(N/A) According to the division algorithm,if $p(x)$ and $g(x)$ are two polynomials with $g(x) \neq 0,$ then we can find polynomials $q(x)$ and $r(x)$ such that $p(x) = g(x) \cdot q(x) + r(x),$ where $r(x) = 0$ or $\operatorname{deg} r(x) < \operatorname{deg} g(x).$
The degree of a polynomial is the highest power of the variable in the polynomial.
To satisfy $\operatorname{deg} p(x) = \operatorname{deg} q(x),$ the degree of the quotient must be equal to the degree of the dividend. This occurs when the divisor $g(x)$ is a constant.
Let us assume the division of $p(x) = 6x^2 + 2x + 2$ by $g(x) = 2.$
Here,$p(x) = 6x^2 + 2x + 2,$
$g(x) = 2,$
$q(x) = 3x^2 + x + 1,$
$r(x) = 0.$
The degree of $p(x)$ is $2$ and the degree of $q(x)$ is $2,$ so $\operatorname{deg} p(x) = \operatorname{deg} q(x).$
Checking the division algorithm:
$p(x) = g(x) \cdot q(x) + r(x)$
$6x^2 + 2x + 2 = 2(3x^2 + x + 1) + 0$
$6x^2 + 2x + 2 = 6x^2 + 2x + 2.$
Thus,the division algorithm is satisfied.