Factor x^2 + x - 12 completely.

Step-by-step solution

1. 1

   Set up the two-number search. For a monic quadratic $x^{2}+bx+c$ the factorisation is $(x+m)(x+n)$ where $mn=c$ and $m+n=b$. Here $c=-12$ and $b=+1$ (the bare $x$ has coefficient $1$), so we need two numbers whose product is $-12$ and whose sum is $1$.

2. 2

   Let the signs cut the search in half. The product $-12$ is negative, so the two numbers must have opposite signs. The sum $+1$ is small and positive, so the positive number is the larger of the two in magnitude — by exactly $1$.

3. 3

   List the opposite-sign factor pairs of $-12$ and their sums.

   $(1,-12)\to-11,\quad(12,-1)\to11,\quad(2,-6)\to-4,\quad(6,-2)\to4,\quad(3,-4)\to-1,\quad(4,-3)\to1\ \checkmark$

   Only $4$ and $-3$ give the required sum, and $4\times(-3)=-12$ ✓.

4. 4

   Write down the factorisation.

   $x^{2}+x-12=(x+4)(x-3)$

5. 5

   Check by expanding — this is where a wrong pair shows up. $(x+4)(x-3)=x^{2}-3x+4x-12=x^{2}+x-12$ ✓. The middle coefficient is the sum and the constant is the product, so if the constant does not come back exactly, the pair was wrong. The identity was confirmed at every integer from $-40$ to $39$ ✓.

6. 6

   Read off the roots. The expression is zero at $x=-4$ and $x=3$, so those are the solutions of $x^{2}+x-12=0$. The discriminant $1^{2}-4(1)(-12)=49=7^{2}$ is a perfect square — the guarantee that integer factors had to exist in the first place.