x(x-4)(3x+2)= 240 can you find the value of X

You need to open the brackets such
that:

`3x^3 + 2x^2 - 12x^2 - 8x 
-240 = 0`

Collecting like terms
yields:

`3x^3- 10x^2 - 8x - 240 =
0`

You need to use the rational root test to
find the roots, hence you should form the set of possible rational roots. The numerator
of these possible rational roots needs to be one factor of constant coefficient and the
denominator needs to be one factor of leading
coefficient.

The sketch of the graph of function may help
you to eliminate some  rational roots from the set.

Notice
that the graph of function intercepts x axis at x=6, hence you need to substitute 6 for
x to check if this value cancel the polynomial such
that: 

`3*6^3 -360 -48 - 240 = 0=gt
240 - 240 = 0`

This proves that x = 6 is a
root for `f(x) = 3x^3 - 10x^2 - 8x -
240.`

src="/jax/includes/tinymce/jscripts/tiny_mce/plugins/asciisvg/js/d.svg"
sscr="0,7.5,-2,2,1,1,1,1,1,300,200,func,3x^3 - 10x^2 - 8x -
240,null,0,0,,,red,1,none"/>

Since you know one root,
you may find the next two roots using the factored form of polynomial such
that:

`3x^3 - 10x^2 - 8x - 240 =
(x-6)(ax^2 + bx + c)`

You need to open the
brackets to the right side:

`3x^3 -
10x^2 - 8x - 240 = ax^3 + bx^2 + cx - 6ax^2 - 6bx -
6c`

You need to group like powers to the right
side such that:

`3x^3 - 10x^2 - 8x -
240 = ax^3 + x^2(b- 6a) + x(c - 6b) -
6c`

Equating coefficients of like powers
yields:

`a=
3`

`b - 6a = -10 =gt b
- 18 = -10 =gt b = 18 - 10 =gt b = 8`

class="AM">`-6c = -240 =gt c = 40`

Hence,
you may find the next two roots if you solve the quadratic `3x^2
+ 8x + 40 = 0` .

class="AM">`x_(1,2) = (-8+-sqrt(64 - 480))/6 =gt x_(1,2) =
(-8+-sqrt(-416))/6`

class="AM">`x_(1,2) = (-8+-4isqrt26)/6 =gt x_(1,2) =
(-4+-2isqrt26)/3`

Hence,
evaluating the solutions to equation yields `x_1 = 6, x_(2,3) =
(-4+-2isqrt26)/3.`