Polynomial over Field is Reducible iff Scalar Multiple is Reducible

Theorem
Let $K$ be a field.

Let $K\left[{X}\right]$ be the polynomial ring over $K$.

Let $P \in K\left[{X}\right]$.

Then for and $\lambda \in K \backslash \left\{ 0 \right\}$, $P$ is irreducible in $K\left[{X}\right]$ iff $\lambda P$ is irreducible.

Proof
Suppose first that $P$ is irreducible.

Suppose further that there is a non-trivial factorization:
 * $\displaystyle \lambda P = Q_1 Q_2$

That is, $Q_1$ and $Q_2$ are not units of $K\left[{X}\right]$.

By Units of Ring of Polynomial Forms over Field it follows that $\deg Q_1 \geq 1$ and $\deg Q_2 \geq 1$.

Now if we let $Q_1' = \lambda^{-1} Q_1$, this implies that:
 * $P = Q_1' Q_2$

with $\deg Q_1' = \deg Q_1 \geq 1$.

But this is a non-trivial factorization of $P$ in $K\left[{X}\right]$, a contradiction.

Therefore $\lambda P$ has no non-trivial factorization, that is, $\lambda P$ is irreducible.

Conversely suppose that $\lambda P$ is irreducible.

Let $Q = \lambda P$.

By the first part; we know that any scalar multiple of $Q$ is irreducible.

In particular,
 * $\lambda^{-1}Q = \lambda^{-1}\lambda P = P$

is irreducible, the required result.