factor the trinomial completely. $x^{2}+x - 6$ $x^{2}+x - 6=\\square$

$factor the trinomial completely. $x^{2}+x - 6$ $x^{2}+x - 6=\\square$$

factor the trinomial completely.
$x^{2}+x - 6$
$x^{2}+x - 6=\square$

We need two numbers that multiply to $-6$ and add to $1$. The numbers are $3$ and $-2$, since $3 \times (-2) = -6$ and $3 + (-2) = 1$.

Substitute the pair into binomial factors of the form $(x + a)(x + b)$.

$(x+3)(x-2)$