I need help solve for x xx43x2240Solutionxx43x2 240 First w

I need help solve for x

x(x-4)(3x+2)=240

x(x-4)(3x+2) = 240

First we will open brackets:

(x^2 - 4x)(3x+2) = 240

==> 3x^3 + 2x^2 -12x^2 - 8x - 240 = 0

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

Now we will try and substitute with 240 factors.

==> x= 1 ==> 3-10-8 -240 = -255

==> x= 3 ==> 27 - 90 - 8 - 240 = -311

==> x= 6 ==> 648 - 360 - 48 - 240 = 0

Then we find that x= 6 is one of the roots ... Then (x-6) is a facto.

Now we will divide the equation by (x-6) to find the other factors.

==> (x-6)(3x^2 +8x + 40) = 0

==> x1= 6

Now we will use the quadratic equation to find the other 2 roots.

==> x2= ( -8 + sqrt( 64-4*3*40)/ 2*3 = ( -8 + sqrt(416)*i)/6= -8+4sqrt26 )/ 6 = (4/3) + (2sqrt26/3)*i

==> x2= (4/3) - (2sqrt26/3)*i

Then the roots are: x= { 6, (-4/3)+(2sqrt26/3)*i , (-4/3 - (sqrt26/3)*i}