Example 4, Section 6.3 of Hoffman’s Linear Algebra 
Example 4: In Example $2$, the operator $T$ also had the characteristic polynomial $f=(x-1)(x-2)^2$. But, this $T$ is not diagonalizable, so we don’t know that the minimal polynomial is $(x-1)(x-2)$. What do we know about
the minimal polynomial in this case? From Theorem 3 we know that its roots are $1$ and $2$, with some multiplicities allowed. Thus we search for $p$ among polynomials of the form $(x-l)^k(x-2)^l$, $k\geq 1$, $l\geq 1$. Try $(x-1) (x-2)$:
$$(A-I)(A-2I)=\begin{bmatrix} 2&0&-1 \\ 2&0&-1 \\ 4&0&-2\\ \end{bmatrix}$$
Thus, the minimal polynomial has degree at least $3$. So, next we should try either $(x-1)^2(x-2)$ or $(x-1)(x-2)^2$. The second, being the characteristic polynomial, would seem a less random choice. One can readily compute that $(A - I)(A - 2I)^2 = 0$. Thus the minimal polynomial for $T$ is its characteristic polynomial.

Que: Why didn’t author consider possibility of minimal polynomial may have irreducible (over $\Bbb{R}$) factor, let say $x^2+1$?
Also we can’t use minimal polynomial divides characteristic polynomial results to confine minimal polynomial possibility to $(x-2)^2$, $(x-1)(x-2)$ and $(x-1)(x-2)^2$, because Cayley-Hamilton theorem is not yet proved in book.
 A: Because the roots of the minimal polynomial and the characteristic polynomial have the same roots. Since the only roots of the characteristic polynomial are $1$ and $2$, then minimal polynomial can have no other roots. Any irreducible factor would have complex roots.
A: The question seems to be about minimal and characteristic polynomials in field extensions.
$\mathbf{Claim:}$ Every irreducible factor of the characteristic polynomial of $T$ is also an irreducible factor of the minimal polynomial of $T$ and vice versa.
Let $k$ be the base field. Choose the splitting field $K$ of the minimal polynomial $p(x)$. We will suppose here that the polynomial has only one irreducible factor $r(x)$ (We can generalize to how many ever we want later). We can say that the irreducible factor $r(x)$ in $K[x]$ will be the product of its Galois conjugates, i.e., $r(x)=(x-c)(x-\sigma_1c) \cdots (x-\sigma_nc)$.
$\mathbf{Fact\ 1:}$ Some $(x-\sigma_ic)$ must divide the characteristic polynomial.
Why?
The minimal polynomial of $T$ as a matrix with entries in $K$ must contain at least one of the $(x-\sigma_ic)$ as factor. Otherwise, $q(x)\in k[x]$ would be a minimal polynomial for $T$ as a matrix with entries in $k$.
$\mathbf{Fact\ 2:}$ Since $(x-\sigma_ic)$ divides the characteristic polynomial, $r(x)$ must divide it as the characteristic polynomial is in $k[x]$.
Combining these two facts shows us that the irreducible factors of the minimal polynomial must also be irreducible factors for the characteristic polynomial.

Now, for the other direction, if $\alpha$ is a root of the characteristic polynomial $c(x)$, then it is a factor of the minimal polynomial of $T$ over $K$, which must divide the minimal polynomial of $T$ over $k$. Therefore, every other Galois conjugate of $\alpha$ must also be a root of the minimal polynomial of $T$ over $k$. So, $r(x)$ divides the minimal polynomial of $T$ over $k$.
