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.
 A: 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*}
A: First , to be at all correct, if you multiply out a suggested factorization, it must equal what you started with. But that is just the starting point. Often, some furher conditions are reqired. So, to give a more detailed answer to your question, I"ll start by asking a number of of other questions.(i)Must every factor have at least one$ 'x' $ in it or is it allowed to have just a plain number as one of the factors? (ii) In a factor, do you allow fractions like 2/3 or do you insist that all the numbers have to be inegers, i.e. whole numbers or their negatives? So, is $(x+3/5)$ allowed, or should it be changed to $(1/5)(5x+3)$?(iii) Do you require that each factor be one that cannot itself be factored further?For example would $x^2+20x+91$ be allowed as a factor or do you insist on breaking it up as $(x+7)(x+13)$? Thre is no general answer to these questions but there is a general principle that a good  question must make it clear exactly what kinds of factorizattions and factors are allowed. Moreover, there is another general principle in marking  an assignment, test or examination,  that if a question is vague as to what kind of answer is required, any reasonable interpretation must be eligible for full marks.
A: 
The above graph shows 3 different quadratics (so, they have different factorisations) that have a common pair of roots.
This is possible precisely when they are vertical scalings of one another; in this case, setting their expressions to $0$ then dividing each equation by the respective coefficient of $x^2$ results in identical equations.
