What is the solution set for $- x^{3} - 2 x^{2} = - x - 2$ $-x^3-2x^2=-x-2$ ?

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Answer 1 · 2015-08-22T04:53:55

The solution set is $x = {- 2, - 1, 1}$ $x={-2,-1,1}$

$- x^{3} - 2 x^{2} = - x - 2$ $-x^3-2x^2=-x-2$

Multiply the equation times $- 1$ $-1$ .

$x^{3} + 2 x^{2} = x + 2$ $x^3+2x^2=x+2$

Move all terms to the left side by subtracting $x + 2$ $x+2$ from both sides.

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

Group the terms into two binomials.

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

Factor out $x^{2}$ $x^2$ from the first group.

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

Factor out $(x + 2)$ $(x+2)$ .

$(x^{2} - 1) (x + 2)$ $(x^2-1)(x+2)$

$(x^{2} - 1)$ $(x^2-1)$ can be factored as the difference of squares.

$(x^{2} - 1) = (x + 1) (x - 1)$ $(x^2-1)=(x+1)(x-1)$

The complete factorization of $- x^{3} - 2 x^{2} = - x - 2$ $-x^3-2x^2=-x-2$ is $(x + 2) (x + 1) (x - 1) = 0$ $(x+2)(x+1)(x-1)=0$ .

Since the factors are equal to zero, we must set each one equal to zero, and solve for $x$ $x$ .

$x + 2 = 0$ $x+2=0$
$x = - 2$ $x=-2$

$x + 1 = 0$ $x+1=0$
$x = - 1$ $x=-1$

$x - 1 = 0$ $x-1=0$
$x = 1$ $x=1$

The solution set is $x = {- 2, - 1, 1}$ $x={-2,-1,1}$

What is the solution set for −x3−2x2=−x−2-x^3-2x^2=-x-2?