Polynomial over Field is Reducible iff Scalar Multiple is Reducible

Theorem
Let $K$ be a field.

Let $K \sqbrk X$ be the ring of polynomial forms over $K$.

Let $P \in K \sqbrk X$.

Let $\lambda \in K \setminus \set 0$.

Then $P$ is irreducible in $K \sqbrk X$ $\lambda P$ is also irreducible in $K \sqbrk X$.

Necessary Condition
Let $P$ be irreducible.

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

that is, such that $Q_1$ and $Q_2$ are not units of $K \sqbrk X$.

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

Let $Q_1' = \lambda^{-1} Q_1$.

This implies that:
 * $P = Q_1' Q_2$

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

But this is a non-trivial factorization of $P$ in $K \sqbrk X$.

This contradicts our supposition that $P$ is irreducible.

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

Sufficient Condition
Let $\lambda P$ be irreducible.

Let $Q = \lambda P$.

From the necessary condition, 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.