How do you factor completely $x^2 - 3x - 24$ ?

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Answer 1 · 2015-11-19T23:29:27

$x^2-3x-24 = (x-(3+sqrt(105))/2)(x-(3-sqrt(105))/2)$

$x^2-3x-24$ is in the form $ax^2+bx+c$ with $a=1$ , $b=-3$ and $c=-3$ and has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(3+-sqrt((-3)^2-(4xx1xx(-24))))/(2xx1)$

$=(3+-sqrt(9+96))/2$

$=(3+-sqrt(105))/2$

Hence:

$x^2-3x-24 = (x-(3+sqrt(105))/2)(x-(3-sqrt(105))/2)$

How do you factor completely x^2 - 3x - 24x^2 - 3x - 24?