<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] Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient. 
In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping:
\begin{align*}
2x^2 - x - 3 &= 2x^2 + (2-3)x - 3 \\ &= 2x^2 + 2x - 3x - 3 \\ &= 2x(x + 1) - 3(x + 1) \\ &= (2x - 3)(x+1).
\end{align*}

The rule comes from the distributive rule: $(ax + b)(cx + d) = (ac)x^2 + (ad + bc)x + bd$. The first coefficient is $ac$ and the last is $bd$, they multiply together to $abcd$, and the middle coefficient is $ad + bc$, the sum of a factor pair of $abcd$.

Another way to do it is to use the quadratic formula to find the roots of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so
$$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm\sqrt{1 - -24}}{4} = \frac{1\pm 5}{4} = \frac{3}{2}, -1,$$
which gives a factorization into the monic polynomials $x+1$ and $2x - 3$.
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