How do you factor x^3+x^2-14x-24?

$\textcolor{red}{{x}^{3} + {x}^{2} - 14 x - 24 = \left(x + 2\right) \left(x + 3\right) \left(x - 4\right)}$

Explanation:

We start from the given 3rd degree polynomial

${x}^{3} + {x}^{2} - 14 x - 24$

Use the monomial $- 14 x$
It is equal to $- 4 x - 10 x$

${x}^{3} + {x}^{2} - 4 x - 10 x - 24$

Rearrange

${x}^{3} - 4 x + {x}^{2} - 10 x - 24$

Regroup

$\left({x}^{3} - 4 x\right) + \left({x}^{2} - 10 x - 24\right)$

Factoring

$x \left({x}^{2} - 4\right) + \left(x + 2\right) \left(x - 12\right)$

$x \left(x + 2\right) \left(x - 2\right) + \left(x + 2\right) \left(x - 12\right)$

Factor out the common binomial factor $\left(x + 2\right)$

$\left(x + 2\right) \left[x \left(x - 2\right) + \left(x - 12\right)\right]$

Simplify the expression inside the grouping symbol [ ]

$\left(x + 2\right) \left[{x}^{2} - 2 x + x - 12\right]$

$\left(x + 2\right) \left({x}^{2} - x - 12\right)$

Factoring the trinomial ${x}^{2} - x - 12 = \left(x + 3\right) \left(x - 4\right)$

We now have the factors

$\left(x + 2\right) \left(x + 3\right) \left(x - 4\right)$

$\textcolor{red}{{x}^{3} + {x}^{2} - 14 x - 24 = \left(x + 2\right) \left(x + 3\right) \left(x - 4\right)}$