Answer
a) $d-2+\frac{{3}}{{d+1}}$
b) $d^2-2d+3+\frac{{-4}}{{d+1}}$
c) $d^3-2d^2+3d-4+\frac{{5}}{{d+1}}$
d) $d^4-2d^3+3d^2-4d+5+\frac{{-6}}{{d+1}}$
e) $d^4-2d^3+3d^2-4d+5+\frac{{-6}}{{d+1}}$
Work Step by Step
Using the long division method shown below, then
$$\begin{aligned}
\text{a) }&
(d^2-d+1)÷(d+1)
\\&=
d-2+\frac{\color{red}{3}}{\color{blue}{d+1}}
,
\\\\\text{b) }&
(d^3-d^2+d-1)÷(d+1)
\\&=
d^2-2d+3+\frac{\color{red}{-4}}{\color{blue}{d+1}}
,
\\\\\text{c) }&
(d^4-d^3+d^2-d+1)÷(d+1)
\\&=
d^3-2d^2+3d-4+\frac{\color{red}{5}}{\color{blue}{d+1}}
.\end{aligned}
$$
d) Based on the pattern above, then the result of $(d^5-d^4+d^3-d^2+d-1)÷(d+1)$ is
$$
d^4-2d^3+3d^2-4d+5+\frac{\color{red}{-6}}{\color{blue}{d+1}}
.$$
e) Using the long division method shown below, then
$$\begin{aligned}
&
(d^5-d^4+d^3-d^2+d-1)÷(d+1)
\\&=
d^4-2d^3+3d^2-4d+5+\frac{\color{red}{-6}}{\color{blue}{d+1}}
.\end{aligned}
$$
