$f(x)=x^{4}+x-2 \cdot f^{(1)}(x)=4 x^{3}+1, f^{(2)}(x)=12 x^{2}, f^{(3)}(x)=24 x, f^{(4)}(x)=24$ and all

other derivatives are zero. Thus

$\begin{aligned} x^{4}+x-2 &=0+(x-1) \times 5+\frac{(x-1)^{2}}{2 !} \times 12+\frac{(x-1)^{3}}{3 !} \times 24+\frac{(x-1)^{4}}{4 !} \times 24 \\ &=5(x-1)+6(x-1)^{2}+4(x-1)^{3}+(x-1)^{4} \end{aligned}$