why using $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to factorize a polynomial degree 2 dose not always work i tried to factorize $5x^3-11x^2+2x$ so i took out $x$ and used $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to find the roots 2 and $\frac{1}{5}$ but to my surprise multiplying the roots like so $x(x-2)\cdot(x-\frac{1}{5})$ produces a fifth of the original polynomial what did i do wrong?

It’s related to the derivation of the quadratic formula, which is done by completing the square. So, we know the formula.
$$ax^2+bx+c = 0 \implies x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
In reality, you aren’t solving for the original equation, which is $ax^2+bx+c = 0$. You’re solving for $x^2+\frac{b}{a}x+\frac{c}{a} = 0$, which gives the same solutions. You might want to check the proof for the quadratic formula (the first step is dividing both sides by $a$).  
For example, looking at the following quadratic polynomials, it’s easy to figure out that their solutions are the same.
$0 = (x-2)(x-3) = x^2-5x+6$ 
$0 = 2(x-2)(x-3) = 2x^2-10x+12$ 
$0 = 3(x-2)(x-3) = 3x^2-15x+18$
$0 = a(x-2)(x-3) = ax^2-5ax+6a; a\neq 0$
For all the quadratic polynomials above, the solutions are $x = 2$ and $x = 3$, but when you use the quadratic function, you end up solving for the FIRST one, but it doesn’t matter, since they share the same solutions. The same applies to any $n^{th}$ degree polynomial, as long as $a$ is a non-zero value.
Another way of thinking about it is that in a polynomial equation $y = a(x-x_a)\cdot(x-x_b)\cdot(x-x_c)\cdot ... \cdot (x-x_n)$, $a$ affects stretch, not the $x$-intercepts, so the solutions remain the same for all values of $a$.
So, for your example, you found the solutions $0$, $\frac{1}{5}$, and $2$. You also know that $a = 5$, so rewrite it using $5$ at the beginning.
$$y = 5(x)(x-\frac{1}{5})(x-2)$$
Now, it’s the correct function!
