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.