2 = 7$.
Since the degree of the dividend $p(x)$ is $6$,and the degree of the product $g(x) \cdot q(x)$ is $7$,it is impossible for $x^{2}-1$ to be the quotient because the degree of the product cannot exceed the degree of the dividend (assuming the remainder is zero or has a lower degree).
Therefore,$x^{2}-1$ cannot be the quotient.