Converting to vertex form, where coefficient of $w^2$ cannot be factored out

Question

I need help on converting this to vertex form:

$$12w^2 + 13w + 3$$

I have tried finding examples online, but every time I find an example where $x^2$ has a coefficient, it is always able to be factored out. I can't do that here because the $13w$ would not factor out cleanly. I did even try that. I was able to get the factored form of $(12)\left(w+\frac{1}{3}\right)\left(w+\frac{3}{4}\right)$, but I cannot for the life of me figure out how to get this into vertex form. I assume that I am supposed to be completing the square, since that is what the chapter was primarily about, but I really just cannot figure out how to do that.

Answer 1

$12w^2+13w+3=12(w^2+\frac{13w}{12}+\frac{1}{4})=12((w+\frac{13}{24})^2-\frac{25}{576}).$

To get the $\frac{13}{24}$, find $(\frac{13}{12 \cdot 2})^2$ and add and subtract it from the expression. In general, when completing the square, find $\frac{b^2}{2a}$ and add and subtract it from the expression.

$$12(w^2+\frac{13w}{12}+\frac{169}{576}-\frac{169}{576}+\frac{1}{4})=12((w+\frac{13}{24})^2-\frac{25}{576})=12(w+\frac{13}{24})^2-\frac{25}{48}$$

Answer 2

Lets continue with your answer $$\begin{array}{lll} 12w^2+13w+3&=&12(w+\frac{1}{3})(w+\frac{3}{4})\\ &=&12(w+\frac{8}{24})(w+\frac{18}{24})\\ &=&12(w+\frac{13}{24}-\frac{5}{24})(w+\frac{13}{24}+\frac{5}{24})\\ &=&12(\color{blue}{(w+\frac{13}{24})}-\frac{5}{24})(\color{blue}{(w+\frac{13}{24})}+\frac{5}{24})\\ \end{array}$$

Can you take it from here?