# How do you factor completely x^2 – 3x – 18?

${x}^{2} - 3 x - 18 = \left(x - 6\right) \left(x + 3\right)$
Find a pair of factors of $18$ which differ by $3$. The pair $6 , 3$ works, hence we find:
${x}^{2} - 3 x - 18 = {x}^{2} - \left(6 - 3\right) x - \left(6 \cdot 3\right) = \left(x - 6\right) \left(x + 3\right)$