Assuming $a, b \in \mathbb{Z}$
$\dfrac{6a \times 10b}{4}=\dfrac{60ab}{4}=15ab \in \mathbb{Z}$
The last statement by closure of multiplication over the integers.
and so $4 \times 15ab = 6a \times 10b$ and 4 is thus a factor of $6a \times 10b$
That's really all there is to it.