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

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Answer 1 · 2016-07-04T03:01:57

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

We start from the given 3rd degree polynomial

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

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

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

Rearrange

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

Regroup

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

Factoring

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

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

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

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

Simplify the expression inside the grouping symbol [ ]

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

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

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

We now have the factors

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

Final answer

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

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

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