Answer
$${P_4}\left( x \right) = {e^2} - {e^2}x + \frac{{{e^2}}}{2}{x^2} - \frac{{{e^2}}}{6}{x^3} + \frac{{{e^2}}}{{24}}{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^{2 - x}} \cr
& {\text{Use the definition of Taylor Polynomial of Degree }}n\,\,\,\left( {{\text{see page 629}}} \right) \cr
& {\text{Let }}f{\text{ be a function that can be differentiated }}n{\text{ times at 0}}{\text{. The Taylor }} \cr
& {\text{polynomial of degree }}n{\text{ for }}f{\text{ at 0 is }} \cr
& {P_n}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \cdots + \frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{n!}}{x^n} = \sum\limits_{i = 0}^n {\frac{{{f^{\left( n \right)}}\left( 0 \right)}}{{i!}}} {x^i} \cr
& {\text{Find the Taylor polynomials of degree 4 at 0}}{\text{. }} \cr
& {\text{then }}n = 4. \cr
& {\text{The }}n{\text{ - th derivatives are}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^{2 - x}}} \right] = - {e^{2 - x}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - {e^{2 - x}}} \right] = {e^{2 - x}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^{2 - x}}} \right] = - {e^{2 - x}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ { - {e^{2 - x}}} \right] = {e^{2 - x}} \cr
& {\text{evaluate }}f\left( 0 \right),{f^{\left( 1 \right)}}\left( 0 \right),{f^{\left( 2 \right)}}\left( 0 \right),{f^{\left( 3 \right)}}\left( 0 \right),{f^{\left( 4 \right)}}\left( 0 \right) \cr
& f\left( 0 \right) = {e^{2 - 0}} = {e^2} \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = - {e^{2 - 0}} = - {e^2} \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = {e^{2 - 0}} = {e^2} \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = - {e^{2 - 0}} = - {e^2} \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = {e^{2 - 0}} = {e^2} \cr
& {\text{Replace the found values into the definition of Taylor Polynomial of Degree }}n \cr
& {\text{for }}n = 4 \cr
& {P_4}\left( x \right) = f\left( 0 \right) + \frac{{{f^{\left( 1 \right)}}\left( 0 \right)}}{{1!}}x + \frac{{{f^{\left( 2 \right)}}\left( 0 \right)}}{{2!}}{x^2} + \frac{{{f^{\left( 3 \right)}}\left( 0 \right)}}{{3!}}{x^3} + \frac{{{f^{\left( 4 \right)}}\left( 0 \right)}}{{4!}}{x^4} \cr
& {P_4}\left( x \right) = {e^2} + \frac{{ - {e^2}}}{{1!}}x + \frac{{{e^2}}}{{2!}}{x^2} + \frac{{ - {e^2}}}{{3!}}{x^3} + \frac{{{e^2}}}{{4!}}{x^4} \cr
& {\text{simplifying}} \cr
& {P_4}\left( x \right) = {e^2} - {e^2}x + \frac{{{e^2}}}{2}{x^2} - \frac{{{e^2}}}{6}{x^3} + \frac{{{e^2}}}{{24}}{x^4} \cr} $$