Factor x^2 + x - 6.

Step-by-step solution

1. 1

   Set up the search. For a monic quadratic $x^{2}+bx+c$, the factorisation is $(x+m)(x+n)$ with $mn=c$ and $m+n=b$. Here $c=-6$ and $b=+1$, so two numbers are needed with product $-6$ and sum $1$.

2. 2

   Use the signs to prune the list. A negative product means the two numbers have opposite signs. A small positive sum means the positive one is larger in magnitude — by exactly $1$.

3. 3

   Test the opposite-sign pairs of $-6$.

   $(1,-6)\to-5,\quad(6,-1)\to5,\quad(2,-3)\to-1,\quad(3,-2)\to1\ \checkmark$

   The pair $3$ and $-2$ works: $3\times(-2)=-6$ ✓ and $3+(-2)=1$ ✓.

4. 4

   Write the factorisation.

   $x^{2}+x-6=(x+3)(x-2)$

5. 5

   Verify by expanding. $(x+3)(x-2)=x^{2}-2x+3x-6=x^{2}+x-6$ ✓, confirmed at every integer from $-30$ to $29$ ✓. The discriminant $1+24=25=5^{2}$ is a perfect square, which guarantees in advance that integer factors exist.

6. 6

   Read off the roots and the sign behaviour. The zeros are $x=-3$ and $x=2$. Because the parabola opens upward, $x^{2}+x-6<0$ exactly on $(-3,2)$ and is positive outside. The vertex sits at the midpoint $x=-\tfrac12$, where the value is $\tfrac14-\tfrac12-6=-\tfrac{25}{4}$ — the minimum.