What are the roots of the equation? 6x3−12x2−48x=0

{0, 2, 4}

{−4, 0, 2}

{−4, −2, 0}

{−2, 0, 4}

Hi! The answer to your question is D.

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f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option

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