Maths question (polynomial)

Using Taylor's series.
f(x) = x^10 - 3x^5 + 1 = A(x - 1)^10 + B(x - 1)^9 + C(x - 1)^8 + .... + J(x - 1) + K
'A' obviously = 1 since coefficient of x^10 term on LHS is 1.
Put x = 1, LHS = 1 - 3 + 1 = - 1 , so K = - 1.
To determine B, C, D,.......I by differentiation.
Differentiate both sides we get
f'(x) = 10x^9 - 15x^4 = 10(x - 1)^9 + 9B(x - 1)^8 + .......+ 2I(x - 1) + J
Again put x = 1, LHS = 10 - 15 = -5, so J = -5.
Repeating the process :
f''(1) = 90(1)^8 - 60(1)^3 = 30 = 2I, so I = 15.
f'''(1) = 720(1)^7 - 180(1)^2 = 540 = 6H, so H = 90
f^4 (1) = 5040(1)^6 - 360(1) = 4680 = 24G, so G = 195
f^5(1) = 30240(1)^5 - 360 = 29880 = 120F, so F = 249
f^6(1) = 151200(1)^4 = 151200 = 720E, so E = 210
f^7(1) = 604800(1)^3 = 604800 = 5040D, so D = 120
f^8(1) = 1814400(1)^2 = 1814400 = 40320C, so C = 45
f^9(1) = 3628800(1) = 3628800 = 362880B, so B = 10
so x^10 - 3x^5 + 1 = (x - 1)^10 + 10(x - 1)^9 + 45(x-1)^8 + 120(x - 1)^7
+ 210(x - 1)^6 + 249(x - 1)^5 + 195(x - 1)^4 + 90(x - 1)^3 + 15(x - 1)^2
-5(x - 1) - 1.

(x-1)^10-3(x-1)^5+1
=(1-10x+45x^2-120x^3+210x^4-252x^5-210x^6-120x^7+45x^8-10x^9+x^10)-3(1-5x+10x^2-10x^3+5x^4-x^5)+1
=1-10x+45x^2-120x^3+210x^4-252x^5-210x^6-120x^7+45x^8-10x^9+x^10-3+5x-10x^2+10x^3-5x^4+x^5+1
=x^10-10x^9+45x^8-120x^7-210x^6-251x^5+205x^4-110x^3+35x^2-5x-1