Cfrgtkky

I tried looking up the "process" of solving that equation, but I couldn't really find the exact way to solve it.

Isolating the $2$ from $2x^2$ might be one of the way, but I couldn't exactly find out what I would have to do after that.

Thanks for helping me.

It is $$2x^2+16x=2(x^2+8x)=2(x^2+8x+16-16)=2(x+4)^2-32$$

While Dr. Sonnhard Graubner's answer is valid, I'd like to present a more intuitive approach.

Recall: the vertex form of a parabola is given by $y = a(x - h)^2 + k$, for vertex $(h,k)$. For the sake of argument, we can expand that form by foiling the squared term:

$$y = ax^2 - 2hax + ah^2 + k$$

We seek to write $y = 2x^2 + 16x$ in this form. Notice, however, that to generate the same parabola, we will need constants $a,h,k$ such that the two equations are equal. That means we set them equal to each other:

$$2x^2 + 16x = ax^2 - 2hax + ah^2 + k$$

In the interest of clarifying my next step, I will add some extra terms and parentheses:

$$(2)x^2 + (16)x + (0) = (a)x^2 + (-2ha)x + (ah^2 + k)$$

What would it mean for these two polynomials to be equal? Well, the constant terms would equal, the coefficients of the linear term $x$ would be equal, and the coefficients of the quadratic term $x^2$ would be equal. That is to say, we would have three equations:

$$begin{align}
2 &= a \
16 &= -2ha \
0 &= ah^2 + k \
end{align}$$

The first equation outright gives us $a = 2$.

Plug that into the second equation and thus $16 = -4h$. Solve for $h$ and you get $h = -4$.

Plug both into the third equation and you get $0 = 32 + k$. Thus, $k = -32$.

Now we just substitute the $a,h,k$ we found into the vertex form:

$$y = a(x - h)^2 + k = 2(x - (-4)^2 + (-32) = 2(x+4)^2 - 32$$