Answer
$8+10(x-2)+6(x-2)^2+(x-2)^3$
Work Step by Step
Differentiate the given function $f(x)$ as follows:
$f'(x)=3x^2-2 \implies f'(2)=10 \\f''(x)=6x \implies f''(2)=12\\ f'''(x)=6 \implies f'''(2)=6$
Now, write the Taylor's series at $x=2$.
$f(x)=f(2)+f'(2) (x-2)+\dfrac{f''(2)(x-2)^2 }{2!}+\dfrac{f'''(2)(x-2)^3 }{3!} =8+10(x-2)+\dfrac{12}{2 \cdot 1} (x-2)^2 +\dfrac{6}{3 \cdot 2 \cdot 1} (x-2)^3 \\=8+10(x-2)+6(x-2)^2+(x-2)^3$
