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

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Answer 1 · 2016-07-04T03:01:57

$x^{3} + x^{2} - 14 x - 24 = (x + 2) (x + 3) (x - 4)$ $color(red)(x^3+x^2-14x-24=(x+2)(x+3)(x-4))$

We start from the given 3rd degree polynomial

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

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

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

Rearrange

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

Regroup

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

Factoring

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

$x (x + 2) (x - 2) + (x + 2) (x - 12)$ $x(x+2)(x-2)+(x+2)(x-12)$

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

$(x + 2) [x (x - 2) + (x - 12)]$ $(x+2)[x(x-2)+(x-12)]$

Simplify the expression inside the grouping symbol [ ]

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

$(x + 2) (x^{2} - x - 12)$ $(x+2)(x^2-x-12)$

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

We now have the factors

$(x + 2) (x + 3) (x - 4)$ $(x+2)(x+3)(x-4)$

Final answer

$x^{3} + x^{2} - 14 x - 24 = (x + 2) (x + 3) (x - 4)$ $color(red)(x^3+x^2-14x-24=(x+2)(x+3)(x-4))$

God bless ....I hope the explanation is useful.

How do you factor x^3+x^2-14x-24x3+x2−14x−24x^3+x^2-14x-24?