We have to solve for x using the equation 3x^3 - 9x^2 - 12x = 0

Now 3x^3 - 9x^2 - 12x = 0

=> 3x ( x^2 - 3x - 4) = 0

=> 3x ( x^2 - 4x + x - 4) = 0

=> 3x [ x(x...

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We have to solve for x using the equation 3x^3 - 9x^2 - 12x = 0

Now 3x^3 - 9x^2 - 12x = 0

=> 3x ( x^2 - 3x - 4) = 0

=> 3x ( x^2 - 4x + x - 4) = 0

=> 3x [ x(x - 4) + 1(x-4)] = 0

=> 3x (x +1) (x-4) = 0

For 3x = 0 , we have x = 0

For x + 1 = 0, we have x = -1

For x - 4 = 0 , we have x = 4.

**Therefore the required values of x that are solutions of 3x^3 - 9x^2 - 12x = 0 are 0, -1 and 4.**

We need to factor the polynomial 3x^3 - 9x^2 - 12x = 0.

We notice that 3x is a common factor for all terms of the polynimial.

Then, we will factor 3x from the polynomial.

First, we will factor 3x from all sides.

==> 3x( x^2 - 3x - 4) = 0

Now we will factor between brackets.

**==> 3x ( x -4)(x+ 1) = 0**

To factor between brackets, we could also use the roots formula to determine the roots of the function then obtain the factors.

x= (-b+- sqrt(b^2 - 4ac) / 2a

==> x1= 4

==> x2= -1

=> ( x- x1) and ( x-x2) are factor of x^2 - 3x -4.

==> x^2 - 3x - 4 = ( x-4) ( x+ 1)