Answer
$${P_4}\left( x \right) = x - {x^2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x{e^{ - 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[ {x{e^{ - x}}} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = x\left( { - {e^{ - x}}} \right) + {e^{ - x}}\left( 1 \right) \cr
& {f^{\left( 1 \right)}}\left( x \right) = - x{e^{ - x}} + {e^{ - x}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = {e^{ - x}}\left( {1 - x} \right) \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x}}\left( {1 - x} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = {e^{ - x}}\left( { - 1} \right) + \left( {1 - x} \right)\left( { - {e^{ - x}}} \right) \cr
& {f^{\left( 2 \right)}}\left( x \right) = {e^{ - x}}\left( { - 1 - 1 + x} \right) \cr
& {f^{\left( 2 \right)}}\left( x \right) = {e^{ - x}}\left( {x - 2} \right) \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x}}\left( {x - 2} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = {e^{ - x}}\left( 1 \right) + \left( {x - 2} \right)\left( { - {e^{ - x}}} \right) \cr
& {f^{\left( 3 \right)}}\left( x \right) = {e^{ - x}}\left( {1 - x + 2} \right) \cr
& {f^{\left( 3 \right)}}\left( x \right) = {e^{ - x}}\left( {3 - x} \right) \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^{ - x}}\left( {3 - x} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = {e^{ - x}}\left( { - 1} \right) + \left( {3 - x} \right)\left( { - {e^{ - x}}} \right) \cr
& {f^{\left( 4 \right)}}\left( x \right) = {e^{ - x}}\left( { - 1 + x - 3} \right) \cr
& {f^{\left( 4 \right)}}\left( x \right) = {e^{ - x}}\left( {x - 4} \right) \cr
& \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) = \left( 0 \right){e^0} = 0 \cr
& {f^{\left( 1 \right)}}\left( 0 \right) = {e^0}\left( {1 - 0} \right) = 1 \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = {e^0}\left( {0 - 2} \right) = - 2 \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = {e^0}\left( {3 - 0} \right) = 3 \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = {e^0}\left( {0 - 4} \right) = - 4 \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) = 0 + \frac{1}{{1!}}x + \frac{{ - 2}}{{2!}}{x^2} + \frac{3}{{3!}}{x^3} + \frac{{ - 4}}{{4!}}{x^4} \cr
& {\text{simplify }} \cr
& {P_4}\left( x \right) = x - {x^2} + \frac{{{x^3}}}{3} - \frac{{{x^4}}}{6} \cr} $$
