$\newcommand{\ch}{\mathrm{ch}}$Let $X \subset \mathbb{P}^4$ be a quintic threefold, and the ground field $\mathbb{C}$. I'm following Wikipedia's method of computing the (total) Chern class of $\mathcal{T}_X$.
We have the short exact sequence $0 \rightarrow \mathcal{T}_X \rightarrow \mathcal{T}_{\mathbb{P}^4}|_X \rightarrow \mathcal{O}_X(5) \rightarrow 0 $ where $\mathcal{O}_X(5)$ is the normal sheaf $\mathcal{N}_{X/\mathbb{P} ^4}$. By the axiom of Chern classes where for short exact sequences $0 \rightarrow E' \rightarrow E \rightarrow E'' \rightarrow 0$ we have $ \ch(E) = \ch(E') \ch(E'')$, we have \begin{align} \ch(\mathcal{T}_X ) \ch( \mathcal{O}_X(5) ) &= \ch( \mathcal{T}_{\mathbb{P}^4}|_X ) \\ &= (1+h)^5 \\ &=1 + 5h+10h^2+10h^3+5h^4+h^5 \\ &= 1 + 5h+10h^2+10h^3. \end{align} where $h$ is the hyperplane class in the Chow ring $A^\bullet(X)$, and we have $\ch( \mathcal{T}_{\mathbb{P}^4}|_X ) = (1+h)^5$ by applying the same axiom, this time to the short exact sequence $0 \rightarrow \mathcal{O}_{\mathbb{P^4}} \rightarrow \mathcal{O}_{\mathbb{P^4}}(1)^{\oplus 5} \rightarrow \mathcal{T}_{\mathbb{P}^4} \rightarrow 0 $, and using that $\ch(\mathcal{O}_{\mathbb{P}^4}(1)) = 1+h$.
I have a question at this point: why are $5h^4$ and $h^5$ zero in the third to fourth line of the computation? I know that $h$ is a hyperplane class in $H^2(\mathbb{P}^4, \mathbb{Z})$, and it generates the ring which is in fact of the form $$ H^2(\mathbb{P}^4, \mathbb{Z}) \cong \mathbb{Z}[h]/h^5. $$ So I can see why the $h^5$ goes to zero, but what about $5h^4$?
For completeness, the computation ends with $$ \ch(\mathcal{T}_X) =\frac{1+5h+10h^2+10h^3}{1+5h} = \cdots = 1+10h^2-40h^3. $$
Many thanks for any answers.