Factoring $3x^2+4x-4=0$ using the quadratic formula The correct answer is $(3x-2)(x+2)$ but I am getting $(x-\frac{2}{3})(x+2)$, why?
My calculations:
$x = \frac{-4\pm\sqrt{16-(-48)}}{6} = \frac{-4\pm8}{6}$, which gives the factors $(x-\frac{4}{6})(x+\frac{12}{6}) = (x-\frac{2}{3})(x+2)$.
 A: Because if the roots of $ax^2+bx+c$ are $r_1$ and $r_2$, then$$ax^2+bx+c=a(x-r_1)(x-r_2).$$You forgot the $a$.
In your case, you get:\begin{align}3x^2+4x-4&=3\left(x-\frac23\right)(x+2)\\&=(3x-2)(x+2).\end{align}
A: Some other people have been saying you made a mistake, but that isn't quite right, at least not how I'm looking at it.  I think the confusion here comes from the fact that the quadratic formula is mainly a tool to help solve quadratic equations, not so much to factor quadratic expressions.  You can still use the quadratic formula to factor $ax^2 + bx + c$, but you'll need to do an extra step or two if $a \ne 1$.
In this case, you (correctly) reduced the equation $3x^2 + 4x - 4 = 0$ to the equation $\left(x - \frac23\right)(x+2) = 0$ by using the quadratic formula.  What happens now if you multiply both sides of your result by $3$?
A: Suppose $Q(x)$ is some quadratic, including yours (for argument's sake, $Q(x)=ax^2+bx+c$).
Then suppose $Q(t)=0$. 
It is then true that the function $\lambda Q(x), \lambda \in \Bbb R$ (so $\lambda Q(x)=a\lambda x^2+b\lambda x+c\lambda$)
has solution $\lambda Q(t)=0$
If we take your factors from the quadratic equation; $(x-\frac23)$ and $(x+2)$, and multiply them, we get $x^2+\frac43 x-\frac43=\frac 13(3x^2+4x-4)$
So we need to multiply this factorisation by $3$, which just means multiplying one of the brackets by $3$. 
A: You are confusing factoring a quadratic with solving a quadratic.
The quadratic $ax^2 + bx + c=0$ will have the same solutions as $x^2 + \frac bax + ca = 0$ (in fact, that was the very first step in developing the quadratic formula) and will have the same as $dx^2 + \frac {db}ax + \frac {dc}a-0$ or any $akx^2 + bkx + ck =0$.
And you will get the same two solutions $r_1 = \frac {-b+\sqrt{b^2 -4ac}}{2a}$ and $r_2 = \frac {-b+\sqrt{b^2-4ac}}{2a}$ for all of those polynomials.
And $x^2 + \frac ba x + \frac ca$ will factor as $(x - r_1)(x - r_2)$.  (If we expand $(x-r_1)(x-r_2)$ we get $x^2 - (r_1+r_2)x + r_1r_2$ and it's not hard to see that $-(r_1 + r_2)= \frac ba$ and $r_1r2 = \frac ca$.)
But that means $ax^2+bx + c = a(x^2 + \frac bax +\frac ca) = a(x-r_1)(x-r_2)$ will factor as $a(x-r_1)(x-r_2)$.  In general $(x-r_1)(x-r_2)$ will expand to a polynomial with leading coefficient $1$.  And $k(x-r_1)(x-r_2)$ will expand to a polynomial with leading coefficient $k$.
And we want leading coefficient $a=3$.  So we need $3x^2 + 4x-4 = 3(x-r_1)(x-r_2)$
Now you figured out $r_1$ and $r_2$ perfectly correctly.
So we have $3x^2 +4x-4 = 3(x-\frac 23)(x+2) = (3x -2)(x+2)$.
....
you know... maybe this will be clearer if we complete the square:
$3x^2 + 4x - 4 =$
$3(x^2 + \frac 43x +\frac {-4}3)=$
$3([x^2 + 2\frac {4}{2*3}x + \frac {4^2}{4*3^2}] - \frac {4^2}{4*3^2}+\frac {-4}3) =$
$3([x+\frac 4{2*3}]^2 - \frac {4^2 -4*(-4)*3}{4*3^2})=$
$3((x+\frac 4{2*3} + \frac {\sqrt{4^2 - 4*(-4)*3}}{2*3})(x+\frac 4{2*3} - \frac {\sqrt{4^2 - 4*(-4)*3}}{2*3}) =$
$3(x- \frac {-4-\sqrt{4^2 -4*(-4)*3}}{2*3})(x-\frac{-4+\sqrt{4^2 -4*(-4)*3}}{2*3})=$.
$3(x-\frac 23)(x+2)= (3x-2)(x+2)$.
Notice how after the first line and until the very last line, the $3$ just hung around for the ride doing absolutely nothing?  That's ... kind of the point.  The leading coefficient $a$ doesn't really do much once you set up to solve (after all, if you set to $0$, which we didn't do this time, we can just divide it out) but it sticks around if we never actually set anything to zero (which in this case we didn't).
