Simple Polynomial Algebra question to complete the square $4y^2+32y = 0$
$4(y^2+8y+64-64) = 0$
$4(y+4)^2 = 64$
Is that correct?

I made a comment, and I would do things slightly differently, so $$4y^2+32y=0$$
divide by $4$$$y^2+8y=0$$ add $\left(\frac 82\right)^2=4^2=16$ to both sides $$y^2+8y+16=16$$Rewrite the left-hand side as a square $$(y+4)^2=16$$If you want the $4$ back you can multiply through by $4$ at the end. If you want a pure square this is $(2y+8)^2=64$. The version you have put in your question does not express the left-hand side as a square.
