A Taylor Series for \(\displaystyle \displaystyle \begin{align*} f(x) \end{align*}\) centred around \(\displaystyle \displaystyle \begin{align*} x = h \end{align*}\) is given by \(\displaystyle \displaystyle \begin{align*} f(x) = \sum_{n = 0}^{\infty} { \frac{ f^{ (n) } (h) }{ n! } \left( x - h \right)^n } \end{align*}\), so calculating the necessary derivatives gives...

\(\displaystyle \displaystyle \begin{align*} f(x) &= x^5 + 3x^2 - x - 1 \\ f(1) &= 1^5 + 3 \cdot 1^2 - 1 - 1 &= 2 \\ \\ f'(x) &= 5x^4 + 6x - 1 \\ f'(1) &= 5 \cdot 1^4 + 6\cdot 1 - 1 \\ &= 10 \\ \\ f''(x) &= 20x^3 + 6 \\ f''(1) &= 20\cdot 1^3 + 6 \\ &= 26 \\ \\ f'''(x) &= 60x^2 \\ f'''(1) &= 60 \cdot 1^2 \\ &= 60 \\ \\ f^{\textrm{iv}} (x) &= 120x \\ f^{\textrm{iv}} (1) &= 120 \cdot 1 \\ &= 120 \\ \\ f^{\textrm{v}} (x) &= 120 \\ f^{\textrm{v}} (1) &= 120 \\ \\ f^{\textrm{vi}}(x) &= 0 \\ \vdots \end{align*}\)

So your Taylor Series is

\(\displaystyle \displaystyle \begin{align*} f(x) &= 2 + 10(x - 1) + \frac{26}{2}(x - 1)^2 + \frac{60}{3!}(x - 1)^3 + \frac{120}{4!}(x - 1)^4 + \frac{120}{5!}(x - 1)^5 \\ &= 2 + 10(x - 1) + 13(x - 1)^2 + 10(x - 1)^3 + 5(x - 1)^4 + (x - 1)^5 \end{align*}\)