<s>[INST] How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
 [/INST] Solution 1:
We usually do some tests by replacing x=0; x=1 and x=-1 in the equation (usual roots, you can take it as habit) here we can notice that -1 is a solution for the equation (root) so :
$$2x^2 - x - 3 =(x+1)\times 2 \times Q$$ and $$Q=ax+b;$$ 
Variables a and b can be extracted easily for this example (sometimes we can even do an euclidean division)
Result is :
$$ 2x^2 - x - 3 =(x+1)\times2\times(x-3)=2(x+1)(x-3)$$
Solution 2: 
The equation is 2nd degree one so either it has 1 double root, 2 roots or no solution in R
(1)  $$2x^2 - x - 3= ax^2 +bx+c=0$$  so a=2, b=-1 and c=-3
We calculate $$Delta=b^2-(4ac)=1-(4\times2\times(-3))=25 >0$$ 
We have two roots for equation 1 so we can write : $$2x^2 - x - 3 = a(x-x1)(x-x2)$$
$$x1=\frac{-b - \sqrt{Delta}}{2a} = \frac{(1-5)}{4}=-1;$$
$$x2=\frac{-b + \sqrt{Delta}}{2a} = \frac{(1+5)}{4}=3;$$
So 
$$2x^2 - x - 3 = a(x-x1)(x-x2) = 2(x+1)(x-3)$$
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