Answer
$${P_4}\left( x \right) = {x^2} + {x^3} + \frac{1}{2}{x^4}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2}{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^2}{e^x}} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 1 \right)}}\left( x \right) = {x^2}{e^x} + 2x{e^x} \cr
& {f^{\left( 1 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 2x} \right) \cr
& {f^{\left( 2 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {{x^2} + 2x} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 2 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 2x} \right) + 2x{e^x} \cr
& {f^{\left( 2 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 4x + 2} \right) \cr
& {f^{\left( 3 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {{x^2} + 4x + 2} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 3 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 4x + 2} \right) + {e^x}\left( {2x + 4} \right) \cr
& {f^{\left( 3 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 6x + 6} \right) \cr
& {f^{\left( 4 \right)}}\left( x \right) = \frac{d}{{dx}}\left[ {{e^x}\left( {{x^2} + 6x + 6} \right)} \right] \cr
& {\text{use product rule}} \cr
& {f^{\left( 4 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 6x + 6} \right) + {e^x}\left( {2x + 6} \right) \cr
& {f^{\left( 4 \right)}}\left( x \right) = {e^x}\left( {{x^2} + 8x + 12} \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( {{0^2} + 2\left( 0 \right)} \right) = 0 \cr
& {f^{\left( 2 \right)}}\left( 0 \right) = {e^0}\left( {{0^2} + 4\left( 0 \right) + 2} \right) = 2 \cr
& {f^{\left( 3 \right)}}\left( 0 \right) = {e^0}\left( {{0^2} + 6\left( 0 \right) + 6} \right) = 6 \cr
& {f^{\left( 4 \right)}}\left( 0 \right) = {e^0}\left( {{0^2} + 8\left( 0 \right) + 12} \right) = 12 \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{0}{{1!}}x + \frac{2}{{2!}}{x^2} + \frac{6}{{3!}}{x^3} + \frac{{12}}{{4!}}{x^4} \cr
& {\text{simplify }} \cr
& {P_4}\left( x \right) = {x^2} + {x^3} + \frac{1}{2}{x^4} \cr} $$
