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

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Answer 1 · 2016-08-22T06:12:50

$x^2-3x-18 = (x-6)(x+3)$

Find a pair of factors of $18$ which differ by $3$ . The pair $6, 3$ works, hence we find:

$x^2-3x-18=x^2-(6-3)x-(6*3)=(x-6)(x+3)$

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