$2 x^{3} + 6 x^{2} + 4 x = 0$ has the following solution set:

{1, 2}

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{- 1, - 2}

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{1, - 2}

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{- 2, - 1, 0}

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{0, 1, 2}

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Solution

The correct option is E ${- 2, - 1, 0}$
The equation $2 x^{3} + 6 x^{2} + 4 x = 0$ may be written as $2 x (x^{2} + 3 x + 2) = 0$
One of the root of the equation $2 x (x^{2} + 3 x + 2) = 0$ is $x = 0$
Now find the roots of the quadratic equation $x^{2} + 3 x + 2 = 0$
$\Rightarrow x^{2} + 3 x + 2 = 0$
$\Rightarrow x^{2} + 2 x + x + 2 = 0$
$\Rightarrow x (x + 2) + 1 (x + 2) = 0$
$\Rightarrow x + 1 = 0$ and $x + 2 = 0$
$\Rightarrow x = - 1$ and $x = - 2$
Hence, the equation $2 x^{3} + 6 x^{2} + 4 x = 0$ has the solution set ${- 2, - 1, 0}$ .

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