[tex]f(x)[/tex] is a polynomial of degree 4, so its Taylor series will also be of degree 4. We have
[tex]f(2)=-10[/tex]
[tex]f'(x)=4x^3-14x\implies f'(2)=4[/tex]
[tex]f''(x)=12x^2-14\implies f''(2)=34[/tex]
[tex]f'''(x)=24x\implies f'''(2)=48[/tex]
[tex]f^{(4)}(x)=24\implies f^{(4)}(2)=24[/tex]
and [tex]f^{(n)}(2)=0[/tex] for all [tex]n\ge5[/tex]. So
[tex]f(x)=f(2)+f'(2)(x-2)+\dfrac{f''(2)}{2!}(x-2)^2+\dfrac{f'''(2)}{3!}(x-2)^3+\dfrac{f^{(4)}(2)}{4!}(x-2)^4[/tex]
[tex]\boxed{f(x)=-10+4(x-2)+17(x-2)^2+8(x-2)^3+(x-2)^4}[/tex]
(Notice that expanding all the binomials and simplifying gives the same [tex]f(x)[/tex] you are given to begin with.)
