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?
 A: Let's take a closer look at that first step:
$$4(y^2+8y+64-64)=0$$
Clearly you are able to see the perfect square, but there isn't one.  Notice that $(y+4)^2=y^2+8y+16$.  So let's change it to $16$.
$$4(y^2+8y+16-16)=0$$
Now we do have a perfect square, but you seem to make a mistake I see happen all the time.
$$\begin{array}{c|c}\text{What you do}&\text{What I do}\\\hline4(y+4)^2-16&4((y+4)^2-16)\end{array}$$
See the big difference some parenthesis make?  Now, we should end up with
$$4(y+4)^2-64=0$$
A: $4y^2+32y = 0$
$4(y^2+8y+16-16) = 0$
$4(y+4)^2 = 64$
$(y+4)^2 = 16$
$y+4= \pm4$
$y_1=-8$, $y_2=0$
A: When you expand $(a\pm b)^2$, look at the coefficient of $x$ and the constant.$$(a\pm b)^2=a^2\pm2ab+b^2\tag1$$
Assuming that it's just $a^2$, we see that to complete the square, you have to divide by $2$ and square the result. Or,$$a^2+ba+\left(\dfrac b2\right)^2=\left(a+\dfrac b2\right)^2\tag2$$
Now, what was wrong with your work is that instead of dividing by $2$, you just squared the coefficient. So really, it should be$$4\left(y^2+8y+\left(\dfrac 42\right)^2-\left(\dfrac 42\right)^2\right)=0$$$$4(y^2+8y+16)-64=0$$$$4(y+4)^2=64$$
A: 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.
