<s>[INST] Quadratics question To solve $-3x^2 +2x +1=0$, I'd normally break the middle term and then factorise. But I was wondering if there was a way to skip the factorising step? The factors I'd use in place of the middle term would be $3$ and $-1$. If I were to flip the sign of each, I would get $-3$ and $1$. And then, were I to divide by the coefficient of $x^2$, I would get $1$ and $-1/3$, which are the correct answers. So, my question is, can all quadratics be solved in this way or was this just a fluke? If they can be, then why? If not, then is there any other quick and easy way of solving quadratics mentally? thanks for your help
 [/INST] Notice, we have $$-3x^2+2x+1=0$$ 
Finding the roots of $-3x^2+2x+1=0$ by quadratic rule as follows
$$x=\frac{-2\pm\sqrt{(2)^2-4(-3)(1)}}{2(-3)}$$
$$x=\frac{-2\pm4}{-6}$$ $$x=\frac{-2+4}{-6}\ \ \vee \ \ x=\frac{-2-4}{-6}$$
$$x=1\ \ \vee \ \ x=-\frac{1}{3}$$
Then, $(x-1)$ & $\left(x+\frac{1}{3}\right)$ will be the factors, thus
 we can factorize the expression as follows $$-3x^2+2x+1=-3(x-1)\left(x+\frac{1}{3}\right)$$ 
Hence, we have $$-3(x-1)\left(x+\frac{1}{3}\right)=0$$ or 
$$(x-1)\left(3x+1\right)=0$$

Alternative Method Factorization can also be done by rearranging the terms as follows $$-3x^2+2x+1=0$$ $$-(3x^2-2x-1)=0$$
  $$-(3x^2-3x+x-1)=0$$  $$-(\underbrace {3x^2-3x}+\underbrace{x-1})=0$$ 
  $$-(3x(x-1)+(x-1))=0$$  $$(x-1)(3x+1)=0$$  

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