There are two methods:
1) Use #f(-1)+f'(-1)(x+1)+(f''(-1))/(2!)(x+1)^2+(f'''(-1))/(3!)(x+1)^3+(f''''(-1))/(4!)(x+1)^4+\cdots#
Here, #f(x)=x^4-x^2+1# so #f'(x)=4x^3-2x#, #f''(x)=12x^2-2#, #f'''(x)=24x#, #f''''(x)=24# and all higher-order derivatives are identically zero. Thus, #f(-1)=1-1+1=1#, #f'(-1)=-4+2=-2#, #f''(-1)=12-2=10#, #f'''(-1)=-24#, #f''''(-1)=24# and all higher-order derivatives at #x=-1# are zero.
Since #2! =2#, #3! =6#, and #4! =24#, this gives the answer above; the Taylor series is:
#1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4#
Since #f(x)# is a polynomial, #f(x)# equals its Taylor series for all #x# and we can write #f(x)=x^4-x^2+1=1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4# for all #x#.
2) Use the substitution #u=x+1# to expand #f(x)=f(u-1)# as a polynomial in powers of #u#. Then replace #u# by #x+1# at the end.
The binomial theorem (Pascal's triangle ) is helpful in doing the expansion:
#f(u-1)=(u-1)^4-(u-1)^2+1#
#=u^4-4u^3+6u^2-4u+1-(u^2-2u+1)+1#
#=1-2u+5u^2-4u^3+u^4#
#=1-2(x+1)+5(x+1)^2-4(x+1)^3+(x+1)^4#, as before.
