Solve by Factoring x^3+3x^2-4x-12=0

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Algebra

$x^{3} + 3 x^{2} - 4 x - 12 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(x^{3} + 3 x^{2}) - 4 x - 12 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x + 3) - 4 (x + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $x + 3$ .

$(x + 3) (x^{2} - 4) = 0$

Rewrite $4$ as $2^{2}$ .

$(x + 3) (x^{2} - 2^{2}) = 0$

Factor.

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Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$(x + 3) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x + 3) (x + 2) (x - 2) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 3 = 0$

$x + 2 = 0$

$x - 2 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x + 3) (x + 2) (x - 2) = 0$ true.

$x = - 3, - 2, 2$

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