I didn't understand the last step please explain how to 2 comes with M1x

Dear Student,
Work done in setting the rigid rod rotating with angular velocity $\omega$ is given as:
$W=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(M_1{x}^2+M_2(l-x{)}^2\right)$
Work done will be minimum when $\frac{dW}{dx}=0$
$$\begin{gathered} \frac{dW}{dx}=\frac{d}{dx}\left[\frac{1}{2}\left(M_1{x}^2+M_2(l-x{)}^2\right)\right]=\frac{1}{2}\left(2M_{1}x+2M_2(l-x\left)\right(-1)\right) \\ \frac{1}{2}\left(2M_{1}x-2M_2(l-x)\right)=0 \\ This implies \\ 2M_{1}x-2M_2(l-x)=0 (\because \frac{1}{2} is not zero) \end{gathered}$$
This is how 2M₁ comes into the expression.
Hope your doubt is cleared.

Regards