# Solve for x if 3x^3 - 9x^2 - 12x = 0.

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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)

To solve 3x^3-9x^2-12x= 0.

Let f(x) = 3x^3-9x^2-12x.

f(x) = 3x(x^2-3x-4).

Therfore f(x) =0 gives 3x = 0, or x^2-3x-4 = 0.

3 x= 0 gives x = 0.

x^2-3x-4 = 0 gives the roots to be found as beow:

x^2-3x-4 = x^2-4x+x-4.

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

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

Therefore x^2-3x-4 = 0 gives:

(x-4)(x+1) = 0.

So x= 4 or x=-1.

Threfore the roots are x= 0, x= 4 orx= -1.