<s>[INST] Confusion while factoring quadratics. I recently factored the quadratic $3x^2+4x-4$, the results where $x = -2$; $x = 2/3$. My question is why is it incorrect to write the factors of the quadratic as $(x+2)(x-2/3)$? The correct way to write it is $(x+2)(3x-2)$, but why? What is the difference in $(x-2/3)$ and $(3x-2)$, both when $x$ is $2/3$ result in the y being zero.
 [/INST] Observe that
\begin{align*}
3x^2 + 4x - 4 & = 3x^2 - 2x + 6x - 4\\
              & = x(3x - 2) + 2(3x - 2)\\
              & = (x + 2)(3x - 2)\\
              & = 3(x + 2)\left(x - \frac{2}{3}\right)\\
              & \neq (x + 2)\left(x - \frac{2}{3}\right)\\
              & = x^2 + \frac{4}{3}x - \frac{4}{3}
\end{align*}
For two polynomials to be equal, they must be equal at each value of $x$ within their domains.
Let $f(x) = 3x^2 + 4x - 4$.  Then $f(0) = -4$.
Let $g(x) = (x + 2)\left(x - \frac{2}{3}\right)$.  Then $g(0) = -4/3 \neq -4 = f(0)$, so the polynomials $3x^2 + 4x - 4$ and $(x + 2)(x - 2/3)$ are not equal.
The graphs of $f$ and $g$ are shown below.

Your confusion stems from the fact these two polynomials have the same roots.
They have the same roots since the two polynomials differ by a scaling factor that can be canceled when you solve for the roots.
\begin{align*}
3x^2 + 4x - 4 & = 0\\
3(x + 2)\left(x - \frac{2}{3}\right) & = 0\\
(x + 2)\left(x - \frac{2}{3}\right) & = 0
\end{align*}
\begin{align*}
x + 2 & = 0 & \text{or} & & x - \frac{2}{3} & = 0\\
x & = -2 & & & x & = \frac{2}{3}
\end{align*}
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